Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.6
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 3.0xl0⁻² T. Obtain the maximum and average emfinduced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
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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 =πr2 = π x (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω= 50 rad/s
Magnetic field strength, B = 3 x 10⁻²T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as: e = Nω AB = 20 x 50 x-π x (0.08)2 x 3 x 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 = 0.0603 A
Average power loss due to joule heating:
P = eI/2 = (0.603×0.0603)/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.