Question

A bar magnet has a magnetic moment equal to $5\times {10}^{-5}$Wb-m. It is suspended in a magnetic field which has a magnetic induction (B) equal to $8\pi \times {10}^{-4}$ T. The magnet vibrates with a period of vibration equal to 15 ms. The moment of inertia of the magnet is:

• A. $7.16\times {10}^{-13}kg{m}^2$
• B. $11.25\times {10}^{-13}kg{m}^2$
• C. $5.62\times {10}^{-13}kg{m}^2$
• D. $0.57\times {10}^{-13}kg{m}^2$

Answer 1

The correct option is A $7.16\times {10}^{-13}kg{m}^2$

Time period of oscillation $T=2\pi \sqrt{\frac{I}{MB}}$
where $I$ is moment of inertia about point of suspension ( mid point of the magnet), $B$ is external magnetic field an $M$ is the magnetic moment.
$\therefore T^2=(2\pi {)}^2\frac{I}{MB}$
Thus moment of inertia of magnet $I=\frac{T^{2}MB}{4{\pi }^2}$

Given : $T=15ms=15\times {10}^{-3}s$, $M=5\times {10}^{-5}/Wb.m$ and $B=8\pi \times {10}^{-4}T$

$\therefore$ $T=\frac{(15\times {10}^{-3}{)}^2\times 5\times {10}^{-5}\times 8\pi \times {10}^{-4}}{4\times {\pi }^2}=7.16\times {10}^{-13}kg{m}^2$