$A$ 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 $3.0 \times 10^{-2}\; T$. 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?

(A) Given:
Radius of the circular coil,$r = 8.0\; cm = 0.08\; m$
Number of turns,$N = 20$
Angular speed,$\omega = 50\; rad\; s^{-1}$
Magnetic field strength,$B = 3.0 \times 10^{-2}\; T$
Resistance of the loop,$R = 10\; \Omega$
Area of the coil,$A = \pi r^2 = \pi \times (0.08)^2 = 0.0201\; m^2$
$1$. Maximum induced $emf$ $(e_{max})$:
$e_{max} = N \omega A B = 20 \times 50 \times 0.0201 \times 3.0 \times 10^{-2} = 0.603\; V$
$2$. Average induced $emf$:
Since the magnetic flux changes sinusoidally over a full cycle,the average induced $emf$ is $0\; V$.
$3$. Maximum current $(I_{max})$:
$I_{max} = \frac{e_{max}}{R} = \frac{0.603}{10} = 0.0603\; A$
$4$. Average power loss $(P_{avg})$:
$P_{avg} = \frac{e_{max} I_{max}}{2} = \frac{0.603 \times 0.0603}{2} \approx 0.018\; W$
$5$. Source of power:
The power dissipated as Joule heating comes from the external agent (rotor) that rotates the coil against the magnetic torque.