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} $
Show Answer
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 $
BETA
