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A bar magnet has a magnetic moment `2.5JT^(-1)` and is placed in a magnetic field of `0.2 T`. Calculate the work done in turning the magnet from parallel to antiparallel position relative to field direction. |
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Answer» Work done in changing the orientation of a dipole of moment `M` in a field `B` from position `theta_(1)` to `theta_(2)` is given by `W=MB(costheta_(1)-costheta_(2))` Here, `theta_(1)=0^(@)` and `theta_(2)=180^(@)` So, `W=2MB=2xx2.5xx0.2=1J` |
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