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Question and Answer

A short bar magnet placed with its axis at \(30°\) with an external field of \(800\;G\)  experiences a torque of \(0.016\;Nm\) . What is the magnetic moment of the magnet? 

Answer

0.40Am2
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Solution

Torque on magnetic dipole:
\(∴\;\tau =m\;B \sin 30{}^{\circ}\)
\(0.016=m\times 800\times 10^{-4}\times \frac 1 2\) 
\(⟹ m=0.40\;\text{Am}^2\)  
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