Q 6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?


 

Maxm emf induced = 0.603 V

Avg emf induced = 0 V

Max current in the coil = 0.0603 A

Power loss (average) = 0.018 W

(Power which is coming from external rotor)

Circular coil radius, r = 8 cm = 0.08 m

Area of the coil,A=πr2=π×(0.08)2m2

Number of turns on the coil, N = 20

Angular speed,

Magnetic field strength,

The resistance of the loop, R = 10 Ω Maximum induced emf is given as: e = NωAB

=20×50×π×(0.08)2×3×102=0.603V

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

The maximum current is given as:

I=eR=0.60310=0.0603A
 Average power because of the Joule heating:

P=eI2=0.603×0.06032=0.018W The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent It must supply torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.