Electro Magnetic Induction And Alternating Currents Question 474

Question: A conducting wire of mass m slides down two smooth conducting bars, set at an angle 0 to the horizintal as shown in Fig. The separation between the bars is $ l $ . The system is located in the magnetic field B, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is

Options:

A) $ \frac{mg,R,\sin \theta }{B^{2}l^{2}} $

B) $ \frac{mg,R,\sin \theta }{Bl^{2}} $

C) $ \frac{mg,R,\sin \theta }{B^{2}l^{5}} $

D) $ \frac{mg,R,\sin \theta }{Bl^{4}} $

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Answer:

Correct Answer: A

Solution:

  • Component of weight along the inclied plane

    $ =mg\sin \theta $

    Again, $ F=BI\ell =B\frac{B\ell v}{R}\ell =\frac{B^{2}{{\ell }^{2}}v}{R} $

    Now, $ \frac{B^{2}{{\ell }^{2}}v}{R}=mg,\sin \theta $

    or $ v=\frac{mgR,\sin ,\theta }{B^{2}{{\ell }^{2}}} $



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