The magnetic susceptibility of a substance is - 1 . 66 × 10 - 5 . Name the type of magnetic substance it represents.

Since the susceptibility of the magnetic material is negative. So the magnetic material is diamagnetic. When the diamagnetic are placed in the uniform magnetic field, it is feebly magnetised opposite to the magnetic field.

Describe the principle and working of a moving coil galvanometer and hence show that the deflection of the coil is directly proportional to the current flowing through it. What is the effect of the radial magnetic field in a moving coil galvanometer?

A moving coil galvanometer is an instrument which is used to measure electric currents. It works on the principle that when a current-carrying coil is placed in external magnetic field experiences magnetic torque. The angle through which the coil is deflected due to the effect of the magnetic torque is proportional to the magnitude of current in the coil. Working: Let P Q R S be a single turn of the coil. A current I flows through the coil. In a radial magnetic field, the plan of the coil is always parallel to the magnetic field. Hence the sides Q R and S P are always parallel to the field. So, they do not experience any force. The sides P Q and R S are always perpendicular to the field. Let, P Q = R S = l , length of the coil and P S = Q R = b , breadth of the coil. Force on the P Q side F = B I ( P Q ) = B I L . The force is normal to the plane coil and acts outwards. Force on R S , F = B I R S = B I l . This force is normal to the plane of the coil and acts inwards. When the equal and opposite forces called a couple, act on the coil, it produces the torque. This torque causes the coil to deflect. If there are n turns in the coil, the moment of the deflecting couple = n B I l b = n B I A The suspension wire twists when the coil deflects. On the account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist. If θ is the angular twist, then the moment of the restoring couple = C θ Where, C is the restoring couple per unit twist. At equilibrium, deflecting couple = restoring couple n I B A = C θ I = C n B A θ C , n , A , B all are constants. Therefore, I ∝ θ Hence, the deflection of the coil is directly proportional to the current flowing through it. Radial magnetic provides a constant torque on the coil. It is applied to measure the relation between the current and the angle which is non-linear such that the current can not be measured. With the help of a radial magnetic field, the angle between the plane of the coil and the magnetic field is maintained at zero in all directions of the coil.

With the help of a labeled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles. Is there an upper limit on the energy acquired by the particle? Given reason.

Principle: Cyclotron works in this principle that when a charged particle is kept in a magnetic field it experiences a force and the perpendicular magnetic field causes the particle to rotate. Working: A dee which is at a negative potential accelerates the positive ions which are produced from the source S . The magnetic field which is perpendicular will move ions in a circular motion inside the Dees. The magnetic field and the frequency of the applied voltage are so chosen that as the ion comes out of a Dee, the Dees change its polarity(positive becoming negative and vice-versa) and the ion is further accelerated. Now, the ions move with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches the periphery of the Dees where an auxiliary negative electrode )deflecting plate) deflects the accelerated ion on the target to be bombarded. Thus the charged particles accelerate. Yes, there is an upper limit on the energy acquired by the particle because the charged particle gains maximum speed when it moves in a path of radius equal to the radius of Dees. v m a x = q B R m where, R is the radius of Dee. So, the maximum kinetic energy K m a x = 1 2 m v 2 = 1 2 q B R m 2 K m a x = q 2 B 2 R 2 2 m

The wire loop P Q R S P formed by joining two semi-circular wires of radii R 1 and R 2 carries a current I as shown in the figure below. Find the magnitude of magnetic field at O .

Magnetic field at a centre of a coil is B = μ ∘ N I 2 r . But according to the given coil type it is a concentric semicircle coil with radii R 1 and R 2 . Since the coil is not a full circle thus the number of turns will be N = 0 . 5 Thus, the magnetic field at O due to the external semicircular coil (with radius R 2 ) part S P is B 2 = μ ∘ I 4 R 2 , field going inside the plane of paper, whereas in case of internal semicircular part of coil (with radius R 1 ) Q R is B 1 = μ ∘ I 4 R 1 , field going out of the plane of paper. The net magnetic field at O is: B n e t = B 1 - B 2 = μ ∘ I 4 R 1 - μ ∘ I 4 R 2 = μ ∘ I 4 1 R 1 - 1 R 2 Hence, the magnetic field at O is μ ∘ I 4 1 R 1 - 1 R 2 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2020>Q 14

HARD

12th Meghalaya Board

IMPORTANT

Earn 100

Derive the torque on a rectangular current-carrying loop placed in a uniform magnetic field. Draw suitable figures. Write one application of this principle.

Important Questions on Moving Charges and Magnetism

EASY

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2020>Q 2

The magnetic susceptibility of a substance is

- 1.66 \times 10^{- 5}

. Name the type of magnetic substance it represents.

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2018>Q 3

If the number of turns, area and current through a coil are

n

,

A

and

I

respectively, then its magnetic moment is

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2018>Q 4

The magnetic field at the centre of a circular coil of radius

5 cm

carrying a current of

1 A

is

1.256 T

. If the radius is made

10 cm

, then the magnetic field at the centre of the loop carrying same current is

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2018>Q 5

Calculate the force per unit length between two parallel long straight wires

2 cm

apart in air, each carrying a current of

4 A

.

HARD

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2018>Q 6

Using Biot-Savart law, find an expression for the magnetic field at a point on the axis of a circular current-carrying loop. Hence, find the expression for the magnetic field at its centre.

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2019>Q 7

A charged particle of mass

m

and charge

q

moves along a circular path with a velocity

v

perpendicular to a magnetic field

\vec{B}

. The radius of the circular path is

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2019>Q 8

To convert a galvanometer into a voltmeter, we must connect a

EASY

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Meghalaya-2019>Q 9

Two parallel wires carrying current in the same direction attract each other while two beams of electrons travelling in the same direction repel each other. Explain why.

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