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JEE Main & Advanced Sample Paper JEE Main Sample Paper-34

  • question_answer
    A metal wire PQ having a mass of 10 g and length 4.9 cm lies at rest on the metal rails as shown in figure. A vertical downward magnetic field of magnitude 0.8T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 200, the wire starts sliding on the rails. The coefficient of fraction is:

    A)  0.25                                      

    B)  0.31

    C)  0.43                                      

    D)  0.12

    Correct Answer: D

    Solution :

    The current through the wire PQ is \[I=\varepsilon /R\] When the wire just starts sliding, the force \[I\ell B\] toward right is equal to the force \[\mu \] mg towards left.             \[\Rightarrow \,\,\mu mg\,=I\ell B\,=(\varepsilon R)\ell B\]             \[\therefore \,\,\mu =\,\frac{\varepsilon \ell B}{mgR}\,=\frac{6\times 4.9\times {{10}^{-2}}\,\times 0.8}{{{10}^{-2}}\,\times 9.8\,\times 20}\,=0.12\]


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