An AC generator consists of coil of 100 turns and cross-sectional area of 3 ${{m}^{2}}$, rotating at a constant angular speed of 60 rad ${{s}^{-1}}$ in a uniform magnetic field 0.04 T. The resistance of the coil is 500 Ω . Calculate (i) maximum current drawn from the generator and (ii) minimum power dissipation of the coil.

Here, total number of turns,

N = 100, A = 3 ${{m}^{2}}$

$\omega$ = 60 rad ${{s}^{-1}}$, B = 0.04 T

(i) Maximum emf produced in the coil,

${ e }*{ 0 }$ = NBAω
= 100 x 0.04 x 3 x 60
${ e }*{ 0 }$ = 720 V

Since,resistance of the coil is 500Ω,the maximum current drawn from the generator is

${ I }*{ 0 }$ =
${ e }*{ 0 }$/R = 720/500 = 1.44A

(ii) Maximum power dissipation in the coil,

P = ${ e }*{ 0 }$
${ I }*{ 0 }$ = 720 x 1.44

=1036.8 W