A coil of 100 turns, each of area 0.02m^{2} is kept in a uniform field of induction 3.5 x10^{-5} T. If the coil rotates with a speed of 6000 r.p.m. about an axis in the plane of the coil and perpendicular to the magnetic induction, calculate peak value of e.m.f. induced in the coil.

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#### Solution

Given: f = 6000 r.p.m = 100 Hz, N = 100, A = 0.02 m^{2}, B = 3.5 10^{-5} T

To find: peak e.m.f (e_{0})

Formula: e_{0} = 2πfBNA

Calculation: From formula,

e_{0} = 2 x 3.14 x 100 x 100 x 0.02 x 3.5 x 10^{-5}

∴ e_{0} = 43.99 x 10^{-3} V

**The peak value of e.m.f. induced in the coil is 43.99 x 10 ^{-3} V.**

Concept: Coil Rotating in Uniform Magnetic Induction

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