Question

A short bar magnet placed with its axis at ${30}^0$ with a uniform external magnetic field of $0.25$ T experiences a torque of magnitude equal to $4.5\times {10}^{-2}J$. What is the magnitude of magnetic moment of the magnet?

• A. $0.36J{T}^{-1}$
• B. $1.36J{T}^{-1}$
• C. $2.36J{T}^{-1}$
• D. $3.36J{T}^{-1}$

Answer 1

The correct option is A $0.36J{T}^{-1}$

Given

Angle:- ${30}^o$

$B=0.25T$

$T=4.5\times {10}^{-2}$J

using

$\tau =M\cdot B\sin \theta$

$M=\frac{\tau }{B\sin \theta }$

$=\frac{4.5\times {10}^{-2}}{0.25\times \sin {30}^o}$

$=\frac{9\times {10}^{-2}}{0.25}$

$M=36\times {10}^{-2}J{T}^{-1}$.