Q. 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 s. The moment of inertia of magnet is
A) $4.54 \times 10^{4}\ kg - m^{2}$
B) 4.54$\ \times 10^{- 5}\ kg - m^{2}$
C) 4.54 $\times 10^{- 4}\ kg - m^{2}$
D) 4.54 $\times 10^{5}\ kg - m^{2}$
Explanation
(d) Here, magnetic moment is given in weber-meter, which is the unit of $\mu_{0}\ M$.
$\therefore\mu_{0}\ M = 5 \times 10^{- 5}$ Wb-m
$\ M = \frac{5 \times 10^{- 5}}{\mu_{0}}Am^{2}$
Also, $B = 8\pi \times 10^{- 4} = \mu_{0}H$
$\therefore H = \frac{8\pi \times 10^{- 4}}{\mu_{0}}$
$T = 2\pi\sqrt{\frac{I}{MH}}$
$I = \frac{MHT^{2}}{4\pi^{2}} = \frac{5 \times 10^{- 5} \times 8\pi \times 10^{- 4} \times 15^{2}}{4\pi^{2}\mu_{0}^{2}}$
$= \frac{5 \times 10^{- 5} \times 8\pi \times 10^{- 4} \times 225}{4\pi^{2}\left( 4\pi \times 10^{- 7} \right)^{2}}$
$I = \frac{2250 \times 10^{- 9}}{\pi(16\pi^{2}) \times 10^{- 14}} = 4.54 \times 10^{5}kg\ m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.