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A short bar magnet placed with its axis ...

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 ?

Text Solution

Verified by Experts

`tau =MB sin theta , theta = 30^(@) , ` hence ` sin theta = 1//2`
Thus `0.016 = m xx ( 800 xx 10^(-4) T) xx (1//2)`
`M= 160 xx 2 // 800 = 0. 40 Am^(2)`
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