A) \[MH(1-cos\theta )\]
B) \[\frac{M}{H}(1-cos\theta )\]
C) \[\frac{M}{H}(cos\theta -1)\]
D) \[MH(cos\theta -1)\]
Correct Answer: A
Solution :
Key Idea: Work done \[W=\tau d\theta \] Work done in rotating a magnet is given by \[W=\int_{0}^{\theta }{\tau \,d\theta }\] where \[\tau \]is torque and \[d\theta \]angular charge Also, \[\tau =MH\sin \theta \] \[\therefore \] \[W=\int_{0}^{\theta }{MH\sin \theta \,d}\theta \] \[\Rightarrow \] \[W=MH\int_{0}^{\theta }{\sin \theta d\theta }\] \[\Rightarrow \] \[W=MH[1-cos\theta ]_{0}^{\theta }\] \[\Rightarrow \] \[W=MH[1-cos\theta +cos0]\] \[\Rightarrow \] \[W=MH[1-cos\theta ]\]
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