question_answer

A conducting rod of length l is hinged at point O. It is free to rotate in vertical plane. There exists a uniform magnetic field \[\vec{B}\]in horizontal direction. The rod is released from position shown in the figure. When rod makes an angle \[\theta \] from released position then potential difference between two ends of the rod is proportional to:         

A) \[{{l}^{1/2}}\]                   

B) The lower end will be at a lower potential   

C) \[\sin \theta \]

D) \[{{\left( \sin \theta  \right)}^{1/2}}\]

Correct Answer: D

Solution :

[d] WET from A to B \[{{\omega }_{mg}}=\Delta K.E.\] \[mg\frac{L}{2}\sin \theta =\frac{1}{2}I{{\omega }^{2}}\]   \[mg\frac{L}{2}\sin \theta =\frac{1}{2}\times \frac{m{{L}^{2}}}{3}{{\omega }^{2}}\] \[\frac{3g}{L}\sin \theta ={{\omega }^{2}}\] Induced emf produced in rod is \[\varepsilon =\frac{1}{2}(B\omega {{L}^{2}})=\frac{1}{2}B\times \sqrt{\frac{3g\,\sin \theta }{L}}\times {{L}^{2}}\]  \[\varepsilon =\frac{1}{2}\times B\times \sqrt{\frac{3g\,\sin \,\theta \times {{L}^{4}}}{L}}=\frac{1}{2}B\sqrt{3g\,\sin \,\theta {{L}^{3}}}\] \[\varepsilon \propto {{L}^{3/2}}\]     \[\varepsilon \propto \sqrt{\sin \theta }\]