Question

A circular coil of radius $8.0\text{cm}$ and $20\text{turns}$ is rotated about its vertical diameter with an angular speed of $50{\text{rad s}}^{-1}in$ 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?

Answer 1

Step 1: Find the maximum and average emf.
Flux through each turn of loop, $\varphi =BA\cos \omega t$ Induced emf in the coil is given by
$ϵ=-N\frac{d\varphi }{dt}$
$=-N\frac{d(BA\cos \omega t}{dt}$
$=NBA\omega \sin \omega t$
Maximum induced emf ${\varphi }_{max}=NBA\omega$
Using the given values,
Radius of the circular coil, $r=8\text{cm}=0.08m$
Area of the coil, $A=\pi r^2=\pi \times (0.08{)}^2{m}^2$
Number of turns , $N=20$
Angular speed, $\omega =50\text{rad/s}$
Magnetic field strength, $B=3\times {10}^{-2}T$
We get,
$ϵ=20\times 3\times {10}^{-2}\times \pi \times (0.08{)}^2\times 50$
$ϵ=0.603V$
Therefore, the maximum emf induced in the coil is $0.603V$.
Over a full cycle, the average emf induced in the coil is zero.

Step 2: Find the maximum value of current.
Formula Used: $I=\frac{emf}{\text{Resistance}}$
We are given,
resistance of the loop, $R=10\Omega$
Maximum current in the coil,
$I=\frac{\text{maximum emf}}{\text{Resistance}}=\frac{0.603}{10}$
$I=0.0603A$
Step 3: Find the power loss.
Average power because of the Joule heating,
$P=\frac{{ϵ}_{max}{I}_{max}}{2}$
$P=\frac{0.603\times 0.0603}{2}=0.018W$
The current induced in the coil produces a torque opposing the rotation of the coil to keep the rotation of the coil continuous, we must find a source of torque which opposes the torque due to emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor.
Final answer: Maximum emf $=0.063V$
Average emf = 0 V
Maximum current $=0.0603A$
Power Loss $=0.018W$