A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter
Question:

A circular coil of radius $8.0 \mathrm{~cm}$ and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad $\mathrm{s}^{-1}$ in a uniform horizontal magnetic field of magnitude $3.0 \times 10^{-2} \mathrm{~T}$. Obtain the maximum and average emf 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?

Solution:

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, $A=\pi r^{2}=\pi \times(0.08)^{2} \mathrm{~m}^{2}$

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3 × 10−2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as:

e = Nω AB

= 20 × 50 × π × (0.08)2 × 3 × 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:

$I=\frac{e}{R}$

$=\frac{0.603}{10}=0.0603 \mathrm{~A}$

Average power loss due to joule heating:

$P=\frac{e I}{2}$

$=\frac{0.603 \times 0.0603}{2}=0.018 \mathrm{~W}$

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.