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 $B=8\pi \times 10^{- 4} \, \text{T}$ . The magnet vibrates with a period of vibration equal to $\text{15 s}$ . The moment of inertia of magnet is

• A. $4.54\times 10^{4} \, kg \, m^{2}$
• B. $ \, 4.54 \times 10^{- 5} \, \text{kg } \text{m}^{2}$
• C. $ \, 4.54 \times 10^{- 4} \, \text{kg } \text{m}^{2}$
• D. $ \, 4.54 \times 10^{5} \, \text{kg} \text{m}^{2}$

Answer 1

Answer: (D)

Here, magnetic moment is given in weber-meter, which is the unit of $\mu _{0} \, M$ .
$∴ \, \mu _{0} \, M=5\times 10^{- 5}$ Wb-m
$M=\frac{5 \times 10^{- 5}}{\mu _{0}}Am^{2}$
Also, $B=8\pi \times 10^{- 4}=\mu _{0}H$
$∴ \, H=\frac{8 \pi \times 10^{- 4}}{\mu _{0}}$
$T=2\pi \sqrt{\frac{I}{M H}}$
$I=\frac{M H T^{2}}{4 \pi ^{2}}=\frac{5 \times 10^{- 5} \times 8 \pi \times 10^{- 4} \times 15^{2}}{4 \pi ^{2} \mu _{0}^{2}}$
$=\frac{5 \times 10^{-5} \times 8 \pi \times 10^{-4} \times 225}{4 \pi^2\left(4 \pi \times 10^{-7}\right)^2}$
$I=\frac{2250 \times 10^{-9}}{\pi\left(16 \pi^2\right) \times 10^{-14}}=4.54 \times 10^5 \mathrm{~kg} \mathrm{~m}^2$