Question

A bar magnet of moment of inertia $1\times {10}^{-2}kg{m}^2$ vibrates in a magnetic field of induction $0.36\times {10}^{-4}T$. The time period of vibration is $10s$. Then the magnetic moment of the bar magnet is (Am${}^2$)

• A. $120$
• B. $111$
• C. $140$
• D. $160$

Answer 1

Answer: (B)

$111$

The time period of oscillation of a bar magnet freely suspended in air is given by
$T=2\pi \sqrt{\frac{I}{m{B}_H}}$
$m=4{\pi }^2\frac{I}{T^2{B}_H}$
$m\approx 110A{m}^2$