In this article, we will derive an expression for induced emf in a rod rotating in a uniform magnetic field. So let’s get started…
Motional EMF in rotating rod Class 12
Consider a metallic rod $OA$ of length $l$, which is rotating with angular velocity $\omega$ in a uniform magnetic field $B$, the plane of rotation is perpendicular to the magnetic field.
Read Also
- Relation between induced charge and change in magnetic flux
- Motional EMF, and energy consideration class 12
A rod may be supposed to be formed of a large number of small elements. Consider a small element of length $dx$ at a distance $X$ from the center. If $v$ is the linear velocity of this element, then the area swept by the element per second is $vdx$. According to Faraday’s law of electromagnetic induction. EMF induced in the rod due to this small area is given as
$$
d \varepsilon=B \frac{d A}{d t}=B v d x
$$
But $v=x \omega$
$\therefore \quad d \varepsilon=B x \omega d x$
$\therefore \quad$ The emf induced across the rod
$$\begin{aligned}
\varepsilon&=\int_{0}^{l} B x \omega d x=B \omega \int_{0}^{l} x d x\\
&=B \omega\left[\frac{x^{2}}{2}\right]_{0}^{l} \\
&=B \omega\left[\frac{l^{2}}{2}-0\right]=\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B} \omega \mathbf{l}^{2}
\end{aligned}
$$
Current induced in the rod
Current induced in rod, $$I=\frac{\varepsilon}{R}=\frac{1}{2} \frac{B \omega l^{2}}{R}$$
The power dissipated through the circuit is
$$
P=\frac{\varepsilon^{2}}{R}=\frac{B^{2} \omega^{2} l^{4}}{4 R}
$$
Stay tuned with Laws Of Nature for more useful and interesting content.
