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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 29

  • question_answer
    A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2s in earths horizontal magnetic field of \[24\mu T\]. When a horizontal field of \[18\mu T\]is produced opposite to the earths field by placing a current carrying wire, the new time period of the magnet will be

    A) 1 s         

    B) 2 s

    C) 3 s                   

    D) 4 s.

    Correct Answer: D

    Solution :

    [d]: The time period T of oscillation of a magnet is given by\[T=2\pi \sqrt{\frac{I}{MB}}\] As I and M remain the same, \[\therefore \]\[T\propto \frac{1}{\sqrt{B}}\]or\[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{B}_{1}}}{{{B}_{2}}}}\] According to given problem, \[{{B}_{1}}=24\mu T,{{B}_{2}}=24\mu T-18\mu T=6\mu T,{{T}_{1}}=2s\] \[\therefore \]\[{{T}_{2}}=(2s)\sqrt{\frac{24\mu T}{6\mu T}}=4s\]


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