Magnetic Effect of Current — Page 2 | JEE Physics

Q11

The current sensitivity of a galvanometer can be increased by : (A) decreasing the number of turns (B) increasing the magnetic field (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below :

A (B) and (C) only

B (C) and (D) only

C (A) and (C) only

D (B) and (D) only

Correct Answer

Option D

Solution

NiAB = k\theta

\Rightarrow {\theta \over i} = {{NAB} \over k}

$\Rightarrow$ Sensitivity increases if

B \uparrow

and

k \downarrow

Q12

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, denuteron and alpha particle are respectively

{r_p},{r_d}

and

{r_\alpha }

. Which one of the following relation is correct?

A

{r_\alpha } = {r_p} = {r_d}

B

{r_\alpha } = {r_p} < {r_d}

C

{r_\alpha } > {r_d} > {r_p}

D

{r_\alpha } = {r_d} > {r_p}

Correct Answer

Option B

Solution

r = {{\sqrt {2mv} } \over {qB}} \Rightarrow r \times v{{\sqrt m } \over q}

Thus we have,

{r_\alpha } = {r_p} < {r_d}

Q13

Two thin, long, parallel wires, separated by a distance

'd'

carry a current of

'i'

A

in the same direction. They will

A repel each other with a force of

{\mu _0}{i^2}/\left( {2\pi d} \right)

B attract each other with a force of

{\mu _0}{i^2}/\left( {2\pi d} \right)

C repel each other with a force

_0{i^2}/\left( {2\pi {d^2}} \right)

D attract each other with a force of

{\mu _0}{i^2}/\left( {2\pi {d^2}} \right)

Correct Answer

Option B

Solution

{F \over \ell } = {{{\mu _0}{i_1}} \over {2\pi d}} = {{{\mu _0}{i^2}} \over {2\pi d}}

(attractive as current is in the same direction)

Q14

If in a circular coil

A

of radius

R,

current

I

is flowing and in another coil

B

of radius

2R

a current

2I

is flowing, then the ratio of the magnetic fields

{B_A}

and

{B_B}

, produced by them will be

A

1

B

2

C

1/2

D

4

Correct Answer

Option A

Solution

KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius

r

at its center is

B = {{{\mu _0}} \over {4\pi }}{I \over r} \times 2\pi

Here

{B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi

and

{B_B} = {{{\mu _0}} \over {4\pi }}{{2I} \over {2R}} \times 2\pi

\Rightarrow {{{B_A}} \over {B{}_B}} = 1

Q15

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

A speed

B mass

C charge

D magnetic induction

Correct Answer

Option A

Solution

KEY CONCEPT : The time period of a charged particle

\left( {m,q} \right)

moving in a magnetic field

(B)

is

T = {{2\pi m} \over {qB}}

The time period does not depend on the speed of the particle.

Q16

A particle of mass

M

and charge

Q

moving with velocity

\overrightarrow v

describe a circular path of radius

R

when subjected to a uniform transverse magnetic field of induction

B.

The network done by the field when the particle completes one full circle is

A

\left( {{{M{v^2}} \over R}} \right)2\pi R

B zero

C

B\,\,Q\,2\pi R

D

B\,Qv\,2\pi R

Correct Answer

Option B

Solution

The work done,

dW = Fds\,\cos \,\theta

The angle between force and displacement is

{90^ \circ }.

Therefore work done is zero.

Q17

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then

A its velocity will increase

B Its velocity will decrease

C it will turn towards left of a direction of motion

D it will turn towards right of direction of motion

Correct Answer

Option B

Solution

Due to electric field, it experiences force and decelerates i.e. its velocity decreases.

Q18

Two long conductors, separated by a distance

d

carry current

{I_1}

and

{I_2}

in the same direction. They exert a force

F

on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to

3d

. The new value of the force between them is

A

- {{2F} \over 3}

B

{F \over 3}

C

-2F

D

- {F \over 3}

Correct Answer

Option A

Solution

Force between two long conductor carrying current,

F = {{{\mu _0}} \over {4\pi }}{{2{I_1}{I_2}} \over d} \times \ell

F' = - {{{\mu _0}} \over {4\pi }}{{2\left( {2{I_1}} \right){I_2}} \over {3d}}\ell

$\therefore$

{{F'} \over F} = {{ - 2} \over 3}

Q19

A particle of charge

- 16 \times {10^{ - 18}}

coulomb moving with velocity

10m{s^{ - 1}}

along the

x

-axis enters a region where a magnetic field of induction

B

is along the

y

-axis, and an electric field of magnitude

{10^4}V/m

is along the negative

z

-axis. If the charged particle continues moving along the

x

-axis, the magnitude of

B

is

A

{10^3}Wb/{m^2}

B

{10^5}Wb/{m^2}

C

{10^{16}}Wb/{m^2}

D

{10^{ - 3}}Wb/{m^2}

Correct Answer

Option A

Solution

The situation is shown in the figure.

{F_E} =

Force due to electric field

{F_B} =

Force due to magnetic field It is given that the charged particle remains moving along

X

-axis (i.e. undeviated). Therefore

{F_B} = {F_E}

\Rightarrow qvB = qE

\Rightarrow B = {E \over v} = {{{{10}^4}} \over {10}}

= {10^3}\,\,weber/{m^2}

Q20

A current

i

ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is

A

{{{\mu _0}} \over {4\pi }},{{2i} \over r}

tesla

B zero

C infinite

D

{{2i} \over r}

tesla

Correct Answer

Option B

Solution

Using Ampere's law at a distance

r

from axis,

B

is same from symmetry.

\int {B.dl = {\mu _0}i}

i.e.,

B \times 2\pi r = {\mu _0}i

Here

i

is zero, for

r < R,

whereas

R

is the radius $\therefore$

B=0