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Magnetics question

2020 · 9 Jan · Shift 2 · Q60
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Magnetics question

2020 · 9 Jan · Shift 2 · Q60

JEE MainPhysicsMagneticsMCQ+4 / −1
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 :
  1. A
    T=2mIBT = \sqrt {{{2m} \over {IB}}}T=IB2m​​
  2. B
    T=πmIBT = \sqrt {{{\pi m} \over {IB}}}T=IBπm​​
  3. C
    T=πm2IBT = \sqrt {{{\pi m} \over {2IB}}}T=2IBπm​​
  4. D
    T=2πmIBT = \sqrt {{{2\pi m} \over {IB}}}T=IB2πm​​
View written solutionFree

Correct answer: D

  1. Magnetic moment of the loop

For a circular loop of radius aaa carrying current III,

μ=IA=Iπa2\mu = IA = I\pi a^2μ=IA=Iπa2

where A=πa2A=\pi a^2A=πa2 is the area of the loop.

  1. Torque on the loop in magnetic field

The loop is placed in a uniform magnetic field BBB perpendicular to its plane. Hence initially the magnetic moment is along BBB.

If the loop is rotated by a small angle θ\thetaθ about its diameter, the magnetic moment also rotates by θ\thetaθ with respect to BBB.

The restoring magnetic torque is

τ=−μBsin⁡θ\tau = -\mu B\sin\thetaτ=−μBsinθ

For small oscillations, sin⁡θ≈θ\sin\theta \approx \thetasinθ≈θ, so

τ≈−μBθ\tau \approx -\mu B\thetaτ≈−μBθ

Thus,

τ=−Iπa2B θ\tau = -I\pi a^2 B\,\thetaτ=−Iπa2Bθ
  1. Moment of inertia about a diameter

The loop is a thin circular ring of mass mmm and radius aaa.

Moment of inertia of a ring about a diameter is

Irot=12ma2I_{\text{rot}} = \frac{1}{2}ma^2Irot​=21​ma2
  1. Equation of SHM

Using rotational dynamics,

Irot θ¨=τI_{\text{rot}}\,\ddot\theta = \tauIrot​θ¨=τ

So,

12ma2 θ¨=−Iπa2B θ\frac{1}{2}ma^2\,\ddot\theta = -I\pi a^2 B\,\theta21​ma2θ¨=−Iπa2Bθ

Cancel a2a^2a2:

12mθ¨=−IπBθ\frac{1}{2}m\ddot\theta = -I\pi B\theta21​mθ¨=−IπBθ θ¨+2IπBmθ=0\ddot\theta + \frac{2I\pi B}{m}\theta = 0θ¨+m2IπB​θ=0

This is SHM with angular frequency

ω2=2πIBm\omega^2 = \frac{2\pi I B}{m}ω2=m2πIB​

Hence,

ω=2πIBm\omega = \sqrt{\frac{2\pi I B}{m}}ω=m2πIB​​
  1. Time period

For SHM,

T=2πωT=\frac{2\pi}{\omega}T=ω2π​

Therefore,

T=2πm2πIBT = 2\pi\sqrt{\frac{m}{2\pi I B}}T=2π2πIBm​​

Simplifying,

T=2πmIBT = \sqrt{\frac{2\pi m}{IB}}T=IB2πm​​
  1. Match with options

Thus the correct option is

T=2πmIB\boxed{T=\sqrt{\frac{2\pi m}{IB}}}T=IB2πm​​​

which is Option D.

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