A) \[\frac{1}{2}B\omega {{l}^{2}}\]
B) \[\frac{3}{4}B\omega {{l}^{2}}\]
C) \[B\omega {{l}^{2}}\]
D) \[2B\omega {{l}^{2}}\]
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 \[\varphi =B\ \left( \frac{1}{2}{{l}^{2}}\theta \right)\cos 0=\frac{1}{2}B{{l}^{2}}\theta =\frac{1}{2}B{{l}^{2}}\omega \,t\] And so \[e=\frac{d\varphi }{dt}=\frac{d}{dt}\left( \frac{1}{2}B{{l}^{2}}\omega t \right)=\frac{1}{2}B{{l}^{2}}\omega \]You need to login to perform this action.
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