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An armature coil consists of 20 turns of...

An armature coil consists of 20 turns of wire each of area `A = 0.09 m^(2)` and total resistance 15.0 ohm. It rotates in a magnetic field of 0.5 T at a constant frequency of `(150)/(pi) Hz`.
Calculate the value of (i) maximum and (ii) average induced e.m.f. produced in the coil.

Text Solution

Verified by Experts

Here,`N = 20, A = 0.09 m^(2)`,
`R = 15.0 ohm, B = 0.5 T`
`v = (150)/(pi) Hz, e_(0) = ?`
`e_(0) = N AB omega = NAB (2 pi v)`
`= 20 xx 0.09 xx 0.5 (2 pi xx (150)/(pi)) = 270 V`
As e.m.f. produced is alternating, average induced e.m.f. produced in the coil is zero.
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