A magnetic needle has magnetic moment of $5 \times 10^{-2} Am^2$ and Moment of inertia $12 \times 10^{-6} Kgm^2$. It performs 60 oscillations in 1 minute when suspended in a magnetic field. What is the magnitude of the magnetic field?

$94.7 \times 10^{-4} T$

The correct answer is Option (2) → $94.7 \times 10^{-4} T$

Period of one oscillation, $T=2\pi\sqrt{\frac{I}{MB}}$

where,

I = Moment of inertia

M = Magnetic moment of needle

B = Magnetic field strength

$T=\frac{60\,seconds}{60\,oscillation}=1s$

Given,

$M=5×10^{-2}A.m^2$

$I=12×10^{-6}kg.m^2$

$T = 1\,second$

$1=2\pi\sqrt{\frac{12×10^{-6}}{5×10^{-2}×B}}$

$1=2\pi^2×\frac{12×10^{-6}}{5×10^{-2}×B}$

$B≃94.8×10^{-4}T$