﻿ Medical Entrance Exam Question and Answers | Moving Charges And Magnetism - Zigya

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# Moving Charges And Magnetism

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1.

Assertion:  We cannot think of a magnetic field configuration with three poles.

Reason:  A bar magnet does exert a torque on itself due to its own field.

• If both the assertion and reason are true and reason is a correct explanation of the assertion.

• If both assertion and reason are true but assertion is not a correct explanation of the assertion.

• If the assertion is true but the reason is false.

• If both assertion and reason are false

D.

If both assertion and reason are false

Magnetic fields are nothing more than result of combining an electric field with the effects of relativity.

Magnetic poles always exist in pairs. However, one can imagine magnetic field configuration with three poles. When north poles of two magnets or south poles of two magnets are glued together, they provide a three pole field configuration. Also a bar magnet does not exert a torque on itself due to its own field.

2.

A particle of unit mass and specific charge s is thrown from the wall perpendicularly to a wall at a distance d from the wall with speed 'v'. The minimum magnetic field produced so that the particle does not touch the wall, is:

• $\frac{\mathrm{v}}{\mathrm{sd}}$

• $\frac{2\mathrm{v}}{\mathrm{sd}}$

• $\frac{\mathrm{v}}{2\mathrm{sd}}$

• $\frac{\mathrm{v}}{4\mathrm{sd}}$

A.

$\frac{\mathrm{v}}{\mathrm{sd}}$

For the particle not to be collided with the wall we have

centripetal force = force in magnetic field

B = $\frac{\mathrm{v}}{\mathrm{sd}}$

3.

The coefficient of mutual inductance, when magnetic flux changes by 2 x 10-2 Wb and current changes by 0.01 A is

• 2 H

• 4 H

• 3 H

• 8 H

A.

2 H

Change in magnetic flux  ( $\mathrm{\varphi }$ ) = 2 × 10-2 Wb

and change in current (I) = 0.01 A

Change in magnetic flux,

$\mathrm{\varphi }$ = 2 × 10-2  = MI

= M × 0.01

=

$\mathrm{\varphi }$  = 2H

( where M  = coefficient of mutual inductance )

4.

The earth's magnetic field at a  given point is 0.5 × 10-5 Wb/m2. This field is to be annulled by magnetic induction at the centre of a circular conducting loop of radius 5.0 cm. The current required to be flown in the loop nearly

• 0.2 A

• 0.4 A

• 4 A

• 40 A

B.

0.4 A

The magnetic field at the centre of a circular loop carrying current  I is given by

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

where r is the radius of the circular lop

Given that

r = 0.5 cm = 0.5 × 10-5

I =

=

= 0.3978

I = 0.4 A°

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5.

A wire length l carries a steady current. It is bent first to form a circular plane loop of one turn. The magnetic field at the centre of the loop is B. The same length is now bent more sharply give a to double loop of smaller radius. The  magnetic field at the centre caused by the same is

• B

• B/4

• 4B

• B/2

C.

4B

The wire of length 'l' is bent to form a circular loop

So     2$\mathrm{\pi }$r = l

⇒  r = $\frac{\mathrm{l}}{2\mathrm{\pi }}$

The magnetic field at the centre of the loop is

B =

=

Now the same length of the wire is bent to from a double loop

∴   2 × 2$\mathrm{\pi }$r' = l

⇒        r' = $\frac{\mathrm{l}}{4\mathrm{\pi }}$

And the magnetic field at the centre

B' =

=

=

∴    $\frac{\mathrm{B}\text{'}}{\mathrm{B}}$ = 4

⇒       B' = 4B

6.

The difference in lengths of a mean solar day and a sideral day is about

• 1 min

• 4 min

• 15 min

• 56 min

B.

4 min

The sidereal day is about 4 min shorter than our solar 24-hour day,  to be precise, the difference is 3 min 56 seconds.

7.

An electric dipole placed in a non-uniform electric field experiences

• both a torque and a net force

• only a force but no torque

• only a torque but no net force

• no torque and no net force

A.

both a torque and a net force

If an electric dipole is placed in a non-uniform electric field, then the positive and negative charges of the dipole will experience a net force. And as one end of the dipole is experiencing a force in one direction and the other end in the opposite direction, so the dipole will have net torque also.

8.

Three charges are placed at the vertices of an equilateral triangle of side a as shown in the given figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is

• Q2 / 4πε0a2

• - Q2 (4πε0a2<)

• zero

• Q2 / 2πε0a2

C.

zero

The force experienced by A due to B is

F1  = 1/(4πε0) x (QQ/a2) along AB  ( attractive)

The force experienced by A due to C is

F2 = 1/(4πε0) x (Q2/a2) along CA   produced   (repulsive )

Now the force F1   is having a component F1 cos30º along and the force   F2   is having a component F2 cos30º along  AE  .   AE   and AF   will both cancel each other. And so the force experienced by the charge at A in the direction normal to  BC   is  zero.

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9.

A metal rod consumes power P on passing current. If it is cut into two half and joined in parallel, it will consume power

• P

• 2P

• P/4

C.

P/4

Let the resistance of the rod be R. If the rod is divided into two equal parts, then each part of the rod will have a resistance of R/2. Now these two parts of the rod are connected in parallel and a voltage Vis applied across them. Hence the total
power consumed is

P' =

= 4 $\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$

P' = 4P

Where P = $\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$ is  the power consumed by the original rod when voltage V is applied across it.

10.

At a place earth's magnetic field, 5 × 105 Wb/m2 is acting perpendicular to a coil of radius R = 5 cm. If  μο / 4$\mathrm{\pi }$ = 107 , then how much current is induced in circular loop?

• 0.2 A

• 0 A

• 4 A

• 40 A

B.

0 A

As the earth's magnetic field is acting perpendicular to the coil and as there is no charge of magnetic flux through the coil, the current induced in the coil is zero.

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