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The work done in rotating a magnet of pole strength `1 A-m` and length `1 cm` through an angle of `60^(@)` from the magnetic meridian is `(H=30A//m)`A. `9.42xx10^(-8)J`B. `3.14xx10^(-8)J`C. `18.84xx10^(-8)J`D. `10xx10^(-8)J` |
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Answer» Correct Answer - C `W=mxx2l B[cos theta_(1)-cos theta_(2)]` |
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