Magnetic Effect of Current — Page 7 | JEE Physics

Q61

A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :

A

T = \sqrt {{{2m} \over {IB}}}

B

T = \sqrt {{{\pi m} \over {IB}}}

C

T = \sqrt {{{\pi m} \over {2IB}}}

D

T = \sqrt {{{2\pi m} \over {IB}}}

Correct Answer

Option D

Solution

$\tau$ = - MBsin $\theta$ I $\alpha$ = - MBsin $\theta$ for small $\theta$ , $\alpha$ =

- {{MB} \over I}\theta

$\therefore$

{\omega ^2}

=

{{MB} \over I}

$\Rightarrow$ $\omega$ =

\sqrt {{{I\left( {\pi {R^2}} \right)B} \over {{{m{R^2}} \over 2}}}}

=

\sqrt {{{2I\pi B} \over m}}

$\therefore$ T =

{{2\pi } \over \omega }

=

\sqrt {{{2\pi m} \over {IB}}}

Q62

A proton, a deuteron and an

\alpha

particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is _________ and their speed is _______, in the ratio.

A 2 : 1 : 1 and 4 : 2 : 1

B 4 : 2 : 1 and 2 : 1 : 1

C 1 : 2 : 4 and 2 : 1 : 1

D 1 : 2 : 4 and 1 : 1 : 2

Correct Answer

Option A

Solution

F = qvB = q

{p \over m}

B F $\propto$

{q \over m}

[as p, B are const.] $\therefore$ F1 : F2 : F3 =

{e \over m}:{e \over {2m}}:{{2e} \over {4m}}

= 2 : 1 : 1 And v1 : v2 : v3 =

{p \over m}:{p \over {2m}}:{p \over {4m}}

= 4 : 2 : 1

Q63

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ________.

A 1.0 m

B 0.15 m

C 0.2 m

D 0.1 m

Correct Answer

Option D

Solution

B = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}

at x1 = 0.05 m,

{B_1} = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {{(0.05)}^2})}^{3/2}}}}

at x2 = 0.2 m,

{B_2} = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {{(0.2)}^2})}^{3/2}}}}

{{{B_1}} \over {{B_2}}} = {{{{({R^2} + 0.04)}^{3/2}}} \over {{{({R^2} + 0.0025)}^{3/2}}}}

{\left( {{8 \over 1}} \right)^{2/3}} = {{{R^2} + 0.04} \over {{R^2} + 0.0025}}

4 (R2 + 0.0025) = R2 + 0.04 3R2 = 0.04 $-$ 0.0100 R2 =

{{0.03} \over 3}

= 0.01 R =

\sqrt {0.01}

= 0.1 m

Q64

A proton and an

\alpha

-particle, having kinetic energies Kp and K

\alpha

respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of

\alpha

-particle is 2 : 1. The ratio of Kp : K

\alpha

is :

A 8 : 1

B 4 : 1

C 1 : 8

D 1 : 4

Correct Answer

Option B

Solution

Radius,

r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}

\Rightarrow K = {{{r^2}{q^2}{B^2}} \over {2m}}

$\therefore$

{{{K_p}} \over {{K_\alpha }}} = {\left( {{{{r_p}} \over {{r_\alpha }}}} \right)^2} \times {\left( {{{{q_p}} \over {{q_\alpha }}}} \right)^2} \times {{{m_\alpha }} \over {{m_p}}}

= {\left( {{2 \over 1}} \right)^2} \times {\left( {{1 \over 2}} \right)^2} \times 4

= 4

Q65

A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let rd and r

\alpha

be their respective radii of circular path. The value of

{{{r_d}} \over {{r_\alpha }}}

is equal to :

A 1

B 2

C

\sqrt 2

D

{1 \over {\sqrt 2 }}

Correct Answer

Option C

Solution

Given, kinetic energy of $\alpha$ -particle (K $\alpha$ ) = kinetic energy of deuteron (Kd) Since, kinetic energy, K =

{1 \over 2}

mv2 $\Rightarrow$ mv2 = 2K $\Rightarrow$ v2 =

{{2K} \over m}

$\Rightarrow$ v =

\sqrt {{{2K} \over m}}

.... (i) We know that, r =

{{mv} \over {Bq}}

.... (ii) where, r = radius of curvature of path of a charged particle, m = mass of the charged particle, q = charge of the particle, v = velocity of charged particle and B = magnetic field.

From Eqs. (i) and (ii), we get

r = {{m\sqrt {{{2K} \over m}} } \over {Bq}}

\Rightarrow r = {{\sqrt {2Km} } \over {Bq}}

.... (iii) Since, m, K and B are same for both deuteron and $\alpha$ -particle. From Eq. (iii), we get

\gamma \propto {{\sqrt m } \over q}

$\therefore$

{{{r_d}} \over {{r_\alpha }}} = \sqrt {{{{m_d}} \over {{m_\alpha }}}} .{{{q_\alpha }} \over {{q_d}}} = \sqrt {{2 \over 4}} \left( {{2 \over 1}} \right)

[ $\because$ md = 2mp and m $\alpha$ = 4mp q $\alpha$ = 2e and qd = e.]

{{{r_d}} \over {{r_\alpha }}} = \sqrt 2

Q66

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take r < a)

A

{3 \over 2}{{{a^2}} \over {{r^2}}}

B

{2 \over 3}{{{a^2}} \over {{r^2}}}

C

{2 \over 3}{{{r^2}} \over {{a^2}}}

D

{3 \over 2}{{{r^2}} \over {{a^2}}}

Correct Answer

Option D

Solution

{B_{axis}} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}

{B_{centre}} = {{{\mu _0}i} \over {2R}}

$\therefore$

{B_{centre}} = {{{\mu _0}i} \over {2a}}

$\therefore$

{B_{axis}} = {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}

$\therefore$ fractional change in magnetic field =

{{{{{\mu _0}i} \over {2a}} - {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}} \over {{{{\mu _0}i} \over {2a}}}} = 1 - {1 \over {{{\left[ {1 + \left( {{{{r^2}} \over {{a^2}}}} \right)} \right]}^{3/2}}}}

\approx 1 - \left[ {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right] = {3 \over 2}{{{r^2}} \over {{a^2}}}

Note :

{\left( {1 + {{{r^2}} \over {{a^2}}}} \right)^{ - 3/2}} \approx \left( {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right)

[True only if r << a] Hence, option (d) is the most suitable option.

Q67

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :

A lighter ion will be deflected less than heavier ion

B lighter ion will be deflected more than heavier ion

C both ions will be deflected equally

D no ion will be deflected

Correct Answer

Option B

Solution

r = {P \over {qB}} = {{\sqrt {2mk} } \over {qB}}

Given they have same kinetic energy

r \propto {{\sqrt m } \over q}

{{{r_1}} \over {{r_2}}} = {{\sqrt 4 } \over 2} \times {3 \over {\sqrt {16} }} = {3 \over 4}

{r_2} = {{4{r_1}} \over 3}

(r2 is for heavier ion and and r1 is for lighter ion)

\sin \theta = {d \over R}

$\theta$ $\to$ Deflection

\theta \propto {1 \over R}

(R $\to$ Radius of path) $\because$ R2 > R1 $\Rightarrow$ $\theta$ 2 < $\theta$ 1

Q68

A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?

A

{{{x^2}} \over {{a^2}}}

B

{{{a^2}} \over {{x^2}}}

C

{{{x^2}} \over {{b^2} - {a^2}}}

D

{{{b^2} - {a^2}} \over {{x^2}}}

Correct Answer

Option A

Solution

when x < a

{B_1}(2\pi x) = {\mu _0}\left( {{{{i_0}} \over {\pi {a^2}}}} \right)\pi {x^2}

B(2\pi x) = {{{\mu _0}{i_0}{x^2}} \over {{a^2}}}

{B_1} = {{{\mu _0}{i_0}x} \over {2\pi {a^2}}}

.... (1) when a < x < b

{B_2}(2\pi x) = {\mu _0}{i_0}

{B_2} = {{{\mu _0}{i_0}} \over {2\pi x}}

..... (2)

{{{B_1}} \over {{B_2}}} = {{{\mu _0}{i_0}{x \over {2\pi {a^2}}}} \over {{{{\mu _0}{i_0}} \over {2\pi x}}}} = {{{x^2}} \over {{a^2}}}

Q69

A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction.)

A 3

\times

10

-

7 T, outside the plane of triangle

B

2\sqrt 3

\times

10

-

7 T, outside the plane of triangle

C

2\sqrt 3

\times

10

-

5 T, inside the plane of triangle

D 3

\times

10

-

5 T, inside the plane of triangle

Correct Answer

Option D

Solution

B = 3\left[ {{{{\mu _0}i} \over {4\pi r}}(\sin 60^\circ + \sin 60^\circ )} \right]

\tan 60^\circ = {{l/2} \over r}

where

r = {{9 \times {{10}^{ - 2}}} \over {2\sqrt 3 }}

M $\therefore$ B = 3 $\times$ 10-5 T Current is flowing in clockwise direction so,

\overrightarrow B

is inside plane of triangle by right hand rule.

Q70

Two long current carrying conductors are placed to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300

\mu

T. The equal current flowing in the two conductors is :

A 30A in the same direction.

B 30A in the opposite direction.

C 60A in the opposite direction.

D 300A in the opposite direction.

Correct Answer

Option B

Solution

As Bnet $\ne$ 0 that is the wires are carrying current in opposite direction.

{{{\mu _0}I \times 2} \over {2\pi (4 \times {{10}^{ - 2}})}} = 30 \times {10^{ - 6}}

T

\Rightarrow I = {{30 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}

A = 30 A in opposite direction.