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Magnetic Effect of Current

JEE Physics · 124 questions · Page 3 of 13 · Click an option or "Show Solution" to reveal answer

Q21
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B.B. It is then bent into a circular loop of nn turns. The magnetic field at the center of the coil will be
A 2n2n BB
B n2 B{n^2}\,B
C nBnB
D 2n2 B2{n^2}\,B
Correct Answer
Option B
Solution

KEY CONCEPT : Magnetic field at the center of a circular coil of radius

RR

carrying current is

B=μ0i2RB = {{{\mu _0}i} \over {2R}}

Given:

n×(2πr′)=2πRn \times \left( {2\pi r'} \right) = 2\pi R
⇒nr′=R           ...(1)\Rightarrow nr' = R\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
B′=n.μ0i2r′           ...(2)B' = {{n.{\mu _0}i} \over {2r'}}\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

from

(1)\left( 1 \right)

and

(2),\left( 2 \right),
B′=nμ0i.n2πR=n2BB' = {{n{\mu _0}i.n} \over {2\pi R}} = {n^2}B
Q22
The magnetic field due to a current carrying circular loop of radius 33 cmcm at a point on the axis at a distance of 44 cmcm from the centre is 54 μT.54\,\mu T. What will be its value at the center of loop?
A 125 μT125\,\mu T
B 150 μT150\,\mu T
C 250 μT250\,\mu T
D 75 μT75\,\mu T
Correct Answer
Option C
Solution

The magnetic field at a point on the axis of a circular loop at a distance

xx

from center is,

B=μ0i a22(x2+a2)3/2B = {{{\mu _0}i\,{a^2}} \over {2\left( {{x^2} + {a^2}} \right)3/2}}
     B′=μ0i2a\,\,\,\,\,B' = {{{\mu _0}i} \over {2a}}

∴\therefore

B′=B.(x2+a2)3/2a3B' = {{B.{{\left( {{x^2} + {a^2}} \right)}^{3/2}}} \over {{a^3}}}

Put

x=4x = 4

&

a=3⇒B′=54(53)3×3×3=250μTa = 3 \Rightarrow B' = {{54\left( {{5^3}} \right)} \over {3 \times 3 \times 3}} = 250\mu T
Q23
If a current is passed through a spring then the spring will
A expand
B compress
C remains same
D none of these
Correct Answer
Option B
Solution

When current is passed through a spring, every current carrying loop of a spring behaves like a tiny magnet and loop of a spring faces another loop are form magnet of different poles which attract one another.

Therefore, spring is compressed.

Q24
If an electron and a proton having same momentum enter perpendicular to a magnetic field, then
A curved path of electron and proton will be same (ignoring the sense of revolution)
B they will move undeflected
C curved path of electron is more curved than that of the proton
D path of proton is more curved.
Correct Answer
Option A
Solution

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius.

r=pqBr = {p \over {qB}}

where

q=q=

Charge of the particle

p=p=

Momentum of the particle

B=B=

Magnetic field Here

p,qp,q

and

BB

are constant for electron and proton, therefore the radius will be same.

Q25
Two concentric coils each of radius equal to 22 π\pi cmcm are placed at right angles to each other. 33 ampere and 44 ampere are the currents flowing in each coil respectively . The magnetic induction in Weber / m2{m^2} at the center of the coils will be (μ=4π×10−7Wb/A.m)\left( {\mu = 4\pi \times {{10}^{ - 7}}Wb/A.m} \right)
A 10−5{10^{ - 5}}
B 12×10−512 \times {10^{ - 5}}
C 7×10−57 \times {10^{ - 5}}
D 5×10−55 \times {10^{ - 5}}
Correct Answer
Option D
Solution

The magnetic field due to circular coil

11

and

22

are

B1=μ0i12r=μ0i12(2π×10−2){B_1} = {{{\mu _0}{i_1}} \over {2r}} = {{{\mu _0}{i_1}} \over {2\left( {2\pi \times {{10}^{ - 2}}} \right)}}
=μ0×3×1024π= {{{\mu _0} \times 3 \times {{10}^2}} \over {4\pi }}
B2=μ0i22(2π×10−2)=μ0×4×1024π{B_2} = {{{\mu _0}{i_2}} \over {2\left( {2\pi \times {{10}^{ - 2}}} \right)}} = {{{\mu _0} \times 4 \times {{10}^2}} \over {4\pi }}
B=B12+B22=μ04π.5×102B = \sqrt {B_1^2 + B_2^2} = {{{\mu _0}} \over {4\pi }}.5 \times {10^2}
⇒B=10−7×5×102\Rightarrow B = {10^{ - 7}} \times 5 \times {10^2}
⇒B=5×10−5 Wb/m2\Rightarrow B = 5 \times {10^{ - 5}}\,Wb/{m^2}
Q26
A charged particle of mass mm and charge qq travels on a circular path of radius rr that is perpendicular to a magnetic field B.B. The time taken by the particle to complete one revolution is
A 2πq2Bm{{2\pi {q^2}B} \over m}
B 2πmqB{{2\pi mq} \over B}
C 2πmqB{{2\pi m} \over {qB}}
D 2πqBm{{2\pi qB} \over m}
Correct Answer
Option C
Solution

Equating magnetic force to centripetal force.

mV2r=qvB sin⁡ 90∘{{m{V^2}} \over r} = qvB\,\sin \,{90^ \circ }

Time to complete one revolution.

T=2πrv=2πmqBT = {{2\pi r} \over v} = {{2\pi m} \over {qB}}
Q27
A charged particle with charge qq enters a region of constant, uniform and mutually orthogonal fields E→\overrightarrow E and B→\overrightarrow B with a velocity v→\overrightarrow v perpendicular to both E→\overrightarrow E and B→,\overrightarrow B, and comes out without any change in magnitude or direction of v→\overrightarrow v . Then
A v→=B→×E→/E2\overrightarrow v = \overrightarrow B \times \overrightarrow E /{E^2}
B v→=E→×B→/B2\overrightarrow v = \overrightarrow E \times \overrightarrow B /{B^2}
C v→=B→×E→/B2\overrightarrow v = \overrightarrow B \times \overrightarrow E /{B^2}
D v→=E→×B→/E2\overrightarrow v = \overrightarrow E \times \overrightarrow B /{E^2}
Correct Answer
Option B
Solution

Here,

E→\overrightarrow E

and

B→\overrightarrow B

are perpendicular to each other and the velocity

v→\overrightarrow v

does not change; therefore

qE=qvB⇒v=EBqE = qvB \Rightarrow v = {E \over B}

Also,

∣E→×B→B2∣=E  Bsin⁡θB2\left| {{{\overrightarrow E \times \overrightarrow B } \over {{B^2}}}} \right| = {{E\,\,B\sin \theta } \over {{B^2}}}
=E  Bsin⁡90∘B2=EB=∣v→∣=v= {{E\,\,B\sin {{90}^ \circ }} \over {{B^2}}} = {E \over B} = \left| {\overrightarrow v } \right| = v
Q28
A charged particle moves through a magnetic field perpendicular to its direction. Then
A Kinetic energy changes but the momentum is constant
B the momentum changes but the kinetic energy is constant
C both momentum and kinetic energy of the particle are not constant
D both momentum and kinetic energy of the particle are constant
Correct Answer
Option B
Solution

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular.

In circular motion the direction of velocity changes at every point (the magnitude remains constant).

Therefore, the tangential momentum will change at every point.

But kinetic energy will remain constant as it is given by

12 mv2{1 \over 2}\,m{v^2}

and

v2{v^2}

is the square of the magnitude of velocity which does not change.

Q29
A current II flows along the length of an infinitely long, straight, thin walled pipe. Then
A the magnetic field at all points inside the pipe is the same, but not zero
B the magnetic field is zero only on the axis of the pipe
C the magnetic field is different at different points inside the pipe
D the magnetic field at any point inside the pipe is zero
Correct Answer
Option D
Solution

There is no current inside the pipe. Therefore

∮B‾.dℓ‾=μ0I\oint {\overline B .\overline {d\ell } } = {\mu _0}I
I=0I=0

∴\therefore

B=0B=0
Q30
A charge Q is moving dl→\overrightarrow {dl} distance in the magnetic field B→\overrightarrow {B} . Find the value of work done by B→\overrightarrow {B} .
A Zero
B −-1
C Infinite
D 1
Correct Answer
Option A
Solution

We know,

F→=q(v→×B→)\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)

∴\therefore

F→⊥v→\overrightarrow F \bot \overrightarrow v

and

F→⊥B→\overrightarrow F \bot \overrightarrow B

Power (p)

=dwdt= {{dw} \over {dt}}
=F→. dl→dt= {{\overrightarrow F .\,\overrightarrow {dl} } \over {dt}}
=F→ . dl→dt= \overrightarrow F \,.\,{{\overrightarrow {dl} } \over {dt}}
=F→ . v→= \overrightarrow F \,.\,\overrightarrow v
=Fvcos⁡θ= Fv\cos \theta
=Fvcos⁡90∘= Fv\cos 90^\circ
=0= 0

As power supply by the field is zero so, total work done also zero.

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