??A metal rod 1.5 m long rotates about its one end in a vertical plane at right angles to the magnetic meridian. If the frequency of rotation is 20 rev/s. Find the emf induced between the ends of rod.BH = 0.32 ´ 10-4T: Horizontal component of earth’s magnetic field.

Dear aakash,

The given information is as follows:

Length of rod     = 1.5 mt.

                             Frequency of rotation    = 20 rev/sec.

                             B_H = 0.32 * 10^-4 T

                             Angular velocity of rod = 20 * 2∏ rad/sec.

                             emf induced = B*dA/dt = 0.32*10^-4 *20*∏*(1.5)².

       Therefore horizontal component of the earth's magnetic feild is  =  4.5 * 10^-3 V.

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