NCERT · Class 12 · Physics · Electromagnetic Induction(a) State Faraday's laws of electromagnetic induction. (b) Derive an expression for the motional electromotive force (emf) induced across the ends of a straight conductor of length $l$ moving with a uniform velocity $v$ perpendicular to a uniform magnetic field $B$. (c) Derive an expression for the emf induced across the ends of a metallic rod of length $l$ rotating with a constant angular velocity $\omega$ about one of its ends in a uniform magnetic field $B$ perpendicular to the plane of rotation.

Faraday's Laws of Electromagnetic Induction

• First Law: Whenever there is a change in the magnetic flux linked with a closed circuit, an electromotive force (emf) is induced in the circuit, which lasts as long as the change in flux continues.
• Second Law: The magnitude of the induced emf in a circuit is directly proportional to the time rate of change of magnetic flux linked with the circuit. $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$ (where the negative sign indicates direction according to Lenz's Law)

Derivation of Motional EMF in a Straight Conductor

\nConsider a straight conducting rod $PQ$ of length $l$ moving perpendicular to a uniform magnetic field $B$ directed into the page with a constant velocity $v$.

• Let a rectangular loop $PQRS$ be placed in the plane, where side $PQ$ is free to slide.
• Let $RS = x$ be the position of the conductor at time $t$.
• The magnetic flux linked with the area $A = l x$ enclosed by the loop is: $$\Phi_B = B A = B l x$$ \nAccording to Faraday's law of electromagnetic induction, the induced emf $\mathcal{E}$ is: $$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(B l x)$$ \nSince $B$ and $l$ are constant: $$\mathcal{E} = -B l \frac{dx}{dt}$$ \nSince the velocity $v = -\frac{dx}{dt}$ (as $x$ decreases with time): $$\mathcal{E} = B l v$$ \nThis induced emf across the moving conductor is called motional emf.

Derivation of EMF in a Rotating Metallic Rod

\nConsider a metal rod of length $l$ rotating with constant angular velocity $\omega$ about a pivot at one end in a plane perpendicular to a uniform magnetic field $B$.

1. Consider a small element of length $dr$ of the rod located at a distance $r$ from the pivot point.
2. Linear velocity of this element is given by $v = r \omega$.
3. The motional emf induced in this small element $dr$ is: $$d\mathcal{E} = B v dr = B (r \omega) dr$$
4. Total induced emf $\mathcal{E}$ across the entire length of the rod from $r = 0$ to $r = l$ is obtained by integration: $$\mathcal{E} = \int_{0}^{l} d\mathcal{E} = \int_{0}^{l} B \omega r dr$$ $$\mathcal{E} = B \omega \left[ \frac{r^2}{2} \right]_{0}^{l}$$ $$\mathcal{E} = \frac{1}{2} B \omega l^2$$ \nThus, the total induced emf across the ends of the rotating rod is $\mathcal{E} = \frac{1}{2} B \omega l^2$.