A short bar magnet placed with its axis at 
with an external field of 800 Gauss, experiences a torque of 
. The work done in moving it from most stable to most unstable position is 
. The value of 
is 
.
Text Solution
Verified by ExpertsThe correct answer is:
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For a magnetic dipole in an external field, torque is 
At 
Work done from most stable (aligned, 
) to most unstable (anti-aligned, 
) position equals the change in potential energy:
Therefore, 
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