Electro Magnetic Induction And Alternating Currents Question 514
Question: A uniform circular loop of radius a and resistance R is placed perpendicular to a uniform magnetic field B. One half of the loop is rotated about the diameter with angular velocity $ \omega $ as shown in Fig. Then, the current in the loop is
Options:
A) $ \frac{\pi a^{2}B\omega }{4R} $ , when $ \theta $ is zero
B) $ \frac{\pi a^{2}B\omega }{2R} $ , when $ \theta $ is zero
C) zero, when $ \theta =\pi /2 $
D) $ \frac{\pi a^{2}B\omega }{2R} $ , when $ \theta =\pi /2 $
Show Answer
Answer:
Correct Answer: D
Solution:
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$ \theta =\omega t $ .
Only half circular part will be involved in inducing emf, so effective area
$ A=\frac{\pi a^{2}}{2} $
$ \phi =BA,\cos ,\theta $
$ e=-\frac{d\phi }{dt} $
$ =+BA,\sin ,\theta ( \frac{d\theta }{dt} ) $
$ \Rightarrow ,e=\frac{B\pi a^{2}}{2}\omega ,\sin ,\theta $
$ I=\frac{e}{R}=\frac{B\pi a^{2}\omega }{2R},\sin ,\theta $
BETA
