Question

A bar magnet of moment of inertia $I$ is vibrated in a magnetic field of induction $0.4\times {10}^{-4}T$. The time period of vibration is $12s$. The magnetic moment of the magnet is $120{Am}^2$. The moment of inertia of the magnet is (in $kg{m}^2$) approximately:

• A. $2.1{\pi }^2\times {10}^{-2}$
• B. $1.57\times {10}^{-2}$
• C. $1728\times {10}^{-2}$
• D. $172.8\times {10}^{-4}$

Answer 1

Answer: (D)

$172.8\times {10}^{-4}$

An SHM takes place due to torque acting on magnet due to presence of magnetic field.

$\tau =\mu Bsin\theta \approx \mu B\theta =I{\omega }^2\theta$

$⟹I=\frac{\mu B}{{\omega }^2}=\frac{\mu B{T}^2}{4{\pi }^2}$

$I=\frac{120\times 0.4\times {10}^{-4}\times {12}^2}{4\times {\pi }^2}$

$=172.8\times {10}^{-4}{kgm}^2$