Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics

Exam Questions Class 12

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 rp and rα are the radii of their circular path, then the ratio rp/rα is.
(a) 1/√2
(b) 1/2
(c) 1
(d) √2

Answer

B

Question. In (i) part, what is the ratio rp/ra 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

Answer

B

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

Answer

C

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

Answer

A

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

Answer

D

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

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)

Answer

D

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.

Answer

A

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

Answer

D

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= (2×105 i + 4 × 105 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

Answer

D

Question. The frequency of revolution of the particle is

Answer

B

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

Motion of charge in magnetic field. An electron with speed r0 << 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 V0→ 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 T0.

Question. If the speed of the electron is now doubled to 2V0. The radius of the circle will change to
(a) 4r0
(b) 2r0
(c) r0
(d) r0/2

Answer

C

Question. If V0 = 2V0, then time required for one revolution of the electron will change to
(a) 4T0
(b) 2T0
(c) T0
(d) T0/2

Answer

C

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

Answer

C

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

Answer

D

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

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

Answer

C

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/r2
(c) decreases as 1/r3
(d) approaches a finite limit as r → ∞

Answer

B

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) μ0i/r
(b) 4μ0i/r
(c) zero
(d) μ0i/4r

Answer

C

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.

Answer

A

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

Answer

B

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.

Answer

D

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

Answer

D

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

Answer

B

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) NA2B2I
(b) NABI2
(c) N2ABI
(d) NABI

Answer

D

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

Answer

B

Case Study Chapter 4 Moving Charges and Magnetism Class 12 Physics