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Class 12 Physics Case Study Questions Chapter 4 Moving Charges and Magnetism

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 12 Physics Chapter 4 Moving Charges and Magnetism Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve NCERT Class 12 Physics Case Study Questions Moving Charges and Magnetism to know their preparation level.

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In CBSE Class 12 Physics Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Moving Charges and Magnetism Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Case Study/Passage-Based Questions

Answer: (a) remains stationary

(ii) A proton is projected with a uniform velocity v along the axis of a current carrying solenoid, then

Answer: (d) the proton will continue to move with velocity v along the axis.

(iii) A charged particle experiences a magnetic force in the presence of the magnetic field. Which of the following statement is correct?

Answer: (b) The particle is moving and the magnetic field is perpendicular to the velocity

(iv) A charge q moves with a velocity 2 ms -1 along the x-axis in a uniform magnetic field B⃗ =(i^+2j^+3k^)TB→=(i^+2j^+3k^)T then a charge will experience a force

Answer: (a) in z-y plane

(v) Moving charge will produce

Answer: (c) both electric and magnetic field

Answer: (d) is a deflection instrument that gives a deflection when a current flows through its coil

(ii) To make the field radial in a moving coil galvanometer

Answer: (d) poles are cylindrically cut

(iii) The deflection in a moving coil galvanometer is

Answer: (b) directly proportional to the number of turns in the coils

(iv) In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is placed in a radial field of strength B. The torque acting on the coil is

Answer: (d) NABI

(v) To increase the current sensitivity of a moving coil galvanometer, we should decrease

Answer: (b) the torsional constant of spring

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Physics Moving Charges and Magnetism Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics

Please refer to below Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics. These Case Study Questions Class 12 Physics will be coming in your examinations. Students should go through the Chapter 4 Moving Charges and Magnetism Case Study based questions in their Class 12 Physics CBSE, NCERT, KVS book as this will help them to secure more marks in upcoming exams.

Case Study Based Questions Physics Class 12 – Chapter 4 Moving Charges and Magnetism

Read the para given below and answer the questions that follow:

The force experienced by a particle of charge q moving with a velocity v given by F → -q [V → ×B → ] which is perpendicular to Both V and B. Since F → is perpendicular to V, no work is done on the charged particle moving in a uniform magnetic field. If V is perpendicular to B, the charged particle follows a circular path whose radius is given by r = mv/qB The frequency of revolution of the particle along the circular path is given by V = qB/2πm

Question. A proton and an alpha particle are projected perpendicular to a uniform magnetic field with equal velocities. The mass of an alpha particle is 4 times that of a proton and its charge is twice that of a proton. If the r p and r α are the radii of their circular path, then the ratio r p /r α is. (a) 1/√2 (b) 1/2 (c) 1 (d) √2

Question. In (i) part, what is the ratio r p /r a if the two particles have equal kinetic energies before entering the region of the magnetic field. (a) 1/2 (b) 1 (c) 2 (d) 4

Question. In (i) part, what is the ratio r p /r a if two particles have equal linear moments before entering the region of the magnetic field. (a) 1 (b) √2 (c) 2 (d) 2√2

Question. In (i) part, what is the is the ratio if two particles are accelerated through the same potential difference before entering the region of the magnetic field. (a) 1/√2 (b) 1 (c) √2 (d) 2

Question. Which of the following in motion cannot be deflected by magnetic field? (a) Protons (b) Beta particles (c) Alpha particles (d) Neutrons

Helical Motion. The path of a charged particle in magnetic field depends upon angle between velocity and magnetic field. If velocity V → is at angle θ to B → , component of velocity parallel to magnetic field (V cos θ) remains constant and component of velocity perpendicular to magnetic field (V sin θ) is responsible for circular motion, thus the charge particle moves in a helical path.

The plane of the circle is perpendicular to the magnetic field and the axis of the helix is parallel to the magnetic field. The charged particle moves along helical path touching the line parallel to the magnetic field passing through the starting point after each rotation. Radius of circular path, r = mv sinθ/qB Hence the resultant path of the charged particle will be helix, with its axis along the direction of B → shown in figure.

Question. When a positively charged particle enters into a uniform magnetic field with uniform velocity, its trajectory can be (I) a straight line (II) a circle (III) a helix. (a) (I) only (b) (I) or (II) (c) (I) or (III) (d) any one of (I), (II) and (III)

Question. Two charged particles A and B having the same charge, mass and speed enter into a magnetic field in such a way that initial path of A makes an angle of 30º and that of B makes an angle of 90º with the field. Then the trajectory of (a) B will have smaller radius of curvature than that of A (b) both will have the same curvature (c) A will have smaller radius of curvature than that of B (d) both will move along the direction of their original velocities.

Question. An electron having momentum 2.4 × 10 –23 kg m/s enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30º with initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be (a) 2 mm (b) 1 mm (c) √3/2 mm (d) 0.5 mm

Question. The magnetic field in a certain region of space is given by B → = 8.35 -2 ×10 -2 i T . A proton is shot into the field with velocity V → = (2×10 5 i + 4 × 10 5 j) m/s. The proton follows a helical path in the field. The distance moved by proton in the x-direction during the period of one revolution in the yz-plane will be (a) 0.053 m (b) 0.136 m (c) 0.157 m (d) 0.236 m

Question. The frequency of revolution of the particle is

Read the para given below and answer the questions that follow:

Motion of charge in magnetic field. An electron with speed r 0 << c moves in a circle of radius r0 in a uniform magnetic field. This electron is able to traverse a circular path as magnetic field is perpendicular to the velocity of the electron. A force acts on the particle perpendicular to both V 0 → and B → This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is T 0 .

Question. If the speed of the electron is now doubled to 2V 0 . The radius of the circle will change to (a) 4r 0 (b) 2r 0 (c) r 0 (d) r 0 /2

Question. If V 0 = 2V 0 , then time required for one revolution of the electron will change to (a) 4T 0 (b) 2T 0 (c) T 0 (d) T 0 /2

Question. If the given electron has a velocity not perpendicular to B, then trajectory of the electron is (a) straight line (b) circular (c) helical (d) zig-zag

Question. If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is (a) Bev (b) Be/ν (c) B/eν (d) zero

Biot-Savart Law. A magnetic field can be produced by moving, charges or electric currents. The basic equation governing the magnetic field due to a current distribution is the Biot-Savart law. Finding the magnetic field resulting from a current distribution involves the vector product, and is inherently a calculus problem when the distance from the current to the field point is continuously changing. According to this law, the magnetic field at a point due to a current element of length dl → carrying current I, at a distance r from the element is

Biot-Savart law has certain similarities as well as difference with Coulomb’s law for electrostatic field e.g., there is an angle dependence in Biot -Savart law which is not present in electrostatic case.

Question. The direction of magnetic field dB → due to a current Id l → at a point of distance r → from it, when a current I passes through a long conductor is in the direction (a) of position vector r → of the point (b) of current element d l → (c) perpendicular to both d l → and r → (d) perpendicular to d l → only

Question. The magnetic field due to a current in a straight wire segment of length L at a point on its perpendicular bisector at a distance r (r >>L) (a) decreases as 1/r (b) decreases as 1/r 2 (c) decreases as 1/r 3 (d) approaches a finite limit as r → ∞

Question. Two long straight wires are set parallel to each other. Each carries a current i in the same direction and the separation between them is 2r. The intensity of the magnetic field midway between them is

(a) μ 0 i/r (b) 4μ 0 i/r (c) zero (d) μ 0 i/4r

Question. A long straight wire carries a current along the z-axis for any two points m the x-y plane. Which of the following is always false? (a) The magnetic fields are equal. (b) The directions of the magnetic fields are the same. (c) The magnitudes of the magnetic fields are equal (d) The field at one point is opposite to that at the other point.

Question. Biot-Savart law can be expressed alternatively as (a) Coulomb’s Law (b) Ampere’s circuital law (c) Ohm’s Law (d) Gauss’s Law

Read the para given below and answer the questions that follow:

Moving Coil Galvanometer. Moving coil galvanometer operates on Permanent Magnet Moving Coil (PMMC) mechanism and was designed by the scientist D’arsonval. Moving coil galvanometers are two types—(i) Suspended coil (ii) Pivoted coil types or tangent galvanometer. Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque. This torque tends to rotate the coil about its axis of suspension in such a way that the magnetic flux passing through the coil the maximum.

Question. A moving coil galvanometer is an instrument which (a) is used to measure emf. (b) is used to measure potential difference . (c) is used to measure resistance. (d) is a deflection instrument which gives deflection when a current flows through its coil.

Question. To make the field radial in a moving coil galvanometer (a) number of turns coil is kept small (b) magnet is taken in the form of horse-shoe (c) poles are of very strong magnets (d) poles are cylindrically cut

Question. The deflection in a moving coil galvanometer is (a) directly proportional to torsional constant of spring (b) directly proportional to the number of turns in the coil (c) inversely proportional to the area of the coil (d) inversely proportional to the current in the coil

Question. In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is placed in a radial field of strength B. The torque acting on the coil is (a) NA 2 B 2 I (b) NABI 2 (c) N 2 ABI (d) NABI

Question. To increase the current sensitivity of a moving coil galvanometer, we should decrease (a) strength of magnet (b) torsional constant of spring (c) number of turns in coil (d) area of coil

Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics

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Class 12 Physics Chapter 4 Case Study Question Moving Charges and Magnetism PDF Download

In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 4 Moving Charges and Magnetism Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Moving Charges and Magnetism to know their preparation level.

Moving Charges and Magnetism Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Case Study/Passage-Based Questions

Question 1:

Answer: (a) remains stationary

(ii) A proton is projected with a uniform velocity v along the axis of a current carrying solenoid, then

Answer: (d) the proton will continue to move with velocity v along the axis.

(iii) A charged particle experiences a magnetic force in the presence of the magnetic field. Which of the following statement is correct?

Answer: (b) The particle is moving and the magnetic field is perpendicular to the velocity

(iv) A charge q moves with a velocity 2 ms -1 along the x-axis in a uniform magnetic field B⃗ =(i^+2j^+3k^)TB→=(i^+2j^+3k^)T then a charge will experience a force

Answer: (a) in z-y plane

(v) Moving charge will produce

Answer: (c) both electric and magnetic field

Question 2:

Answer: (d) is a deflection instrument that gives a deflection when a current flows through its coil

(ii) To make the field radial in a moving coil galvanometer

Answer: (d) poles are cylindrically cut

(iii) The deflection in a moving coil galvanometer is

Answer: (b) directly proportional to the number of turns in the coils

(iv) In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is placed in a radial field of strength B. The torque acting on the coil is

Answer: (d) NABI

(v) To increase the current sensitivity of a moving coil galvanometer, we should decrease

Answer: (b) the torsional constant of spring

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Physics Moving Charges and Magnetism Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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Chapter-4: Moving Charges and Magnetism of NCERT class 12 Physics note is available here.

Moving Charges and Magnetism is one of the important chapters of CBSE class 12 Physics. So, students must prepare this chapter thoroughly. The notes provided here will be very helpful for the students who are going to appear in CBSE class 12 Physics board exam .

Oersted’ Experiment

During a lecture demonstration in 1820, the Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He further investigated this phenomenon and confirmed the phenomenon of magnetic field around current carrying conductor.

Magnetic Field

It is the space around a magnet or current carrying conductor around which magnetic effects can be experienced. It is a vector quantity and its SI unit is tesla (T) or Wbm ‒2 .

Moving Charge & Magnetic Field

A charge can produce magnetic field if it is in motion. Magnetic field can also interact with a moving charge.

Lorentz Force

Assume a point charge q (moving with a velocity v and, located at r at a given time t ) in presence of both the electric field E ( r ) and the magnetic field B ( r ). The force on an electric charge q due to both of them is given by,

F = q [ E ( r ) + v × B ( r )] ≡ F Electric + F Magnetic

Careful analysis of this expression shows that:

• Lorentz Force depends on q , v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge.

• The magnetic force q [ v × B ] includes a vector product of velocity & magnetic field. Vector product makes the force due to magnetic field become zero if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as shown in figure given below

Direction of magnetic field by right hand rule for vector

Image Source: NCERT Books

• The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force

F = q [ v × B ] = q |v||B| sin θ ň, where θ is angle between v and B.

Magnetic force on a current-carrying conductor

Magnetic force on a conductor of length l carrying a current I placed in a uniform magnetic field B is given by

F = I ( l × B ) or |F| = I | l| | B | sin θ.

The direction of F is perpendicular to both l and B and can be obtained with the help of Fleming’s Left hand rule.

Motion of a charged particle in Magnetic Field

A force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle.

In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).

Generally two types of cases are possible:

Case 1 st : When v is perpendicular to B

The perpendicular force, q v × B , acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other.

Charge particle describing a circle if v and B are mutually perpendicular

In this case, radius described by charge particle is given by, r = m v / q B

If ω is the angular frequency, then ω = 2π v = q B / m , where, v is frequency of rotation

The time taken for one revolution is T= 2π/ω ≡ 1/ν.

Case 2 nd : When v is making an angle with B other than 0 o

In this case, velocity has a component along B , this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion

Helical Path of a moving charge particle in magnetic field

The distance moved along the magnetic field in one rotation is called pitch p and,

p = v || T = 2π m v || / q B

The radius of the circular component of motion is called the radius of the helix.

Motion of a charge in Combined Electric and Magnetic Fields

A charge q moving with velocity v in presence of both electric and magnetic fields experiences a force given by F = q ( E + v × B ) = F E + F B

Motion of a charge in Cross Electric and Magnetic Fields

Consider the situation shown in figure given above, in this particular case we have:

Force on charge particle in cross electric and magnetic fields

Here, electric and magnetic forces are in opposite directions as shown in the figure.

If we adjust the value of E and B such that magnitude of the two forces are equal. Then, total force on the charge is zero and the charge will move in the fields undeflected.

This happens when, qE = qvB or v = E/B

This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed E and B fields, therefore, serve as a velocity selector.

Only particles with speed E/B pass undeflected through the region of crossed fields.

It is a machine to accelerate charged particles or ions to high energies.

The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other they are called crossed fields.

A schematic sketch of the cyclotron is shown in the figure given above. There is a source of charged particles or ions at P which move in a circular fashion in the dees, D1 and D2, on account of a uniform perpendicular magnetic field B. An alternating voltage source accelerates these ions to high speeds. The ions are eventually ‘extracted’ at the exit port.

In case of cyclotron,

cyclotron frequency and maximum kinetic energy

Magnetic Field Due to a Current Element: Biot-Savart Law

Image Source: NCERT Textbooks

Magnetic field strength at a point P due to a small length dl of the conductor carrying current I is given by,

CBSE Class 12th Chemistry Notes: Polymers

Magnetic Field on the Axis of a Circular Current Loop:

Magnetic field at a point on the axis of a coil, having radius R, distance x from the centre of the coil is given by,

Magnetic field at the centre of a current carrying coil:

To find the magnetic field at the centre of the coil, we can put x = 0 is above equation.

So, B at the centre of a current carrying coil on N turn is given by,

The magnetic field lines due to a circular wire form closed loops and are shown in the figure given above. The direction of the magnetic field is given by right-hand thumb rule which is if we curl the palm of our right hand around the circular wire with the fingers pointing in the direction of the current then; the right-hand thumb gives the direction of the magnetic field.

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Ampere’s Circuital Law

According to this law, the line integral of magnetic field B around any closed path in vacuum is μ o times the net current ( I ) threading through the area enclosed by the curve.

Magnetic field due to a solenoid

Solenoid is a device used to generate magnetic fields. It consists of a long conducting wire wound in the form of a helix where the neighboring turns are closely spaced. When current flows in the solenoid, then each turn can be regarded as conducting circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enameled wires are used for winding so that turns are insulated from each other.

Image Source: wikipedia.org

With the help of Ampere’s circuital law we can calculate the magnetic field due to a solenoid.

Magnetic field at the centre of a long solenoid having n turns per unit length and carrying a current I is given by: B = μ o n I

The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field.

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Magnetic field due to a toroid

Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a solenoid which has been bent into a circular shape to close on itself.

The magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero. The field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns and is given by, B = μ o n I .

Force between two Parallel Current Carrying Wires

Image Source: NCERT textbooks

Two long parallel conductors a and b kept at a distance d apart in vaccum and carrying a current I a and I b respectively. The force per unit length experienced by each of them is given by

Expression for Force between two Parallel Currents

This force will be equal and opposite on both wires.

Nature of force of interaction between the current carrying conductors can be explained on the basis of Fleming’s left hand rule.

In general, two long parallel conductors carrying current

in same direction will attract each other
in different direction will repel each other

Definition of Ampere

Definition of ampere follows from the formula for force between two parallel current carrying conductors.

When 1 ampere each is passing through two long, parallel conductors kept 1 m apart in vacuum, a force of magnitude 2 × 10 ‒7 N is experienced by a meter length of each conductor.

Torque on a Rectangular Current loop in a Uniform Magnetic Field

A rectangular loop having N turns of area A each, carrying a steady current I and placed in a uniform magnetic field B such that the normal to the plane of the loop makes an angle θ with the direction of the magnetic field B, then a torque τ experienced by the loop whose magnitude is given by, τ = NIAB sin θ .

This equation can be expressed in vector form as shown below:

Here, | m | = NIA is called magnetic dipole moment of the current loop.

One can define magnetic dipole moment of the current carrying loop as product of current in the loop and total area of the loop, i.e., M = I ( NA ).

Magnetic dipole moment is a vector quantity and its direction is along the direction of magnetic field due to the current in the loop. We can also find the direction of dipole moment vector with the help of right hand grip rule.

Magnetic Dipole Moment of a Revolving Electron

Any charge in uniform circular motion would have an associated magnetic moment given by,

Here, e = 1.6 × 10 ‒19 C, m e = 9.1 × 10 ‒31 kg, l is the magnitude of the angular momentum of the electron about the central nucleus (“orbital” angular momentum) and | l | = ( n h )/2π, where, n is a natural number ( n = 1, 2, 3, ….) and h is a constant named after Max Planck (Planck’s constant) with a value h = 6.626 × 10 –34 J s.

The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment. Instead of electron with charge (– e), if we had taken a particle with charge (+ q), the angular momentum and magnetic moment would be in the same direction.

Potential Energy of a current loop in a Magnetic Field

If a current loop of magnetic moment, M = NIA is held in a uniform magnetic field B in such a way that direction of magnetic dipole moment makes an angle θ with the direction of magnetic field, then the potential energy of the dipole is given by,

Moving Coil Galvanometer

It is an instrument used for detection and measurement of small current.

The galvanometer consists of a light rectangular coil of N turns each having area A wound on an aluminium frame.

The coil is free to rotate free to rotate about a fixed axis (as shown in figure given below), in a uniform radial magnetic field.

There is a cylindrical soft iron core which is used to make the field radial and also to increase the strength of the magnetic field.

When a current flows through the coil, a torque acts on it. This torque is given by, τ = NIAB where the symbols have their usual meaning.

Since the field is radial by design, we have taken sin θ = 1 in the above expression for the torque. The magnetic torque NIAB tends to rotate the coil. A spring S p provides a counter torque kϕ that balances the magnetic torque NIAB ; resulting in a steady angular deflection ϕ . In equilibrium

kϕ = NI AB

Here, k is the torsional constant of the spring; i.e. the restoring torque per unit twist.

The deflection ϕ is indicated on the scale by a pointer attached to the spring. We have

The quantity in brackets is a constant for a given galvanometer.

Sensitivity of a Galvanometer

Current Sensitivity:

Current sensitivity of a galvanometer is defined as the deflection per unit current

Mathematically,

Voltage Sensitivity:

Voltage sensitivity of a galvanometer is defined as the deflection per as the deflection per unit voltage.

Conversion of a Galvanometer into Voltmeter

A galvanometer of coil resistance R G , showing full scale deflection for a current I G can be converted into a voltmeter for measuring potential differences having values greater than I G R G by connecting high resistance R in series with the galvanometer where,

Expression for resistance in series with the galvanometer

An ideal voltmeter has infinite resistance.

Conversion of a Galvanometer into Ammeter:

A galvanometer of coil resistance R G , showing full scale deflection for a current I G can be converted into an ammeter for measuring current having values more than I G (i.e., I > I G ) by putting a low resistance r S in parallel with the galvanometer where,

Expression for Shunt Resistance in Galvanometer

Here, r s is also called shunt resistance. An ideal ammeter has zero resistance.

CBSE Class 12 Physics Important Questions Chapter 4 – Moving Charges and Magnetism

1 mark questions.

1. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer? Ans. (a) Non-Brittle conductor (b) Restoring Torque per unit Twist should be small.

2. What will be the path of a charged particle moving along the direction of a uniform magnetic field? Ans. The path of a charged particle will be a straight line path as no force acts on the particle.

4. A cyclotron is not suitable to accelerate electron. Why? Ans. A cyclotron is not suitable to accelerate electron because its mass is less due to which they gain speed and step out of the dee immediately.

2 Marks Questions

3. Give one difference each between diamagnetic and ferromagnetic substances. Give one example of each? Ans. Diamagnetic substances are weakly repelled by a magnet eg. Gold. Ferromagnetic materials are strongly attracted by a magnet eg. Iron.

3 Marks Questions

18. Answer the following questions: (a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle? (b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment? (c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path. Ans. (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. (b) Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude. (c) An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.

5 Marks Questions

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NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

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NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism: If you're a Class 12 student in search of NCERT solutions , you've come to the right place. On this page, you'll find comprehensive class 12 physics chapter 4 ncert solutions from questions 1 to 28. Our class 12 physics ch 4 ncert solutions have been meticulously crafted by subject experts, providing detailed explanations for each step. Also, you have the option to download these solutions in PDF format, enabling you to work on them offline as well.

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Class 12 Physics ch 4 Exercise solutions

Moving Charges And Magnetism Class 12 Main Topics-

Ncert solutions subject wise, importance of ncert solutions for class 12 physics chapter 4 moving charges and magnetism in exams:.

A moving charge in a magnetic field can produce a force and a current-carrying conductor in a uniform magnetic field will experience a force. The solutions of NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism cover problems based on these two basic concepts. Class 12 chapter 4 physics ncert solutions will help you to boost the concepts studied in this chapter of the NCERT syllabus .

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Two main laws discussed in the NCERT chapter Moving charges and Magnetism Class 12 are Ampere's law and Biot-savant law. Based on these laws many problems are discussed in CBSE NCERT solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism. The galvanometer is discussed in Moving charges and Magnetism Class 12 in detail with its working principle. NCERT Solutions for Class 12 Physics Chapter 4 Moving charges and Magnetism helps students for solving homework problems.

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The right-hand thumb rule, Flemings right-hand rule and Flemings left-hand rule are three important concepts of Moving charges and Magnetism chapter of NCERT textbook . The problems related to finding the directions can be solved using these three rules or by using the concepts of vectors. The NCERT solutions for Class 12 Physics also helps in preparing for competitive exams like NEET and JEE Mains.

Free download class 12 chapter 4 physics ncert solutions PDF for CBSE exam.

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C lass 12 Physics ch 4 Exercise solutions

1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

The magnitude of the magnetic field at the centre of a circular coil of radius r carrying current I is given by,

$|B|=\frac{\mu _{0}I}{2r}$

For 100 turns, the magnitude of the magnetic field will be,

$|B|=100\times \frac{\mu _{0}I}{2r}$

2. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,

$|B|=\frac{\mu _{0}I}{2\pi r}$

In this case

$|B|=\frac{4\pi \times 10^{-7}\times 35}{2\pi\times 0.2}=3.5\times 10^{-5}T$

3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

$|B|=\frac{4\pi \times 10^{-7}\times 50}{2\pi\times 2.5}$

4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

$|B|=\frac{4\pi \times 10^{-7}\times 90}{2\pi\times 1.5}=1.2\times10^{-5} T$

The current in the overhead power line is going from the East to West direction and the point lies below the power line. Applying Maxwell's right-hand thumb rule we can see that the direction of the magnetic field will be towards the South.

$30 \degree$

For a straight wire of length l in a uniform magnetic field, the Force equals to

$\\\vec{F}=\int_{0}^{l}I\vec{dl}\times \vec{B}\\ |\vec{F}|=BIlsin\theta$

In the given case the magnitude of force per unit length is equal to

$\times$

6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

In the given case the magnitude of the force is equal to

The direction of this force depends on the orientation of the coil and the current-carrying wire and can be known using the Flemings Left-hand rule.

7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

The magnitude of magnetic field at a distance r from a long straight wire carrying current I is given by,

In this case the magnetic field at a distance of 4.0 cm from wire B will be

$|B|=\frac{4\pi \times 10^{-7}\times 5}{2\pi\times 0.04}$

The force on a straight wire of length l carrying current I in a uniform magnetic field B is given by

$F=BIlsin\theta$

The force on a 10 cm section of wire A will be

$F=2.5\times10^{-5}\times8\times0.1 \times sin90^\circ$

8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

The magnitude of the magnetic field at the centre of a solenoid of length l, total turns N and carrying current I is given by

$B=\frac{\mu _{o}NI}{l}$

The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by

$\tau =nBIAsin\theta$

The coil, therefore, experiences a torque of magnitude 0.96 Nm.

$M_1$

we know V=IR

$\phi k=\frac{B_{1}A_{1}N_{1}V}{R_{1}}$

Their ratio of voltage sensitivity of coil M 2 to that of coil M 1

$=\frac{B_{2}A_{2}N_{2}R_{1}}{B_{1}A_{1}N_{1}R_{2}}$

Since the velocity of the shot electron is perpendicular to the magnetic field, there is no component of velocity along the magnetic field and therefore the only force on the electron will be due to the magnetic field and will be acting as a centripetal force causing the electron to move in a circular path. (if the initial velocity of the electron had a component along the direction of the magnetic field it would have moved in a helical path)

$-1.6\times 10^{-19}C$

The angle between the direction of velocity and the magnetic field = 90 o

Since the force due to the magnetic field is the only force acting on the particle,

$\\\frac{mV^{2}}{r}=q\vec{V}\times \vec{B}\\ \frac{mV^{2}}{r}=|qVBsin\theta| \\ r=|\frac{mV}{qBsin\theta }|$

12) In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain

$r=\frac{eB}{mv}$

From the above equation, we can see that this frequency is independent of the speed of the electron.

$\nu =\frac{1.6\times 10^{-19}\times 4.8\times 10^{6}}{2\pi \times 9.1\times 10^{-31}}=18.2 MHz$

Number of turns in the coil(n)=30

The radius of the circular coil(r)=8.0 cm

Current flowing through the coil=6.0 A

Strength of magnetic field=1.0 T

The angle between the field lines and the normal of the coil=60 o

The magnitude of the counter-torque that must be applied to prevent the coil from turning would be equal to the magnitude of the torque acting on the coil due to the magnetic field.

$\\\tau =nBIAsin\theta\\ \tau = 30\times 1\times 6\times \pi \times (0.08)^{2}\times sin60^{o}$

A torque of magnitude 3.13 Nm must be applied to prevent the coil from turning.

13 b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

NCERT solutions for moving charges and magnetism additional exercise:

14) Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Using the right-hand thumb rule we can see that the direction of the magnetic field due to coil X will be towards the east direction and that due to coil Y will be in the West direction.

We know the magnetic field at the centre of a circular loop of radius r carrying current I is given by

$B=\frac{\mu _{o}I}{2r}$

The net magnetic field at the centre of the coils,

B net =B y - B x

The direction of the magnetic field at the centre of the coils is towards the west direction.

$100 G ( 1G = 10 ^{-4}) T$

Therefore keeping the number of turns per unit length and the value of current within the prescribed limits such that their product is approximately 8000 we can produce the required magnetic field.

e.g. n=800 and I=10 A.

16.(a) For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

Show that this reduces to the familiar result for the field at the centre of the coil.

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

For finding the field at the centre of coil we put x=0 and get the familiar result

$B = \frac{\mu _0 IN}{2R}$

16. (b) For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

$B = 0.72 \frac{\mu _0 NI}{R}$

Let a point P be at a distance of l from the midpoint of the centres of the coils.

The distance of this point from the centre of one coil would be R/2+l and that from the other would be R/2-l.

The magnetic field at P due to one of the coils would be

$B_{1}= \frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}$

The magnetic field at P due to the other coil would be

$B_{2}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}$

Since the direction of current in both the coils is same the magnetic fields B 1 and B 2 due to them at point P would be in the same direction

B net =B 1 +B 2

$\\B_{net}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}+\frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( (R/2-l)^2 + R^2 )^{-3/2} + ( (R/2+l)^2 + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{R^{2}}{4}-Rl+l^{2} + R^2 )^{-3/2} + ( \frac{R^{2}}{4}+Rl+l^{2} + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{5R^{2}}{4}-Rl+l^{2} )^{-3/2} + ( \frac{5R^{2}}{4}+Rl+l^{2} )^{-3/2}\right ]\\$

Since l<<R we can ignore term l 2 /R 2

$\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{5}{4})^{-3/2}\left [ ( 1-\frac{4l}{5R} )^{-3/2} + ( 1+\frac{4l}{5R} )^{-3/2}\right ]\\$

Since the above value is independent of l for small values it is proved that about the midpoint the Magnetic field is uniform.

17.(a) A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field outside the toroid

Outside the toroid, the magnetic field will be zero.

17.(b) A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field inside the core of the toroid?

The magnetic field inside the core of a toroid is given by

Total number of turns(N)=3500

Current flowing in toroid =11 A

Length of the toroid, l=

$\\l=2\pi \left (\frac{ r_{1}+r_{2}}{2} \right )\\ \\l=\pi ( r_{1}+r_{2})\\ \\l=\pi (0.25+0.26)\\ \\l=0.51\pi$

17.(c) A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field in the empty space surrounded by the toroid?

The magnetic field in the empty space surrounded by the toroid is zero.

18. (a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

The charged particle is not deflected by the magnetic field even while having a non zero velocity, therefore, its initial velocity must be either parallel or anti-parallel to the magnetic field i.e. It's velocity is either towards the east or the west direction.

18 b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

Yes, its final speed will be equal to the initial speed if it has not undergone any collision as the work done by the magnetic field on a charged particle is always zero because it acts perpendicular to the velocity of the particle.

18 c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

The electron would experience an electrostatic force towards the north direction, therefore, to nullify its force due to the magnetic field must be acting on the electron towards the south direction. By using Fleming's left-hand rule we can see that the force will be in the north direction if the magnetic field is in the vertically downward direction.

Explanation:

$q\vec{V}$

19 (a) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field is transverse to its initial velocity

(a) The electron has been accelerated through a potential difference of 2.0 kV.

$1.6$

Since the electron initially has velocity perpendicular to the magnetic field it will move in a circular path.

The magnetic field acts as a centripetal force. Therefore,

$\\\frac{mv^{2}}{r}=evB\\ r=\frac{mv}{eB}\\ r=\frac{9.1\times 10^{-31}\times 2.67\times 10^{7}}{1.6\times 10^{-19}\times 0.15}=1.01mm$

The electron has been accelerated through a potential difference of 2.0 kV.

The component of velocity perpendicular to the magnetic field is

$\\v_{p}=vsin30^{o}\\ v_{p}=1.33\times 10^{7}\ ms^{-1}$

The electron will move in a helical path of radius r given by the relation,

$\\\frac{mv^{2}_{p}}{r}=ev_{p}B\\ r=\frac{mv_{p}}{eB}\\ r=\frac{9.1\times 10^{-31}\times 1.33\times 10^{7}}{1.6\times 10^{-19}\times 0.15}$

The component of velocity along the magnetic field is

$\\v_{t}=vcos30^{o}\\ v_{t}=2.31\times 10^{7}\ ms^{-1}$

The electron will move in a helical path of pitch p given by the relation,

$\\p=\frac{2\pi r}{v_{p}}\times v_{t}\\ p=\frac{2\pi \times 5\times 10^{-4}}{1.33\times 10^{7}}\times 2.31\times 10^{7}$

The electron will, therefore, move in a helical path of radius 5 mm and pitch 5.45 mm.

$9.0 \times 10 ^{-5} V m ^{-1}$

Let the beam consist of particles having charge q and mass m.

After being accelerated through a potential difference V its velocity can be found out by using the following relation,

$\\\frac{1}{2}mv^{2}=qV\\ \\v=\sqrt{\frac{2qV}{m}}$

Using the value of v from equation (ii) in (i) we have

$\\E=B\sqrt{\frac{2qV}{m}}\\ \frac{q}{m}=\frac{E^{2}}{2VB^{2}}\\ \frac{q}{m}=\frac{(9\times 10^{-5})^{2}}{2\times 15\times 10^{3}\times (0.75)^{2}}=4.8\times 10^{-13}$

21. (a) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

In order for the tension in the wires to be zero the force due to the magnetic field must be equal to the gravitational force on the rod.

$mg=BIl$

mass of rod=0.06 g

length of rod=0.45m

the current flowing through the rod=5 A

$\\B=\frac{mg}{Il}\\ B=\frac{0.06\times 9.8}{5\times 0.45}\\ B=0.261\ T$

A magnetic field of strength 0.261 T should be set up normal to the conductor in order that the tension in the wires is zero

$g = 9.8 m s ^{-2}$

If the direction of the current is reversed the magnetic force would act in the same direction as that of gravity.

Total tension in wires(T)=Gravitational force on rod + Magnetic force on rod

$\\T=mg+BIl\\ \\T=0.06\times 9.8+0.261\times 5\times 0.45\\ \\T=1.176\ N$

The total tension in the wires will be 1.176 N.

22. The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Since the distance between the wires is much smaller than the length of the wires we can calculate the Force per unit length on the wires using the following relation.

$F=\frac{\mu _{o}I_{1}I_{2}}{2\pi d}$

Current in both wires=300 A

Distance between the wires=1.5 cm

$F=\frac{4\pi \times 10^{-7}\times 300\times 300}{2\pi \times 0.015}$

F=1.2 Nm -1

23.(a) A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, the wire intersects the axis,

The length of wire inside the magnetic field is equal to the diameter of the cylindrical region=20.0 cm=0.2 m.

Magnetic field strenth=1.5 T.

Current flowing through the wire=7.0 A

The angle between the direction of the current and magnetic field=90 o

Force on a wire in a magnetic field is calculated by relation,

$\\F=BIlsin\theta \\ F=1.5\times 7\times 0.2$

This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.

23.(b) A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, the wire is turned from N-S to northeast-northwest direction,

The angle between the direction of the current and magnetic field=45 o

The radius of the cylindrical region=10.0 cm

$\\l=\frac{2r}{sin\theta }\\$

This force will be independent of the angle between the wire and the magnetic field as we can see in the above case.

Note: There is one case in which the force will be zero and that will happen when the wire is kept along the axis of the cylindrical region.

23 c) A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

The wire is lowered by a distance d=6cm.

In this case, the length of the wire inside the cylindrical region decreases.

Let this length be l.

$\\(\frac{l}{2})^{2}+d^{2}=r^{2}\\ \\(\frac{l}{2})^{2}=0.1^{2}-0.06^{2}\\ \\(\frac{l}{2})^{2}=0.01-0.0036\\ \\\\(\frac{l}{2})^{2}=0.0064\\ \\\frac{l}{2}=0.08\\ \\l=0.16m$

This force acts in the vertically downward direction.

24.(a) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

The magnetic field is

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

A=0.005 m 2

$\vec{A}=0.005\ m^{2}\hat{i}$

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero.

24.(b) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero. This was exactly the case in 24. (a) as well.

24 (c). A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-x-direction. The force on the loop is zero.

24 (d) . A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

The torque on the loop has a magnitude of 0.018 Nm and at an angle of 240 o from the positive x-direction. The force on the loop is zero.

24. (e) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Since the area vector is along the direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.

24 (f) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Since the area vector is in the opposite direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.

The force on the loop in all the above cases is zero as the magnetic field is uniform

25. (a) A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the total torque on the coil,

As we know the torque on a current-carrying loop in a magnetic field is given by the following relation

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\$

It is clear that the torque, in this case, will be 0 as the area vector is along the magnetic field only.

25. (b) A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, total force on the coil.

The total force on the coil will be zero as the magnetic field is uniform.

$10 ^{-5} m^2$

The average force on each electron in the coil due to the magnetic field will be eV d B where V d is the drift velocity of the electrons.

The current is given by

$I=neAV_{d}$

where n is the free electron density and A is the cross-sectional area.

$\\V_{d}=\frac{I}{neA}\\ \\V_{d}=\frac{5}{10^{29}\times 1.6\times 10^{-19}\times 10^{-5}}$

The average force on each electron is

$F=eV_dB$

The magnetic field inside the solenoid is given by

$B=\mu _{0}nI$

n is number of turns per unit length

$n=\frac{3\times 300}{0.6}$

n=1500 m -1

Current in the wire I w = 6 A

Mass of the wire m = 2.5 g

Length of the wire l = 2 cm

The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances weight of the wire

$BI_wl=mg$

Therefore a current of 108.37 A in the solenoid would support the wire.

$12 \Omega$

The galvanometer can be converted into a voltmeter by connecting an appropriate resistor of resistance R in series with it.

At the full-scale deflection current(I) of 3 mA the voltmeter must measure a Voltage of 18 V.

$I$

The galvanometer can be converted into an ammeter by connecting an appropriate resistor of resistance R in series with it.

At the full-scale deflection current(I) of 4 mA, the ammeter must measure a current of 6 A.

Since the resistor and galvanometer coil are connected in parallel the potential difference is the same across them.

$IG=(6-I)R$

Understanding moving charges and magnetism class 12 ncert solutions is like building the foundation of a tall building. It helps you do well in your regular school exams (like the bricks at the bottom) and is also crucial for clearing tough entrance exams like JEE and NEET (like the support structure that holds up the building). So, it's important for both your regular studies and future career goals.

Class 12 physics chapter 4 exercise solutions: Important Formulas and Diagrams

Important formulas of physics chapter 4 class 12 ncert solutions are listed below:

Force Applied To A Moving Charge

Neet previous year papers with solutions.

Solve NEET previous years question papers & check your preparedness

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

$\vec{F}=q(\vec{v}\times \vec{B})$

Biot-Savart’s Law

Ampere Circuital Law

$\int B.dl=\mu_0I$

The following topics are covered in ch 4 Physics Class 12:

Magnetic Force- The force on a moving charge in a uniform magnetic field and also a force on a current-carrying conductor in a magnetic field are discussed in chapter 4 Physics Class 12. Some questions based on these concepts are discussed in the Class 12 Physics Chapter 4 NCERT solutions. For example questions, 5, 11, 18 and 23 of NCERT solutions for Class 12 Physics Chapter 4 uses the concepts of force on a charge/conductor in a uniform magnetic field.

Motion in a magnetic field- This part of Moving charges and Magnetism Class 12 discusses the trajectory of a charge in a magnetic field. Questions based on this are discussed in the Moving Charges and Magnetism solutions given above.

The motion of combined electric and magnetic field- The motion of charge in the presence of both electric and magnetic fields and also the topic cyclotron is discussed in this portion of Class 12 NCERT Physics.

Biot-Savart law, Amperes circuital law- These two laws and their applications are discussed in Physics chapter 4 Class 12. Other topics discussed in class 12 physics chapter 4 question answer are solenoid, toroid, the force between parallel conductors, torques on a rectangular loop in a magnetic field, the angular momentum of the electron and the concept of moving coil galvanometer and conversion of galvanometer to ammeter and voltmeter. Understanding the formulas and concepts in these topics are important to get a better idea while solving the problems.

Key Features of moving charges and magnetism class 12 ncert solutions

Comprehensive Coverage: The class 12 physics chapter 4 exercise solutions cover all the topics and questions presented in the Class 12 Physics Chapter "Moving Charges and Magnetism."

Step-by-Step Explanations: Each class 12 physics chapter 4 question answer provides detailed step-by-step explanations, helping students understand complex concepts.

Clarity and Simplicity: The physics chapter 4 class 12 ncert solutions are written in clear and simple language, making it easier for students to comprehend the content.

Practice Questions: Exercise questions are included to help students practice and assess their understanding.

Exam Preparation: These class 12 physics ch 4 exercise solutions aid students in preparing effectively for board exams and competitive exams like JEE and NEET.

Foundation for Advanced Topics: The concepts covered in this chapter are essential for more advanced topics in physics and electrical engineering.

Free Access: These solutions are available for free, ensuring accessibility to all students.

These features make the Class 12 "Moving Charges and Magnetism" solution a valuable resource for students, facilitating their success in exams and future studies.

Also Check NCERT Books and NCERT Syllabus here:

NCERT Books Class 12 Physics
NCERT Syllabus Class 12 Physics
NCERT Books Class 12
NCERT Syllabus Class 12
NCERT Exemplar Class 12 Solutions

NCERT Solutions For Class 12 Physics Chapter - Wise

NCERT solutions for Class 12 Mathematics
NCERT solutions for Class 12 Chemistry
NCERT solutions for Class 12 Physics
NCERT solutions for Class 12 Biology

As CBSE board exam is concerned, the solutions of NCERT Class 12 Physics chapter 4 Moving Charges and Magnetism is important. In 2019 CBSE board exam 12 % of questions are asked from chapter 4 and 5. Same questions discussed in the chapter 4 Physics Class 12 NCERT solutions can be expected in the board exams. NCERT Exemplar Class 12 Physics Solutions

Frequently Asked Question (FAQs)

The chapter Moving Charges and Magnetism have 8 to 10 percentage weightage. The questions asked from the chapter can be of a numerical, derivation or theory questions. CBSE board follows NCERT Syllabus. To practice problems refer to NCERT text book, NCERT syllabus and previous year board papers of Class 12 Physics.

No, you can not skip it. Since from NCERT Class 12 Physics chapter 4 you can expect 2 questions for NEET exam.

No, you have to practise more questions for doing well in JEE main exams. The questions of Moving Charges and Magnetism need a good base of vectors and a thorough understanding of concepts

Electromagnetic theory is a theory that explains the relationship between electric fields and magnetic fields. It is based on the idea that electric and magnetic fields are two different aspects of the same underlying electromagnetic field.

Magnetic resonance imaging (MRI) is a medical technique that uses a strong magnetic field and radio waves to create detailed images of the inside of the body. It is used to diagnose a wide range of medical conditions.

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Questions related to CBSE Class 12th

Dear Aspirant !

Hope you are doing well !

These questions can be asked which are as follows;-

What are liquid assets? ...
What is the purpose of audits? ...
What is a value-added tax? ...
What is a capital asset? ...
What is a subsidy? ...
What is the role of a public relations department? ...
What do you understand about equilibrium? ...
What is a futures market?.

Hope it helps you!

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Colleges After 12th

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Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive.

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

Geotechnical engineer

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The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art , science , and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

Product Manager

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Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

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Operations manager.

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

Bank Probationary Officer (PO)

Investment director.

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities.

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.

Naval Architect

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Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

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Pathologist.

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

Speech Therapist

Gynaecologist.

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

Videographer

Multimedia specialist.

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism , Advertising , Marketing Management . Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French , German , Italian .

Public Relation Executive

Travel journalist.

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

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NCERT Solutions for Class 12 Physics Chapter 4: Moving Charges and Magnetism

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For the benefit of CBSE Class 12 Science students, the NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism are provided here. The concepts of Moving Charges and Magnetism are covered in Chapter 4 of the NCERT Solutions for Class 12 Physics . This chapter also covers several topics with major markings, such as the magnetic field and direction of a circular coil, magnetic force, and so on.

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The ideas covered in this chapter of NCERT Solutions for Class 12 Physics are vital not only for the CBSE Term I test but also for competitive exams such as JEE and NEET. Subject experts generate the solutions by the most recent term – I CBSE Syllabus 2021-22 and its requirements. Students may now access the NCERT Solutions for Class 12 Physics Chapter 4 in PDF format using the link provided below.

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CBSE Class 12 Physics Chapter 4 Revision Notes – Moving Charges and Magnetism

Moving charges and magnetism class 12 cbse revision notes.

Chapter 4 – Moving charges and magnetism, is from the class 12 textbook of physics. In this chapter, the students will learn how the magnetic field exerts forces on moving charged particles, like electrons, protons, and some wires carrying current. Moreover, various topics will be covered in moving charges and magnetism class 12 notes which will be clearing all the doubts of the students on this topic. Further, there are many derivations and exercises that will let you practice more.

Topics like a magnetic force, motion in a magnetic field, motion in a combined electric field and magnetic fields and many other topics are covered in this chapter. In addition, many laws are there through which you have to solve the derivations and the questions. Thus, there are a lot of questions in the exercise portion as well in moving charges and magnetism class 12 notes.

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Sub-topics covered under Moving Charges and Magnetism:

Ampere’s Circuital Law
Magnetic Field Due to a Current Element, Biot-Savart Law
Magnetic Force and Magnetic Field
Motion in Combined Electric and Magnetic Field
Moving Coil Galvanometer
The Solenoid and the Toroid
Torque on Current Loop, Magnetic Dipole

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Moving Charges and Magnetism Class 12 Notes CBSE Physics Chapter 4 (Free PDF Download)
Revision Notes

Revision Notes for CBSE Class 12 Physics Chapter 4 (Moving Charges and Magnetism) - Free PDF Download

Magnetism and moving charges, i.eElectricity has been studied for more than a century now. The precursor to the relationship between the two was the phenomena that were noted with the alignment of a needle. It was seen that its alignment remains tangent to an imaginary circle where the centre has a straight wire and the plane of the circle is perpendicular to the wire. However, when the current is passed, the needle's orientation changes. It was theorized that the flow of charges leads to the creation of a magnetic field. The related developments on this topic have been included in Moving Charges And Magnetism Class 12 Notes by Vedantu.

Important Topics Covered in This Chapter

The important topics that are covered under the chapter on Moving Charges and Magnetism are listed as follows:

Magnetic Field Caused by a Current Element, Biot-Savart Law.

Magnetic Field and Magnetic Force.

Ampere's Circuital Law.

Motion in Combined Electric and Magnetic Field.

The Toroid and the Solenoid.

Torque on Current Loop, Magnetic Dipole.

The Moving Coil Galvanometer.

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Moving Charges and Magnetism Class 12 Notes Physics - Basic Subjective Questions

Section-a (1 marks questions).

1. Is any work done on a moving charge by a magnetic field?

Ans. Since the magnetic force is perpendicular to the displacement of the moving charge, therefore the work done by the magnetic force is zero.

2. What is the gyromagnetic ratio?

Ans. Gyromagnetic ratio of an electron is the ratio of its magnetic moment to its orbital angular momentum.

$\mu _{l}=\dfrac{e}{2m_{c}}l$

μ l is the magnetic moment of the electron, l is the oriental angular momentum of the electron.

$g(gyromagnetic\;ratio)=\dfrac{\mu _{l}}{l}=\dfrac{e}{2m_{c}}$

3. If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis, which way would the Magnetic force be for :

(a) an electron (negative charge),

(b) a proton (positive charge)?

(a) For electron, the force will be along –z axis;

(b) For a positive charge (proton), the force is along the +z axis.

4. Why should the spring/suspension wire in a moving coil galvanometer have a low torsional constant?

Ans. Low torsional constant is basically required to increase the current/charge sensitivity in a moving coil ballistic galvanometer.

5. State one limitation of Ampere’s Circuital Law.

Ans. Ampere’s Circuital Law is only valid for steady currents. If the current changes with time then Ampere’s Circuital Law is not applicable.

Section – B (2 Marks Questions)

6. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer?

Ans. Two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer

(a) Non - Brittle conductor

(b) Restoring Torque per unit Twist should be small.

7. Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons?

Ans. We know that, Torque = NAIB

Since the area of circular loops is more than of a square loop torque experienced by a circular loop is greater.

8. A cyclotron is not suitable to accelerate electrons. Why?

Ans. A cyclotron is not suitable to accelerate electrons because its mass is less due to which they gain speed and step out of the dee immediately.

9. Write the two measures that can be taken to increase the sensitivity of a galvanometer.

Ans. Measures that can be taken to increase the sensitivity of a galvanometer,

Increasing the no. of turns and the area of the coil,

Increasing the magnetic induction and,

Decreasing the couple per unit twist of the suspension wire.

10. Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.

Ans. The electric field and magnetic field should be perpendicular to each other. The velocity of the moving charge should be, v = E/B.

PDF Summary - Class 12 Physics Moving Charges and Magnetism Notes (Chapter 4)

1. Force on a moving charge:

The source of the magnetic field is a moving charge.

Suppose a positive charge $\text{q}$ is in motion in a uniform magnetic field $\overset{\to }{\mathop{\text{B}}}\,$ with velocity $\overset{\to }{\mathop{\text{v}}}\,$ .

$\text{n}$

$\therefore \,\,\text{F}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{qBvsin }\!\!\theta\!\!\text{ }\Rightarrow \text{F=kqBvsin }\!\!\theta\!\!\text{ }\,\,\left[ \text{k}=\text{constant} \right]$

Where in S.I. system, $\text{k}=1$

$\therefore \,\,\text{F=qBsin }\!\!\theta\!\!\text{ }\,\,\text{and}\,\,\overset{\to }{\mathop{\text{F}}}\,\text{=q}\left( \overset{\to }{\mathop{\text{v}}}\,\times \overset{\to }{\mathop{\text{B}}}\, \right)$

2. Magnetic field strength $\left( \overset{\to }{\mathop{\text{B}}}\, \right)$ :

We can see that in the equation,

$\text{F}=\text{qBvsin }\!\!\theta\!\!\text{ }$, if $\text{q}=1,\,\text{v}=1$,

$\text{sin }\!\!\theta\!\!\text{ }=1$ i.e. $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$ then $\text{F}=\text{B}$.

Therefore magnetic field strength can be known as the force felt by a unit charge in motion with unit velocity perpendicular to the direction of the magnetic field.

There are some special cases for this:

If $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$ or

${{180}^{\circ}}$, $\text{sin }\!\!\theta\!\!\text{ =0}$

$\therefore \,\,\text{F}=0$

A charged particle that is in motion parallel to the magnetic field will be not experiencing any force.

When $\text{v}=0,\text{F}=0$

At rest, a charged particle in a magnetic field will be not experiencing any force.

When $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }=1$ then the force will be maximum

${{\text{F}}_{\text{max}\text{.}}}=\text{qvB}$

A charged particle in a motion perpendicular to the magnetic field will be experiencing maximum force.

3. S.I. unit of magnetic field intensity:

The S.I unit has been found to be tesla (T).

$\text{B}=\frac{\text{F}}{\text{qvsin }\!\!\theta\!\!\text{ }}$

When $\text{q}=1\text{C},\,\text{v}=\text{1m/s},\,\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$ That is, $\text{sin }\!\!\theta\!\!\text{ }=1$ and $\text{F}=\text{1N}$

Then $\text{B}=\text{1T}$.

At a point, the strength of the magnetic field can be called as \[\text{1T}\] if a charge of \[\text{1C}\] which have a velocity of $1$ m/s while in motion at the right angle to a magnetic field experiences a force of \[\text{1N}\] at that point.

4. Biot-Savart’s law:

The strength of magnetic flux density or magnetic field at a point \[\text{P}\](dB) because of the current element \[\text{dl}\] will be dependent on,

$\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{I}$

$\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{dl}$

$\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{sin }\!\!\theta\!\!\text{ }$

$\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\frac{\text{1}}{{{\text{r}}^{\text{2}}}}$,

When we combine them, $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}\Rightarrow \text{dB=k}\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}$ [$\text{k}=$Proportionality constant]

In S.I. units, $\text{k}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}$ where ${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}$ can be called as permeability of free space.

${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}=\text{4 }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{\text{-7}}}\,\text{T}{{\text{A}}^{\text{-1}}}\text{m}$

$\therefore \,\,\text{dB}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}$ and \[\overset{\to }{\mathop{\text{dB}}}\,=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\text{I}\frac{\left( \overset{\to }{\mathop{\text{dl}}}\,\times \overset{\to }{\mathop{\text{r}}}\, \right)}{{{\text{r}}^{\text{3}}}}\]

$\overset{\to }{\mathop{\text{dB}}}\,$ will be perpendicular to the plane containing $\overset{\to }{\mathop{\text{dl}}}\,$ and $\overset{\to }{\mathop{\text{r}}}\,$ and will be directed inwards.

5. Applications of Biot-Savart’s law:

Magnetic field $\left( \text{B} \right)$ kept at the Centre of a Current Carrying Circular Coil of radius $r$.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{2r}}$

If there are $n$ turns, then the magnetic field at the centre of a circular coil of n turns will be,

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}}{\text{2r}}$

Here \[\text{n}\] will be the number of turns of the coil. \[\text{I}\] will be the current in the coil and \[\text{r}\] will be the radius of the coil.

Magnetic field because of a straight conductor carrying current.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{4 }\!\!\pi\!\!\text{ a}}\left( \sin {{\phi }_{2}}+\sin {{\phi }_{1}} \right)$

Here \[\text{a}\] will be the perpendicular distance of the conductor from the point where the field is to the measured.

${{\phi }_{1}}\,\,\text{and}\,\,{{\phi }_{2}}$ will be the angles created by the two ends of the conductor with the point.

In case of an infinitely long conductor, ${{\phi }_{1}}={{\phi }_{2}}=\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

$\therefore \,\,\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\text{.}\frac{\text{2I}}{\text{a}}$

At a point on the axis, magnetic field of a Circular Coil Carrying Current.

If point \[\text{P}\] is lying far away from the centre of the coil.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\frac{\text{2M}}{{{\text{x}}^{\text{3}}}}$

Where $\text{M}=\text{nIA}=\text{magnetic}\,\text{dipole}\,\text{moment}\,\text{of}\,\text{the}\,\text{coil}$.

$\text{x}$ be the distance of the point where the field is needed to be measured, \[\text{n}\] be the number of turns, \[\text{I}\] be the current and \[\text{A}\] be the area of the coil.

Magnetic field at the centre of a semi-circular current-carrying conductor will be,

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{4a}}$

Magnetic field at the centre of an arc of circular current-carrying conductor which is subtending an angle 0 at the centre will be,

$\text{B=}\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I }\!\!\theta\!\!\text{ }}{\text{4 }\!\!\pi\!\!\text{ a}}$

6. Ampere’s circuital law:

Around any closed path in vacuum line integral of magnetic field \[\overset{\to }{\mathop{\text{B}}}\,\] will be ${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}$ times the total current through the closed path. that is, $\oint{\overset{\to }{\mathop{\text{B}}}\,\text{.}\overset{\to }{\mathop{\text{dl}}}\,}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}$

7. Application of Ampere’s circuital law:

Magnetic field because of a current carrying solenoid, $\text{B}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}$

$\text{n}$ be the number of turns per unit length of the solenoid.

In the edge portion of a short solenoid, $\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}}{\text{2}}$

Magnetic field because of a toroid or endless solenoid

$\text{B}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}$

8. Motion in uniform electric field of a charged particle:–

Parabola is the path of a charged particle in an electric field.

Equation of the parabola be ${{\text{x}}^{\text{2}}}=\frac{\text{2m}{{\text{v}}^{\text{2}}}}{\text{qE}}\text{y}$

Where $\text{x}$ be the width of the electric field.

$\text{y}$ be the displacement of the particle from its straight path.

$\text{v}$ be the speed of the charged particle.

$\text{q}$ be the charge of the particle

$\text{E}$ be the electric field intensity.

$\text{m}$ be the mass of the particle.

9. In a magnetic field $(\overset{\to }{\mathop{B}}\,)$ which is uniform, the path of a particle which is charged in motion with a velocity $\overset{\to }{\mathop{\text{v}}}\,$ creating an angle $\text{ }\!\!\theta\!\!\text{ }$ with $\overset{\to }{\mathop{\text{B}}}\,$ will be a helix.

Velocity is inclined to the magnetic field

The component of velocity \[\text{vcos }\!\!\theta\!\!\text{ }\] will not be given a force to the charged particle, hence under this velocity in the direction of \[\overset{\to }{\mathop{\text{B}}}\,\], the particle will move forward with a fixed velocity. The other component $\text{vsin }\!\!\theta\!\!\text{ }$ will create the force $\text{F}=\text{qBvsin }\!\!\theta\!\!\text{ }$, which will be supplying the needed centripetal force to the charged particle in the motion along a circular path having radius $\text{r}$.

$\because \,\,\text{Centripetal}\,\text{force}=\frac{\text{m}{{\left( \text{vsin }\!\!\theta\!\!\text{ } \right)}^{\text{2}}}}{\text{r}}=\text{Bqvsin }\!\!\theta\!\!\text{ }$

$\therefore \,\,\text{vsin }\!\!\theta\!\!\text{ }=\frac{\text{Bqr}}{\text{m}}$

Angular velocity of rotation$=\text{w=}\frac{\text{vsin }\!\!\theta\!\!\text{ }}{\text{r}}=\frac{\text{Bq}}{\text{m}}$

Frequency of rotation\[=\text{v}=\frac{\text{ }\!\!\omega\!\!\text{ }}{\text{2 }\!\!\pi\!\!\text{ }}=\frac{\text{Bq}}{\text{2 }\!\!\pi\!\!\text{ m}}\]

Time period of revolution$=\text{T=}\frac{\text{1}}{\text{v}}=\frac{\text{2 }\!\!\pi\!\!\text{ m}}{\text{Bq}}$

10. Cyclotron:

This can be defined as a device we use for accelerating and therefore energize the positively charged particle. This can be created by keeping the particle, in an oscillating perpendicular magnetic field and a electric field. The particle will be moving in a circular path.

$\therefore \,\,\,\text{Centripetal}\,\text{force}=\text{magnetic}\,\text{Lorentz}\,\text{force}$

$\Rightarrow \frac{\text{m}{{\text{v}}^{\text{2}}}}{\text{r}}=\text{Bqv}\Rightarrow \frac{\text{mv}}{\text{Bq}}=\text{r}$ $\leftarrow $ radius of the circular path

Time for travelling a semicircular path$=\frac{\text{ }\!\!\pi\!\!\text{ r}}{\text{v}}=\frac{\text{ }\!\!\pi\!\!\text{ m}}{\text{Bq}}=\text{constant}$.

When ${{\text{v}}_{\text{0}}}$ be the maximum velocity of the particle and ${{\text{r}}_{\text{0}}}$ be the maximum radius of its path then we can say that,

$\frac{\text{mv}_{\text{0}}^{\text{2}}}{{{\text{r}}_{\text{0}}}}=\text{Bq}{{\text{v}}_{\text{0}}}\Rightarrow {{\text{v}}_{\text{0}}}=\frac{\text{Bq}{{\text{r}}_{\text{0}}}}{\text{m}}$

Maximum kinetic energy of the particle$=\frac{\text{1}}{\text{2}}\text{mv}_{\text{0}}^{\text{2}}=\frac{\text{1}}{\text{2}}\text{m}{{\left( \frac{\text{Bq}{{\text{r}}_{\text{0}}}}{\text{m}} \right)}^{\text{2}}}\Rightarrow {{\left( \text{K}\text{.E}\text{.} \right)}_{\text{max}\text{.}}}=\frac{{{\text{B}}^{\text{2}}}{{\text{q}}^{\text{2}}}\text{r}_{\text{0}}^{\text{2}}}{\text{2m}}$

Time period of the oscillating electric field$\Rightarrow \text{T}=\frac{\text{2 }\!\!\pi\!\!\text{ m}}{\text{Bq}}$.

Time period be the independent of the speed and radius.

Cyclotron frequency $=\text{v}=\frac{\text{1}}{\text{T}}=\frac{\text{Bq}}{\text{2 }\!\!\pi\!\!\text{ m}}$

Cyclotron angular frequency$={{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}=\text{2 }\!\!\pi\!\!\text{ v}=\frac{\text{Bq}}{\text{m}}$

11. Force acting on a current carrying conductor kept in a magnetic field will be,

\[\overset{\to }{\mathop{\text{F}}}\,=\text{I}\left| \overset{\to }{\mathop{\text{l}}}\,\times \overset{\to }{\mathop{\text{B}}}\, \right|\] or $\text{F}=\text{IlBsin }\!\!\theta\!\!\text{ }$

Here $\text{I}$ be the current through the conductor

$\text{B}$ be the magnetic field intensity.

$\text{l}$ be the length of the conductor.

$\text{ }\!\!\theta\!\!\text{ }$ be the angle between the direction of current and magnetic field.

If $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$ or ${{180}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }\Rightarrow 0\Rightarrow \text{F}=0$

$\therefore $ If a conductor is kept along the magnetic field, no force will be acting on the conductor.

If $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }=1$, $\text{F}$ will be maximum.

${{\text{F}}_{\text{max}}}\text{=IlB}$

If the conductor has been kept normal to the magnetic field, it will be experiencing maximum force.

12. The force between two parallel current-carrying conductors:–

If the current will be in a similar direction the two conductors will be attracting each other with a force

$\text{F}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}.\frac{\text{2}{{\text{I}}_{\text{1}}}{{\text{I}}_{\text{2}}}}{\text{r}}$ per unit length of the conductor

If the current is in opposite direction the two conductors will be repelling each other with an equal force.

S.I. unit of current is \[1\] ampere. \[\left( \text{A} \right)\text{.}\]

$\text{1A}$ can be defined as the current which on flowing through each of the two parallel uniform linear conductor kept in free space at a distance of $\text{1m}$ from each other creates a force of $\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}\,\text{N/m}$ along their lengths.

13. Torque experienced on a current carrying coil kept in a magnetic field:

$\overset{\to }{\mathop{\tau }}\,=\overset{\to }{\mathop{\text{M}}}\,\times \overset{\to }{\mathop{\text{B}}}\,\Rightarrow \tau =\text{MBsin }\!\!\alpha\!\!\text{ }=\text{nIBAsin }\!\!\alpha\!\!\text{ }$ where $\text{M}$ be the magnetic dipole moment of the coil.

$\text{M}=\text{nIA}$

Where $\text{n}$ be the number of turns of the coil.

$\text{I}$ be the current through the coil.

$\text{B}$ be the intensity of the magnetic field.

\[\text{A}\] be the area of the coil.

\[\text{ }\!\!\alpha\!\!\text{ }\] will be the angle in between the magnetic field $\left( \overset{-}{\mathop{\text{B}}}\, \right)$ and normal to the plane of the coil.

Special Cases will be:

When the coil has been kept parallel to magnetic field $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$, $\text{cos }\!\!\theta\!\!\text{ }=1$ then torque will be maximum.

${{\text{ }\!\!\tau\!\!\text{ }}_{\text{max}\text{.}}}=\text{nIBA}$

When the coil is kept perpendicular to magnetic field, $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{cos }\!\!\theta\!\!\text{ }=0$

$\therefore \,\,\text{ }\!\!\tau\!\!\text{ }=0$

14. Moving coil galvanometer:

This has been on the basis on the principle that if a coil carrying current has been kept in a magnetic field it is experiencing a torque. There is a restoring torque because of the phosphor bronze strip which is bringing back the coil to its normal position.

In equilibrium,

\[\text{Deflecting torque }=\text{ Restoring torque}\]

$\text{nIBA}=\text{k }\!\!\theta\!\!\text{ }$ [$\text{k}=\text{restoring torque/unit twist of the phosphor bronze strip}]\text{ }$

$\text{I}=\frac{\text{k}}{\text{nBA}}\text{ }\!\!\theta\!\!\text{ }=\text{G }\!\!\theta\!\!\text{ }$ where $\text{G}=\frac{\text{k}}{\text{nBA}}=\text{Galvanometer}\,\,\text{constant}$

$\therefore \,\,\text{I}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{ }\!\!\theta\!\!\text{ }$

Current sensitivity of the galvanometer can be defined as the deflection made if the unit current has been passed through the galvanometer.

${{\text{I}}_{\text{s}}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{I}}=\frac{\text{nBA}}{\text{k}}$

Voltage sensitivity can be explained as the deflection created if unit potential difference has been applied across the galvanometer.

${{\text{V}}_{\text{s}}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{V}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{IR}}=\frac{\text{nBA}}{\text{kR}}$ $\left[ \text{R = Resistance of the galvanometer} \right]$

15. The maximum sensitivity of the galvanometer is having some conditions:-

The galvanometer has been defined to be sensitive if a small current develops a large deflection.

$\because \,\,\text{ }\!\!\theta\!\!\text{ }=\frac{\text{nBA}}{\text{k}}\text{I}$

$\because \,\,\text{ }\!\!\theta\!\!\text{ }$ will be large if (i) $\text{n}$ is large, (ii) $\text{B}$ is large (iii) $\text{A}$ is large and (iv) $\text{k}$ is small.

16. Conversion of galvanometer into voltmeter and ammeter:

A galvanometer has been converted to voltmeter by putting a high resistance in series with it.

\[\text{Total resistance of voltmeter }=\text{ }{{\text{R}}_{\text{g}}}\text{ + R}\] where ${{\text{R}}_{\text{g}}}$ be the galvonometer resistance.

$\text{R}$ be the resistance added in series.

Current through the galvanometer$={{\text{I}}_{\text{g}}}=\frac{\text{V}}{{{\text{R}}_{\text{g}}}\text{+R}}$

Here $\text{V}$ is the potential difference across the voltmeter.

Conversion of galvanometer into voltmeter

$\therefore \,\,\text{R}=\frac{\text{V}}{{{\text{I}}_{\text{g}}}}-\text{G}$

Range of the voltmeter: \[0\text{V volt}\text{. }\]

A galvanometer can be converted into an ammeter by the connection of a low resistance in parallel with it (shunt)

$\text{Shunt}=\text{S}=\left( \frac{{{\text{I}}_{\text{g}}}}{\text{I}-{{\text{I}}_{\text{g}}}} \right){{\text{R}}_{\text{g}}}$ where ${{\text{R}}_{\text{g}}}$ be the galvanometer’s resistance.

$\text{I}$ be the total current through the ammeter.

${{\text{I}}_{\text{g}}}$ be the current through the ammeter.

Effective resistance of the ammeter will be,

$\text{R}=\frac{{{\text{R}}_{\text{g}}}}{{{\text{R}}_{\text{g}}}+\text{S}}$

The range of the ammeter will be $0-\text{IA}$. An ideal ammeter will be having zero resistance.

Class 12th Physics Chapter 4 Notes Free PDF Download

Physics Class 12 Chapter 4 Notes illustrate the exertion of the magnetic field on the forces surrounding the current-carrying wires. It also shows how magnetic fields are produced with the help of currents. It is seen that within a cyclotron, there can be an acceleration of particles at relatively high energies.

The notes of Physics Class 12 Chapter 4 includes the sources, magnetic fields, motion within a magnetic field or in a combination of electric and magnetic fields. Apart from the laws that are discussed, there are ample derivations to take note of as well as mathematical sums on the topic. Solving this needs a clear understanding of the principles in this chapter. Students may download a free PDF of Class 12 Physics Chapter 4 Notes to prepare the topic better.

Moving Charges and Magnetism

The key concepts that are covered in the notes of Class 12 Physics Chapter 4 are:

1. Force on a Moving Charge

(image will be uploaded soon)

Magnetic field imposes a force on moving charges. The direction of the force that is exerted on a moving charge remains perpendicular to the plane that is formed which is denoted by 'v' and 'B' respectively. The 'Right-Hand Rule' is followed in this case. Moreover, the force's magnitude is in direct proportion to the sine of the angle between 'B' and 'v'.

2. Strength of Magnetic Field

Magnetic field strength is also known as magnetic field intensity. It includes such a component of a magnetic field which is a result of external current. It is denoted as a vector, H and the unit for measurement is amperes per metre. The equation is,

H = B/μ − M

B = magnetic flux density

μ = magnetic permeability

M = magnetisation]

3. Biot Savart's Law

The law relates to current sources and magnetic fields. The determination of the magnetic field arises from the distribution of current involving vector products. It includes a problem in calculus where the distance among field points and current is constantly changing. Its major application includes calculating the magnetic impact on a molecular and atomic level and determining velocity in aerodynamics.

4. Ampere's Circuital Law

Ampere's Circuital Law encompasses the relationship between the magnetic field and its source current. The magnetic field density on an imaginary enclosed path amounts to be same as the product of current in the medium and permeability of the latter.

Moving charges and Magnetism Class 12 Notes also discuss cyclotron. It is engaged in the acceleration of charged atomic particles within a uniform magnetic field. It results in the production of radioactive isotopes which are utilised for imaging procedures.

Benefits of CBSE Class 12 Revision Notes on Physics Chapter 4 - Moving Charges and Magnetism

The following are the benefits of Vedantu’s Revision Notes on Class 12 CBSE Physics Chapter 4 - Moving Charges and Magnetism:

Our revision notes cater to students' need for readability as well as detailed content as a part of their preparation experience.

These notes are best suited for students who wish to cover the syllabus in a short period of time, without actually missing out on any important concept or idea in the chapter.

Vedantu’s revision notes prepare students to gain the ability to solve any kind of question that is asked from the specific chapter.

Students will be greatly benefitted from Vedantu’s revision notes on CBSE Class 12 Physics Chapter 4 - Moving Charges and Magnetism in terms of proper internalization and conceptualisation of the topics covered in the chapter. They will be able to quickly identify the questions that demand explanation or elaboration of the ideas covered in these notes, present in the chapter.

FAQs on Moving Charges and Magnetism Class 12 Notes CBSE Physics Chapter 4 (Free PDF Download)

1. What are the Different Applications of Ampere's Circuital Law?

Class 12 Physics Moving Charges and Magnetism Notes include a concise explanation of the various applications of Ampere's Circuital Law. It includes – (1) long current-carrying wire induced magnetism, (2) long current transmitting cylinder creating a magnetic field and (3) current carrying hollow cylinder creating a magnetic field. The relevant derivation of all these applications is elaborated in the notes of Class 12 Chapter Moving Charges and Magnetism.

2. What is Understood by a Cyclotron?

As discussed in Moving Charges and Magnetism Notes, the cyclotron is one of the first particle accelerators. Even though it has subsequently undergone a host of modification, prototypes still find application in the initial stage of specific multi-stage particle accelerators. NCERT Class 12 Physics Chapter 4 notes also discuss that the property of a magnetic force's impact on a moving charge is utilised for bending the latter onto a semi-circular pathway.

3. What is the Relationship Between Moving Charges and Magnetism?

Moving charges or flow of charge causes magnetism. Notes of Moving Charges and Magnetism explain that magnetic fields further exert forces on the flow of charge, which in turn, exerts a force on other magnets. Such a phenomenon takes place because of the presence of consistent moving charges.

Given that electricity is a flow of moving charge, the relationship between electricity and magnetism involves both attraction and repulsion among various charged particles, as well as exertion of force within such charges. As indicated in the Magnetic Effect of Electric Current Class 12 Notes such interaction between electricity and magnetism is termed as electromagnetism.

4. What are the Applications of Biot Savart Law?

The Biot Savart Law has been mentioned in Moving Charges and Magnetism class 12 Notes. The major application of the law includes – (1) calculation of magnetic reactions on an atomic and molecular level, (2) determination of velocity within the theory of aerodynamics. The derivations from Biot Savart Law explained in CBSE Class 12 Physics Chapter 4 Notes show that elements of current can be calculated with the help of this theory.

5. How are NCERT Solutions helpful in preparation for Physics of Class 12?

The essential principles are defined well in NCERT Solutions and Notes for all the chapters in Physics Class 12. A comprehensive reading of these works improves one's grasp of the fundamental principles. They also provide you with answers to all the questions in the textbook as well as extra questions designed by subject experts. Once you've mastered the fundamentals and practiced well, you will be ready to ace your Physics exam.

6. Where can I download Notes for Class 12th Physics Chapter 4?

Physics of Class 12 is a subject that students often find difficult to cope up with. Right guidance can assist them in their exam preparations. The sources, magnetic fields, motion inside a magnetic field, or motion in a mix of electric and magnetic fields are all covered in the Physics Class 12 Chapter 4 notes. There are several derivations and numericals on the subject in the notes. You can download the notes from the official website of Vedantu and the Vedantu app.

7. Should I practice all the questions in NCERT Solutions for Chapter 4 of Physics class 12th?

It is critical to understand all of the ideas and strategies used to answer the problems in Chapter 4 of Physics Class 12th. The chapter contains numerous numerical sums that require regular practice to help you remember the formula and comprehend how to solve them. The theory also requires thorough learning and revision. Hence, it is suggested that you solve all the questions in the NCERT Solutions before the exams to score good marks.

8. What do you understand about the Right-Hand Rule according to Chapter 4 of Class 12 Physics?

The Right-Hand Rule is applied to calculate the direction of the magnetic force on a (+ive) moving charge. In this case, we stretch our right hand's thumb and first two fingers such that the thumb points to the conductor's motion, the first finger points to the magnetic field, and the middle finger points to the induced current. It is important to learn the theory, definition as well as the explanation, along with knowing the application of the formula for numericals.

9. What is the equation for the strength of a magnetic field according to Chapter 4 of Class 12 Physics?

The strength for a magnetic field, also known as its intensity, is denoted by a vector (H). The unit used for its measurement is ampere/metre. It is calculated using magnetic flux density (B), magnetic permeability (μ), and magnetisation (M) forming the equation: H = B/μ − M. It is critical to understand the theory, definition, and explanation, as well as the application of the numerical formula for the equation of a magnetic field.

Previous Year Question Papers CBSE Class 12

CBSE Notes For Class 12
Physics Notes Class 12
Chapter 4: Moving Charges And Magnetism

CBSE Class 12 Physics Notes Chapter 4 Moving Charges and Magnetism

What is lorentz force.

When the total force on a charge c which is moving with a velocity v in the presence of electric field E and magnetic field B is termed as the lorentz force.

F= q(v * B + E) (This acts normal to v and the work done by it is zero)

Cyclone Frequency

A charge c completes a circular orbit on a plane normal to B which is the uniform magnetic field . The uniform circular motion frequency is known as cyclone frequency. This frequency is unaffected by the radius and speed of the particle. It can be determined with the help of a machine known as a cyclotron, which is used to accelerate the particles which are charged.

Biot-Savart law

According to this law, the magnetic field dB due to the presence of an element dl which is carrying a continuous flow of current I at the point P at a distance r from the current element, is to obtain a total field at P and integration of this vector expression over the entire length of the conductor.

Ampere’s Circuital Law

According to this law, an open surface S, if bounded by a loop C, then the current I passing through S is determined by

The sign of I is determined by the right-hand rule. Non-parallel currents repel and parallel currents attract each other.

For more information on Ampere’s Circuital Law and the Magnetic Effects of Electric Current, watch the below videos

To know more about electric charge and magnetism, keep visiting BYJU’S website.

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

Moving Charges and Magnetism Study Notes
Moving Charges and Magnetism MCQ Practice Questions
Moving Charges and Magnetism MCQ Practice Solution

Important Questions

Calculate the magnetic field at the coil’s centre if the radius and current flowing in a circular coil of wire having 200 turns are 20 cm and 2 A, respectively.
Examine the magnitude of the field at a point 60 cm from the wire if the Current flowing in a straight wire is 45 A.
Find out the direction and magnitude at a point 6 m east of the wire in which the current of 70 A is flowing and is passing in the north and south direction.

Frequently Asked Questions on Moving Charges and Magnetism

An electron is projected into a uniform magnetic field of 3t and moves along a helical path of radius 1 cm with pitch 2π cm. find the angle of projection of the electron with the magnetic field..

The angle of projection of the electron with the magnetic field is 45 0 .

Can a cyclotron accelerate?

Cyclotron is an apparatus used to accelerate atomic and subatomic particles by an alternating electric field in a constant magnetic field.

A beam of electrons passes, un-deflected, through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move in which direction?

The direction of motion of an electron must be perpendicular to both electric and magnetic fields. When an electric field is switched off, then the magnetic field is perpendicular to the direction of motion of the charged particle, and therefore, the electrons start exhibiting circular motion.

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NCERT Solutions for Class 12 Physics Chapter 4 PDF Download

Some of the students often waste their time searching for the right study material; for them, the NCERT Solutions for Class 12 Physics Chapter 4 are the right ones. It is considered to be the right one as it provides accurate answers for the Class 12 Physics Chapter 4; accordingly, students can improve their accuracy level.

NCERT Solutions for Class 12 Physics Chapter 4 PDF

To have success in Class 12 Physics Chapter 4, students need to take some effort or steps. One of the important steps is to access the NCERT Solutions for Class 12 Physics Chapter 4 PDF and need to start solving questions. Students can access the Class 12 Physics Chapter 4 questions through the Selfstudys website from their comfort zone.

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It is important for students to download the NCERT Solutions for Class 12 Physics Chapter 4 so that they can easily refer to the answers; steps to download are:

Open the Selfstudys website.

NCERT Solutions for Class 12 Physics Chapter 4, NCERT Solutions for Class 12 Physics Chapter 4 PDF, Download NCERT Solutions for Class 12 Physics Chapter 4 PDF, NCERT Solutions for Class 12 Physics Chapter 4 Revision, NCERT Solutions for Class 12 Physics Chapter 4 Theory

Click the navigation button and select NCERT Books & Solutions.

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Select Class 12th from the list

And select Physics from the list of subjects.
A new page will appear, now click Chapter 4 from the list of Chapters.

Everything Students Need to Know About NCERT Solutions for Class 12 Physics Chapter 4 PDF

The NCERT Solutions for Class 12 Physics Chapter 4 PDF are considered to be important study material for students as it provides answers to all the textbook questions; here is that students need to know:

Keywords or Formulas are Presented: Formulas provide a way to present the substance using the symbol of elements; these keywords or formulas are given in the NCERT Solutions for Class 12 Physics Chapter 4.
Explained in Step Wise Manner: The answers in the NCERT Solutions for Class 12 Physics Chapter 4 revision are explained in a stepwise manner so that students can understand each point and step.
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Tables are Included: Inside the Class 12 Physics NCERT Solutions, tables are included for Chapter 4 so that students can have organised information about the important themes.
Available in the PDF: The Class 12 Physics questions of Chapter 4 from the NCERT Solutions are available in the PDF so that students can access accurate answers from their comfort zone.

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Identify the Patterns of Mistakes: Patterns of mistakes are considered to be those errors which are done by most students; in that case, they need to identify the pattern of mistakes while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 revision.
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Class 12 Physics Case Study Questions Chapter 4 Moving Charges and
Here, we have provided case-based/passage-based questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism. Case Study/Passage-Based Questions. Case Study 1: A charged particle moving in a magnetic field experiences a force that is proportional to the strength of the magnetic field, the component of the velocity that is perpendicular ...
Case Study Question for Class 12 Physics Chapter 4 Moving Charges and
Case Study Question for Class 12 Physics Chapter 4 Moving Charges and Magnetism Case Study Question 1: Moving coil galvanometer operates on Permanent Magnet Moving Coll (PMMC) mechanism and was designed by the scientist Darsonval. Moving coil galvanometers are of two types(i) Suspended coll(ii) Pivoted coil type or tangent galvanometer, Its working is based on … Continue reading Case Study ...
Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics
To increase the current sensitivity of a moving coil galvanometer, we should decrease. (a) strength of magnet. (b) torsional constant of spring. (c) number of turns in coil. (d) area of coil. Answer. Please refer to below Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics. These Case Study Questions Class 12 Physics will be ...
CBSE Class 12 Physics Case Study Questions PDF Download
Chapter-wise Solved Case Study Questions for Class 12 Physics. Chapter 1 Electric Charges and Fields. Chapter 2 Electrostatic Potential and Capacitance. Chapter 3 Current Electricity. Chapter 4 Moving Charges and Magnetism. Chapter 5 Magnetism and Matter. Chapter 6 Electromagnetic Induction. Chapter 7 Alternating Current.
Class 12 Physics Chapter 4 Case Study Question Moving Charges and
In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 4 Moving Charges and Magnetism Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Moving ...
Case Study on Moving Charges And Magnetism Class 12 Physics PDF
The PDF file of the Moving Charges And Magnetism Case Study for Class 12 Physics with Solutions is a very important study resource that can help students better prepare for the exam and boost conceptual learning. The solutions are in the hint manner as well as contain full examples too, refer to the link to access the Case Study on Moving ...
NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism
NCERT Solutions for Class 12 Physics Chapter 4 - Free PDF Download. The NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism is given here for the benefit of CBSE Class 12 Science students. In Chapter 4, NCERT Solutions for Class 12 Physics, students will learn the concepts of Moving Charges and Magnetism.Also, this chapter covered several topics with significant marks ...
CBSE Class 12 Physics
Chapter-4: Moving Charges and Magnetism of NCERT class 12 Physics note is available here. Moving Charges and Magnetism is one of the important chapters of CBSE class 12 Physics. So, students must prepare this chapter thoroughly. The notes provided here will be very helpful for the students who are going to appear in CBSE class 12 Physics board ...
NCERT solutions for Physics Class 12 chapter 4
A magnetic field of 100 G (1 G = 10 −4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10 −3 m 2.The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m −1.Suggest some appropriate design particulars of a solenoid for the ...
NCERT Solutions for Class 12 Physics Chapter 4
The smallest value of M is called the Bohr magneton MB and it is MB = 9.27×10-24 J/T. Access NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism. 1. A Circular Coil of Wire Consisting of $100$ Turns, Each of Radius $8.0cm$ Carries a Current of $0.40A$.
NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges ...
Two main laws discussed in the NCERT chapter Moving charges and Magnetism Class 12 are Ampere's law and Biot-savant law. Based on these laws many problems are discussed in CBSE NCERT solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism. The galvanometer is discussed in Moving charges and Magnetism Class 12 in detail with its working principle.
NCERT Exemplar for Class 12 Physics Chapter-4 (Book Solutions)
Free PDF download of NCERT Exemplar for Class 12 Physics Chapter 4 - Moving Charges And Magnetism solved by expert Physics teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 4 - Moving Charges And Magnetism exercise questions with solutions to help you to revise complete syllabus and score more marks in your examinations.
NCERT Solutions for Class 12 Physics Chapter 4: Moving Charges and
The concepts of Moving Charges and Magnetism are covered in Chapter 4 of the NCERT Solutions for Class 12 Physics. This chapter also covers several topics with major markings, such as the magnetic field and direction of a circular coil, magnetic force, and so on. Fill Out the Form for Expert Academic Guidance! Grade. Target Exam.
Category: Case Study Based Questions for Class 12 Physics
August 6, 2021 October 7, 2022 Physics Gurukul Leave a Comment on Case Study Question for Class 12 Physics Chapter 4 Moving Charges and Magnetism. ... March 23, 2021 October 7, 2022 Physics Gurukul Leave a Comment on Case Study Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter.
Physics Case Study for Class 12 (Download Free PDF)
Physics Case Study for Class 12: Here, you will get class 12 case study questions and answers for Physics pdf at free of cost. Along with you can also download Physics case study questions for class 12 chapter wise for getting higher marks in board exams.
Case Study Questions for Class 12 Physics
Case Study Questions for Class 12 Physics. Chapter 1: Electric Charges and Fields. Chapter 2: Electrostatic Potential and Capacitance. Chapter 3: Current Electricity. Chapter 4: Moving Charges and Magnetism. Chapter 5: Magnetism and Matter. Chapter 6: Electromagnetic Induction. Chapter 7: Alternating Current. Chapter 8: Electromagnetic Waves.
CBSE Class 12 Physics Chapter 4
The set of important questions for CBSE Class 12 Physics Chapter 4 - Moving Charges and Magnetism for the academic year 2023-24 prove to be a valuable tool for students' exam preparation. These questions cover vital concepts such as magnetic fields, Ampere's law, and applications of moving charges in magnetic fields.
Class 12 Physics Chapter 4 Notes
Chapter 4 - Moving charges and magnetism, is from the class 12 textbook of physics. In this chapter, the students will learn how the magnetic field exerts forces on moving charged particles, like electrons, protons, and some wires carrying current. Moreover, various topics will be covered in moving charges and magnetism class 12 notes which ...
Important Questions for Class 12 Physics Chapter 4
Important Questions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism. Magnetism is a physical characteristic caused by magnetic fields that produce attractive and repulsive effects on other objects. A magnetic field is generated by electric currents and magnetic moments of particles and influences other currents and magnetic moments.
Moving Charges and Magnetism Class 12 Notes CBSE Physics Chapter 4
PDF Summary - Class 12 Physics Moving Charges and Magnetism Notes (Chapter 4) 1. Force on a moving charge: The source of the magnetic field is a moving charge. Suppose a positive charge q is in motion in a uniform magnetic field →B with velocity →v . n.
CBSE Class 12 Physics Chapter 4 Notes Moving Charges And Magnetism
The direction of motion of an electron must be perpendicular to both electric and magnetic fields. When an electric field is switched off, then the magnetic field is perpendicular to the direction of motion of the charged particle, and therefore, the electrons start exhibiting circular motion. CBSE Notes for Class 12 Physics Charges and ...
NCERT Solutions for Class 12 Physics Chapter 4 PDF Download
To have success in Class 12 Physics Chapter 4, students need to take some effort or steps. One of the important steps is to access the NCERT Solutions for Class 12 Physics Chapter 4 PDF and need to start solving questions. Students can access the Class 12 Physics Chapter 4 questions through the Selfstudys website from their comfort zone.

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