A) \[0.904\times {{10}^{-5}}wb\]
B) \[5.43\times {{10}^{-5}}wb\]
C) \[2\times {{10}^{-2}}{{m}^{2}}\]
D) \[{{10}^{6}}\]
Correct Answer: A
Solution :
Component of weight along the inclined plane \[=mg\,\sin \theta \] Again, \[F=Bil=B\frac{Bhv}{R}l=\frac{{{B}^{2}}{{l}^{2}}v}{R}\] Now, \[\frac{{{B}^{2}}{{l}^{2}}v}{R}=mg\,\sin \theta \] or \[v=\frac{mgR\,\sin \theta }{{{B}^{2}}{{l}^{2}}}\]
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