Question

A bar magnet suspended in a field of induction $1.5\times {10}^{-6}T$ has a time period $\frac{\pi }{5}$ sec.The moment of inertia about the axis of suspension is $1.5\times {10}^{-4}kg{m}^2$. The magnetic moment of the magnet is

• A. $10000A{m}^2$
• B. $\frac{1}{16}A{m}^2$
• C. $10A{m}^2$
• D. $1A{m}^2$

Answer 1

The correct option is A $10000A{m}^2$

$T=2\pi \sqrt{\frac{I}{MB}}$

$T^2=4{\pi }^2\frac{I}{MB}$

$\therefore M=\frac{4{\pi }^{2}I}{B{T}^2}$

$=\frac{4{\pi }^2\times 1.5\times {10}^{-4}}{1.5\times {10}^{-6}\times \frac{{\pi }^2}{25}}=10000A{m}^2$