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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 work done in moving it from its most stable to most unstable position ?

Text Solution

Verified by Experts

From `E_(q).U_(m) = -m`. B, the most stable position is `theta = 0^(@)` and the most unstable position is `q = 180^(@)`. Work done is given by
`W = U_(m) (theta = 180^(@)) - U_(m) (theta = 0^(@))`
`=2 m B = 2 xx 0.40 xx 800 xx 10^(-4) = 0.064 J`
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