Electro Magnetic Induction And Alternating Currents Question 306

Question: A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity $ \omega $ . The induced e.m.f. between the two ends is [MP PMT 1992; Orissa JEE 2003]

Options:

A) $ \frac{1}{2}B\omega l^{2} $

B) $ \frac{3}{4}B\omega l^{2} $

C) $ B\omega l^{2} $

D) $ 2B\omega l^{2} $

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Answer:

Correct Answer: A

Solution:

If in time t. the rod turns by an angle q, the area generated by the rotation of rod will be

$ =\frac{1}{2}l\times l\theta $ $ =\frac{1}{2}l^{2}\theta $

So the flux linked with the area generated by the rotation of rod

$ \varphi =B\ ( \frac{1}{2}l^{2}\theta )\cos 0=\frac{1}{2}Bl^{2}\theta =\frac{1}{2}Bl^{2}\omega ,t $

And so $ e=\frac{d\varphi }{dt}=\frac{d}{dt}( \frac{1}{2}Bl^{2}\omega t )=\frac{1}{2}Bl^{2}\omega $



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