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, \[\tau =MH\sin \theta \] \[\therefore \] \[W=\int_{0}^{\theta }{MH}\sin \theta \,\,d\theta \] \[\Rightarrow \] \[W=MH\int_{0}^{\theta }{\sin \theta }\,d\theta \] \[\Rightarrow \] \[W=MH[1-\cos \theta ]_{0}^{\theta }\] \[\Rightarrow \] \[W=MH[1-\cos \theta +\cos 0]\] \[\Rightarrow \] \[W=MH[1-\cos \theta ]\]
You need to login to perform this action.
You will be redirected in
3 sec
