Question

A magnetic needle having moment of inertia of $40\, g\, cm ^{2}$ has time period of $3\, s$ in earth's horizontal field of $3.6 \times 10^{-5}\, Wb / m ^{2}$. Its magnetic moment will be

• A. $0.5\, A\, m ^{2}$
• B. $5\, A\, m ^{2}$
• C. $0.250\, A\, m ^{2}$
• D. $5 \times 10^{2} A\, m ^{2}$

Answer 1

Answer: (A)

$T=2 \pi \sqrt{\frac{I}{M B_{H}}}$
$I=40\, g\, cm ^{2}=400 \times 10^{-8}\, kg\, m ^{2}$
$\therefore 3=2 \pi \sqrt{\frac{400 \times 10^{-8}}{36 \times 10^{-6} \times M}}$
$\Rightarrow \frac{1}{M}=\frac{9}{4 \pi^{2}} \times \frac{36}{4}$
$\Rightarrow M=0.5\, A\, m ^{2}$