Question

Deduce an expression for the induced emf in the coil of the generator.

Hint:

Peak emf $e_0 = N B A \omega$ can be increased by increasing the number of turns $N$, magnetic field strength $B$, coil area $A$, or speed of rotation $\omega$.

Answer 1

Concept:
• Magnetic flux linked with a single turn of area $A$ inclined at angle $\theta$ to magnetic field $\vec{B}$ is $\Phi = B A \cos \theta$.

• If the coil rotates with uniform angular speed $\omega$, $\theta = \omega t$.

• By Faraday's Law, induced emf is $e = -N \frac{d\Phi}{dt}$.

Step 1: Magnetic Flux Expression
Let $N$ = total number of turns in the rectangular coil, $A$ = area of each turn, $B$ = magnitude of magnetic field.
At time $t = 0$, let the normal to the coil be parallel to $\vec{B}$ ($\theta = 0$).
At time $t$, the coil rotates through angle $\theta = \omega t$.
The magnetic flux linked with all $N$ turns of the coil is:
\[ \Phi = N (\vec{B} \cdot \vec{A}) = N B A \cos(\omega t) \]

Step 2: Applying Faraday's Law
According to Faraday's law of electromagnetic induction, induced emf $e$ is:
\[ e = -\frac{d\Phi}{dt} = -\frac{d}{dt} \left[ N B A \cos(\omega t) \right] \]
Differentiating $\cos(\omega t)$ with respect to $t$:
\[ \frac{d}{dt} [\cos(\omega t)] = -\omega \sin(\omega t) \]
Substitute back into the expression:
\[ e = -N B A \left( -\omega \sin(\omega t) \right) \]
\[ e = N B A \omega \sin(\omega t) \]

Step 3: Peak Value and Final Relation
Let $e_0 = N B A \omega$ be the maximum or peak value of induced emf.
Then the instantaneous induced emf equation becomes:
\[ e = e_0 \sin(\omega t) \]

Step 4: Conclusion
The induced emf varies sinusoidally with time according to $e = N B A \omega \sin(\omega t) = e_0 \sin(\omega t)$.