A) \[\sqrt{\frac{mv_{0}^{2}}{L}}\]
B) \[\sqrt{\frac{mv_{0}^{2}}{2L}}\]
C) \[\sqrt{\frac{mv_{0}^{2}}{4L}}\]
D) zero
Correct Answer: B
Solution :
[b] \[I(t)=\sqrt{\frac{m}{L}}{{v}_{0}}\,\sin \,\omega t\] where \[\omega =\frac{B\ell }{\sqrt{mL}}\] \[\Rightarrow \,\,I=\sqrt{\frac{mv_{0}^{2}}{2L}}\]You need to login to perform this action.
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