A) \[\frac{Rmg}{{{B}^{2}}{{L}^{2}}}\]
B) \[\frac{Rmg}{{{B}^{2}}{{L}^{2}}}.\sec \theta .\tan \theta \]
C) \[\frac{Rmg}{{{B}^{2}}{{L}^{2}}}.\tan \theta \]
D) \[\frac{Rmg}{{{B}^{2}}{{L}^{2}}}.cot\theta \]
Correct Answer: B
Solution :
Due to motion of bar in magnetic field, emf induced, \[E=Bl\upsilon =BL\upsilon \cos \theta \] |
Current in metal bar, \[I=\frac{E}{R}=\frac{BL\upsilon }{R}\cos \theta \] |
Force on bar up the inclined plane is \[F=BIL\cos \theta =\frac{{{B}^{2}}{{L}^{2}}\upsilon }{R}{{\cos }^{2}}\theta \] |
When this force is equal to downwards gravitational pull, bar reaches terminal speed. |
\[\frac{{{B}^{2}}{{L}^{2}}\upsilon {{\cos }^{2}}\theta }{R}=mg\sin \theta \] |
\[\upsilon =\frac{Rmg}{{{B}^{2}}{{L}^{2}}}=\sec \theta .\tan \theta \] |
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