A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microteslas. When a horizontal field of 18 microteslas is produced opposite to the earth's 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

1 Answer

  1. The time period T of oscillation of a magnet is given as –

    T = 2π√I/MB where,

    l = Moment of inertia of the magnet about the axis of rotation

    M = Magnetic moment of the magnet

    B = Uniform magnetic field As l, B remains the same


    T∝1/√B or T2/T1 = √B1/B2

    = B1 = 24ut

    B2 = 24ut – 18ut

    B2 = 6ut

    T1 = 2 seconds

    Hence, T2 = (2)√ 24ut/6ut

    = 4

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