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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.

6 years ago


Answers : (1)


Dear aakash,

The given information is as follows:

Length of rod     = 1.5 mt.

                             Frequency of rotation    = 20 rev/sec.

                             BH = 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. provides online iit jee courses and IIT JEE Test Series with IITians. Click here to get free online test series and check your status timely or you can join us as our registered user for getting best iit jee study material or iit jee test series.

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thanks and regards.

Akhilesh Shukla



6 years ago

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