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 $

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Answer:

Correct Answer: D

Solution:

  • $ \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 $



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