Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 25

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Question 1
1 / -0

Best method to increase the sensitivity of the moving coil galvanometer is to decrease

A
radius of the coil

B
number of turns of the coil

C
external magnetic field

D
couple per unit twist

Solution

Sensitivity of a galvanometer increases when radius of the coil, number of turns of the coil and external magnetic field increases whereas couple per unit twist decreases.
Question 2
1 / -0

A moving coil type of galvanometer is based upon the principle that

A
coil carrying current experiences a torque in magnetic field.

B
a coil carrying current produces a magnetic field.

C
a coil carrying current experiences impulse in a magnetic field.

D
a coil carrying current experiences a force in a magnetic field.

Solution

In a moving coil type of galvanometer needle deflects when torque in magnetic field is non-zero. So, a moving coil galvanometer is based upon the principle that coil carrying current experiences a torque in a magnetic field.
Question 3
1 / -0

Identify the quantity which changes when a charged particle moves through a magnetic field ?

A
Energy

B
Mass

C
Speed

D
Direction of motion

Solution

When a charged particle moves through a magnetic field it suffers a change in its direction of motion.
Question 4
1 / -0

There is $$10$$ units of charge at the centre of a circle of radius $$10m$$. The work done in moving one unit of charge around the circle once is

A
$$0$$

B
$$10$$ units

C
$$100$$ units

D
$$1$$ unit

Solution

Work done is zero because the direction of movement of charge and direction of force are perpendicular to each other.
work= $$F.r. \cos(\theta)$$
so if the angle is $$90^o$$ then work done$$=0$$
Question 5
1 / -0

Calculate the work done on electron beam which is moving in a magnetic field ?

A
2 times of product of electronic charge and the magnetic field

B
Product of Square of charge and magnetic field

C
Product of electronic charge and the magnetic field

D
Zero

Solution

Electron beam is moving in a magnetic field. Work done is given by:
$$W=\vec{F}.\vec{s}$$
or, $$W=\vec{F}.\vec{v}t=0$$
Since, magnetic force is always perpendicular to the velocity of the particles. Particle actually moves in a circle.
Question 6
1 / -0

A charged particle is moving through uniform magnetic field, then in which case will magnetic field exert force on charged particle ?

A
Always exerts a force on the particle

B
Never exerts a force on the particle.

C
Exerts a force, if the particle is moving along the field.

D
Exerts a force, if the particle is moving perpendicular to the direction of the field.

Solution

Answer is D.

A charged particle moving through uniform magnetic field, experiences a force due to magnetic field and moves perpendicular to the direction of magnetic field.
The expression is given as,
$$F=qV\times B$$
where,
$$F -$$ Magnetic force
$$q -$$ charge
$$V -$$ Velocity of the moving charge
$$B -$$ Magnetic field
The force exerted is always perpendicular to the plane formed by $$V$$ and $$B$$. Oppositely directed forces exerted on oppositely charged particles will cause the particles to move in opposite directions. Therefore, option D is the answer.
Question 7
1 / -0

The magnetic force acting on a charged particle of charge $$-2\mu c$$ in a magnetic field of 2T acting in y direction, when the particle velocity is $$(2\hat{i}+3\hat{j})\times10^6\;ms^{-1}$$, is :

A
8N in z direction

B
8N in -z direction

C
4N in z direction

D
8N in y direction

Solution

$$\vec{F}=q(\vec{v}\times\vec{B})$$
$$=-2\times10^{-6}[(2\hat{i}+3\hat{j})\times10^6\times2\hat{j}$$
$$=-(8N)\hat{k}$$
Question 8
1 / -0

A proton is fired with a speed of $$2\times10^6$$ m/s at an angle of $$60^o$$ to the X- axis. If a uniform magnetic field of $$0.1T$$ is applied along the Y- axis, the force acting on the proton is

A
$$1.63 \times 10^{-14} N$$

B
$$1.6 \times 10^{-14} N$$

C
$$3.23 \times 10^{-14} N$$

D
$$3.2 \times 10^{-14} N$$

Solution

$$F=q (\vec {V}\times \vec {B})$$
$$\theta=30^o$$ angle between velocity & field
$$=1\cdot 6\times 10^{-19}\times 2\times 10^6\times 0\cdot 1\times \sin 30^o$$
$$=1\cdot 6\times 10^{-14} N$$
Question 9
1 / -0

Force between the two parallel wires carrying currents has been used to define

A
ampere

B
coulomb

C
volt

D
watt

Solution

Ampere is defined as the constant current that will produce an attractive force of $$2 \times 10^{-7}$$ newton per metre of length between two straight, parallel conductors( carrying the constant current) of infinite length and negligible circular cross section placed one metre apart in a vacuum. This is the Ampere's Force Law.
Question 10
1 / -0

In a moving coil galvanometer, the magnetic pole pieces are made cylindrical and a soft iron core is placed at the centre of the coil,the purpose for doing so is:

A
to make the magnetic field strong

B
to make the magnetic field strong and radial

C
to make the magnetic field uniform

D
to make the magnetic field strong and uniform

Solution

In a moving coil galvanometer, the magnetic pole pieces are made cylindrical and a soft iron core is placed at the centre of the coil to make the magnetic field strong and radial.

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Moving Charges and Magnetism Test - 25

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Moving Charges and Magnetism Test - 25

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Question 1

radius of the coil

number of turns of the coil

external magnetic field

couple per unit twist

Solution

Question 2

coil carrying current experiences a torque in magnetic field.

a coil carrying current produces a magnetic field.

a coil carrying current experiences impulse in a magnetic field.

a coil carrying current experiences a force in a magnetic field.

Solution

Question 3

Energy

Mass

Speed

Direction of motion

Solution

Question 4

$$0$$

$$10$$ units

$$100$$ units

$$1$$ unit

Solution

Question 5

2 times of product of electronic charge and the magnetic field

Product of Square of charge and magnetic field

Product of electronic charge and the magnetic field

Zero

Solution

Question 6

Always exerts a force on the particle

Never exerts a force on the particle.

Exerts a force, if the particle is moving along the field.

Exerts a force, if the particle is moving perpendicular to the direction of the field.

Solution

Question 7

8N in z direction

8N in -z direction

4N in z direction

8N in y direction

Solution

Question 8

$$1.63 \times 10^{-14} N$$

$$1.6 \times 10^{-14} N$$

$$3.23 \times 10^{-14} N$$

$$3.2 \times 10^{-14} N$$

Solution

Question 9

ampere

coulomb

volt

watt

Solution

Question 10

to make the magnetic field strong

to make the magnetic field strong and radial

to make the magnetic field uniform

to make the magnetic field strong and uniform

Solution

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