A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 50 `rads^(-1)` in a uniform horizontal magnetic field of magnitude `3 xx 10^(-2) T`. Obtain the maximum and average e.m.f. induced in the coil . If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Explanation

Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W
(Power comes from the external rotor)
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A = `pir^(2) = pi xx (0.08)^(2) m^(2)`
Number of turns on the coil, N = 20
Angular speed, `omega` = 50 rad/s
Magnetic field strength, B =` 3 xx10^(−2)` T
Resistance of the loop, R = 10 `omega`
Maximum induced emf is given as:
e = `Nomega` AB
= `20 xx 50 xx pi xx (0.08)^(2) xx 3 xx 10^(−2)`
= 0.603 V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:
`I=e/R`
`=0.603/10`=0.0603A
Average power loss due to joule heating:
`P=(el)/2`
`=(0.603xx0.0603)/2=0.018 W`
The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.