Question

A short bar magnet placed with its axis at ${30}^0$ with a uniform external magnetic field of $0.35T$ experiences a torque of magnitude equal to $4.5\times {10}^{-2}$N m. The magnitude of magnetic moment of the given magnet is

• A. $26J{T}^{-1}$
• B. $2.6J{T}^{-1}$
• C. $0.26J{T}^{-1}$
• D. $0.026J{T}^{-1}$

Answer 1

The correct option is C $0.26J{T}^{-1}$

Here, $\theta ={30}^0,B=0.35T$
$T=4.5\times {10}^{-8}Nm$

then, $m=\frac{T}{Bsin\theta }=\frac{4.5\times {10}^{-2}}{0.35\times sin{30}^0}$

$m=\frac{4.5\times {10}^{-2}}{0.35\times \frac{1}{2}}=\frac{2\times 4.5\times {10}^{-2}}{0.35}=0.26J{T}^{-1}$