The loop is rotated with a constant angular speed \[\omega \] about an axis passing through the centre O, and perpendicular to the page. Then the induced current in the wire loop is
A) zero
B) \[{{\operatorname{Br}}^{2}}\omega /R\]
C) \[{{\operatorname{Br}}^{2}}\omega /2R\]
D) \[\operatorname{B}\pi {{r}^{2}}\omega /R\]
Correct Answer: C
Solution :
The area swept by radius OC in one half circle is \[\pi {{r}^{2}}/2\] The flux change in time T/2 is thus \[\left( \pi {{r}^{2}}B/2 \right)\]. The induced emfis then \[e=\pi {{r}^{2}}B/T=B\omega {{r}^{2}}/2\,\,\left[ \because \,\,T=\frac{2\pi }{\omega } \right]\] The induced current is then \[\operatorname{I} = e/R = B\omega {{r}^{2}}/2R\]
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