Question

A magnetic needle is pivoted through its centre of mass and is free to rotate in a plane containing uniform magnetic field $200\times {10}^{-4}T$ . When it is displaced slightly from the equilibrium, it makes $2$ oscillations per second. If the moment of inertia of the needle about the axis of oscillation is $0.75\times {10}^{-5}kg{m}^2$ , the magnetic moment of the needle is

• A. $6J{T}^{-1}$
• B. $0.6J{T}^{-1}$
• C. $0.06J{T}^{-1}$
• D. $0.006J{T}^{-1}$

Answer 1

Answer: (C)

$0.06J{T}^{-1}$

The time period of oscillation of a bar magnet freely suspended in air is given by
$T=2\pi \sqrt{\frac{I}{m{B}_H}}$
$m=4{\pi }^2\frac{I}{T^2{B}_H}$
$m\approx 0.06J{T}^{-1}$