Magnetic Effect of Current — Page 6 | JEE Physics

Q51

A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field

\overrightarrow B

. The magnetic moment of this moving particle:

A

{{m{v^2}\overrightarrow B } \over {2{B^2}}}

B -

{{m{v^2}\overrightarrow B } \over {2{B^2}}}

C -

{{m{v^2}\overrightarrow B } \over {{B^2}}}

D -

{{m{v^2}\overrightarrow B } \over {2\pi {B^2}}}

Correct Answer

Option B

Solution

Magnetic dipole moment M = iA $\Rightarrow$ M =

\left( {{q \over T}} \right) \times \pi {R^2}

=

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

=

{{qvR} \over 2}

$\Rightarrow$ M =

{{qv} \over 2} \times {{vm} \over {qB}}

As you can see from the figure, direction of magnetic moment(M) is opposite to magnetic field. $\therefore$

\overrightarrow M = - {{m{v^2}} \over {2B}}\widehat B

=

- {{m{v^2}} \over {2B}}\left( {{{\overrightarrow B } \over B}} \right)

=

- {{m{v^2}} \over {2{B^2}}}\overrightarrow B

Q52

A particle of charge q and mass m is moving with a velocity

- v\widehat i

(v

\ne

0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field

\overrightarrow B = {B_0}\widehat k

, the maximum value of v for which the particle will not hit the screen is :

A

{{2qd{B_0}} \over m}

B

{{qd{B_0}} \over {3m}}

C

{{qd{B_0}} \over {2m}}

D

{{qd{B_0}} \over {m}}

Correct Answer

Option D

Solution

In uniform magnetic field particle moves in a circular path, if the radius of the circular path is 'd', particle will not hit the screen. r =

{{mv} \over {q{B_0}}}

To not collide, r < d $\Rightarrow$

{{mv} \over {q{B_0}}}

< d $\Rightarrow$ v <

{{q{B_0}d} \over m}

$\therefore$ vmax =

{{q{B_0}d} \over m}

Q53

An electron is moving along +x direction with a velocity of 6

\times

106 ms–1. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be :

A 3

\times

10–4 T, along –z direction

B 5

\times

10–3 T, along –z direction

C 5

\times

10–3 T, along +z direction

D 3

\times

10–4 T, along +z direction

Correct Answer

Option C

Solution

\overrightarrow B

must be in +z axis.

\overrightarrow V = 6 \times {10^6}\widehat i

\overrightarrow E = 300\widehat j

V/cm = 3 $\times$ 104

\widehat j

V/m

\overrightarrow F

=

q\overrightarrow E + q\overrightarrow V \times \overrightarrow B

$\therefore$

q\overrightarrow E + q\overrightarrow V \times \overrightarrow B

= 0 $\Rightarrow$ qE = qVB $\Rightarrow$ B =

{E \over V}

=

{{3 \times {{10}^4}} \over {6 \times {{10}^6}}}

= 5 $\times$ 10–3 T

Q54

An iron rod of volume 10–3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

A 5

\times

102 Am2

B 0.5

\times

102 Am2

C 500

\times

102 Am2

D 50

\times

102 Am2

Correct Answer

Option A

Solution

Given, V = 10–3 m3 = Al I = 0.5A $\mu$ r = 1000 n = 10 turns/cm =

{{10} \over {{{10}^{ - 2}}}}

turn/m = 1000 turn/m Magnetic moment, M = NIA( $\mu$ r - 1) = (nl)IA( $\mu$ r - 1) = nI(Al)( $\mu$ r - 1) = 1000 × 0.5 × 10–3 (1000 – 1) = 0.5 × (999) = 499.5

\simeq

500 = 5 $\times$ 102 Am2

Q55

A square loop of side 2

a

, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about zaxis is given by :

A

{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

B

{{{\mu _0}{I^2}{a^2}} \over {2\pi b}}

C

{{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}

D

{{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}

Correct Answer

Option A

Solution

We know,

\tau = MB\sin \theta

Here M = IA = I(2a)2 = 4a2I B =

{{{\mu _0}I} \over {2\pi b}}

Angle between B and M = 90o $\therefore$

\tau = \left( {4{a^2}I} \right)\left( {{{{\mu _0}I} \over {2\pi b}}} \right)\sin 90^\circ

=

{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

Q56

A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying current to produce a magnetic moment of 20 Am2 . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60o will be:

A 10

\pi

rad s–1

B 20

\pi

rad s–1

C

10{\left( 3 \right)^{1/4}}

rad s–1

D 20 rad s–1

Correct Answer

Option C

Solution

By energy conservation Ui + Ki = Uf + Kf $\Rightarrow$

- MB\,\cos 90^\circ + 0 = - MB\,\cos 30^\circ + {1 \over 2}I{\omega ^2}

$\Rightarrow$

MB{{\sqrt 3 } \over 2}

= {1 \over 2}I{\omega ^2}

$\Rightarrow$

\omega =

\sqrt {{{MB\sqrt 3 } \over I}}

=

\sqrt {{{20 \times 4 \times \sqrt 3 } \over {0.8}}}

=

10\sqrt {\sqrt 3 } = 10{\left( 3 \right)^{1/4}}

Q57

A charged particle carrying charge 1

\mu

C is moving with velocity

\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)

ms–1. If an external magnetic field of

\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)

× 10–3 T exists in the region where the particle is moving then the force on the particle is

\overrightarrow F

× 10–9 N. The vector

\overrightarrow F

is :

A

{ - 0.30\widehat i + 0.32\widehat j - 0.09\widehat k}

B

{ - 300\widehat i + 320\widehat j - 90\widehat k}

C

{ - 30\widehat i + 32\widehat j - 9\widehat k}

D

{ - 3.0\widehat i + 3.2\widehat j - 0.9\widehat k}

Correct Answer

Option C

Solution

Given,

{\overrightarrow V }

=

\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)

ms–1

{\overrightarrow B }

=

\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)

× 10–3 T q = 1 $\mu$ C

\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)

=

{10^{ - 6}} \times {10^{ - 3}} \times \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & 3 & 4 \\ 5 & 3 & { - 6} \end{array} \right|

= (

{ - 30\widehat i + 32\widehat j - 9\widehat k}

) $\times$ 10-9

Q58

A particle of mass m and charge q has an initial velocity

\overrightarrow v = {v_0}\widehat j

. If an electric field

\overrightarrow E = {E_0}\widehat i

and magnetic field

\overrightarrow B = {B_0}\widehat i

act on the particle, its speed will double after a time:

A

{{3m{v_0}} \over {q{E_0}}}

B

{{\sqrt 2 m{v_0}} \over {q{E_0}}}

C

{{\sqrt 3 m{v_0}} \over {q{E_0}}}

D

{{2m{v_0}} \over {q{E_0}}}

Correct Answer

Option C

Solution

Electric field will increase the speed of particle in x direction. Fx = qE $\therefore$ a =

{{qE} \over m}

Also vx = at =

{{qE} \over m}

t

v_x^2 + v_y^2 = {v^2}

$\Rightarrow$

v_x^2 + v_0^2 = {\left( {2{v_0}} \right)^2}

$\Rightarrow$ vx =

\sqrt 3

v0 $\therefore$

{{qE} \over m}

t =

\sqrt 3

v0 $\Rightarrow$ t =

{{\sqrt 3 m{v_0}} \over {q{E_0}}}

Q59

Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10–27 kg) :

A 0.71mT

B 7.1mT

C 0.071mT

D 71mT

Correct Answer

Option A

Solution

K.E =

{1 \over 2}m{v^2}

$\Rightarrow$ 1 MeV =

{1 \over 2}m{v^2}

$\Rightarrow$ 1.6 $\times$ 10-19 $\times$ 106 =

{1 \over 2}

$\times$ 1.6 × 10–27 $\times$ v2 $\Rightarrow$ v =

\sqrt 2 \times {10^7}

m/s Fm = qvB sin $\theta$ $\Rightarrow$ Fm = qvB [as $\theta$ = 90o] $\Rightarrow$ ma = Bev $\Rightarrow$ 1.6 × 10–27 $\times$ 1012 = 1.6 $\times$ 10-19 $\times$

\sqrt 2 \times {10^7}

$\times$ B $\Rightarrow$ B = 0.71 mJ

Q60

A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance

{a \over 3}

and 2

a

, respectively from the axis of the wire is :

A 2

B

{1 \over 2}

C

{3 \over 2}

D

{2 \over 3}

Correct Answer

Option D

Solution

Let current density be J. $\therefore$ Applying Ampere's law. For point P

\oint {\overrightarrow {B.} d\overrightarrow l } = {\mu _0}i

$\Rightarrow$ BP2 $\pi$

{a \over 3}

=

{\mu _0}J\pi {{{a^2}} \over 9}

...(1) $\therefore$ Applying Ampere's law. For point Q BQ2 $\pi$

{(2a)}

=

{\mu _0}J\pi {a^2}

...(2) Dividing (1) by (2) we get

{{{B_P}2\pi {a \over 3}} \over {{B_Q}4\pi a}} = {{{\mu _0}J\pi {{{a^2}} \over 9}} \over {{\mu _0}J\pi {a^2}}}

$\Rightarrow$

{{{B_P}} \over {{B_Q}}} = {2 \over 3}