question_answer

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

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 \[\theta =\pi /2\]

Correct Answer: D

Solution :

[d]: \[\theta =\omega t\]. Only half circular part will be involved in inducing emf, so effective area \[A=\frac{\pi {{a}^{2}}}{2}\] \[\phi =BA\,\cos \,\theta \] \[e=-\frac{d\phi }{dt}\]                  \[=+BA\,\sin \,\theta \left( \frac{d\theta }{dt} \right)\]  \[\Rightarrow \,\,\,e=\frac{B\pi {{a}^{2}}}{2}\omega \,\sin \,\theta \]   \[I=\frac{e}{R}=\frac{B\pi {{a}^{2}}\omega }{2R}\,\sin \,\theta \]