A) \[\lambda BK{{r}^{2}}\]
B) \[2\pi \lambda BK{{r}^{2}}\]
C) \[2\pi \lambda BKr\]
D) \[\pi \lambda BK{{r}^{2}}\]
Correct Answer: B
Solution :
\[e=\frac{d\phi }{dt}=\frac{d}{dt}\,(B\pi {{r}^{2}})=2\pi Br\,\frac{dr}{dt}\] \[=2\pi BKr.\,\,\left( \sin ce\,\frac{dr}{dt}=K \right)\] \[E.2\pi r=2\pi BKr\]\[E=BK\] \[\tau =\lambda \,BKr\,\,\int_{{}}^{{}}{ds\,=\lambda \,BKr.\,2\pi r=2\pi \lambda \,BK{{r}^{2}}}\]You need to login to perform this action.
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