Question

A rectangular loop of sides $l$ and $b$ and resistance ‘R’ is kept in a region in which the magnetic field varies as $B = B_0 \sin \omega t$. Derive expression for the emf induced in the loop.

Hint:

The negative sign expresses Lenz's Law, indicating that the induced emf produces an induced current whose magnetic field opposes the time rate of change of external magnetic flux.

Answer 1

Concept:
• According to Faraday's Law of Electromagnetic Induction, induced electromotive force (emf) is equal to the negative rate of change of magnetic flux: $e = -\frac{d\Phi}{dt}$.

• Magnetic flux linked with a loop of area $A$ in a uniform magnetic field $B$ perpendicular to loop plane is $\Phi = B A$.

Step 1: Calculate magnetic flux linked with the loop
Area of rectangular loop $A = l \times b$.
Time-varying magnetic field $B(t) = B_0 \sin \omega t$.
Assuming magnetic field is perpendicular to the plane of the loop ($\theta = 0^\circ$):
\[ \Phi(t) = B(t) \cdot A = (B_0 \sin \omega t) (l b) = l b B_0 \sin \omega t \]

Step 2: Differentiate flux with respect to time
Apply Faraday's law of induction:
\[ e = -\frac{d\Phi}{dt} = -\frac{d}{dt} \left( l b B_0 \sin \omega t \right) \]
Since $l, b, B_0, \omega$ are constants:
\[ e = -l b B_0 \frac{d}{dt} (\sin \omega t) \]
\[ e = -l b B_0 \omega \cos \omega t \]

Step 3: Conclusion
The instantaneous induced emf in the rectangular loop is $e(t) = -l b B_0 \omega \cos \omega t = e_0 \cos(\omega t + \pi)$, where peak emf $e_0 = l b B_0 \omega$.