A small circular loop of conducting wire has radius a and carries current I.

Question:

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. (1) $\mathrm{T}=\sqrt{\frac{2 \mathrm{~m}}{\mathrm{IB}}}$

  2. (2) $\mathrm{T}=\sqrt{\frac{\pi \mathrm{m}}{2 \mathrm{IB}}}$

  3. (3) $\mathrm{T}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}$

  4. (4) $\mathrm{T}=\sqrt{\frac{\pi \mathrm{m}}{\mathrm{IB}}}$


Correct Option: , 3

Solution:

(3) Torque on circular loop, $\tau=M B \sin \theta$

where, $M=$ magnetic moment

$B=$ magnetic field

Now, using $\tau=I \alpha$

$\therefore \quad \tau=M B \sin \theta=I \alpha$

$\Rightarrow \pi R^{2} I B \theta=\frac{m R^{2} \alpha}{2}$

$(\because m=I A$ and moment of inertia of circular loop,

$I=\frac{m R^{2}}{2}$ )

$\Rightarrow \pi R^{2} I B \theta=\frac{m R^{2}}{2} \omega \theta$

$\Rightarrow \omega=\sqrt{\frac{2 \pi I B}{m}} \Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{2 \pi I B}{m}}$

$\Rightarrow T=\sqrt{\frac{2 \pi m}{I B}}$

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