Electro Magnetic Induction And Alternating Currents Question 481
Question: 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 . When rod makes an angle $ \theta $ from released position then potential difference between two ends of the rod is proportional to:
Options:
A) $ {{l}^{1/2}} $
B) The lower end will be at a lower potential
C) $ \sin \theta $
D) $ {{( \sin \theta )}^{1/2}} $
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Answer:
Correct Answer: D
Solution:
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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{mL^{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 } $
