A) \[\frac{{{B}^{2}}{{a}^{2}}\omega }{2R}\]
B) \[\frac{{{B}^{2}}\omega {{a}^{4}}}{8R}\]
C) \[\frac{{{B}^{2}}a{{\omega }^{2}}}{4R}\]
D) \[\frac{{{B}^{2}}\omega {{a}^{4}}}{4R}\]
Correct Answer: D
Solution :
[d] Emf induced between O and A \[E=\frac{1}{2}B\omega {{a}^{2}}\] so \[i\,\,(in\,\,loop)=\frac{E}{R}=\frac{B\omega {{a}^{2}}}{2R}\] \[\tau =(Bia)\frac{a}{2}=B\left( \frac{B\omega {{a}^{2}}}{2R} \right)\frac{{{a}^{2}}}{2}=\frac{{{B}^{2}}\omega {{a}^{4}}}{4R}\]You need to login to perform this action.
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