Question

An L shape object is made of two rods. Rod $AB$ is made of insulating material while $BC$ is made of conducting material. This object is rotated about end A with angular speed $\omega$ in a uniform magnetic field B as shown. Find emf induced across the ends of the conducting rod $BC$.

Answer 1

$\frac{1}{2} B \omega \ell^2$

Answer 2

The induced emf across the conducting rod BC is calculated by integrating the effect of motional emf along the rod. For a small element $d\vec{l}$ at position $\vec{r}$ from the pivot A, its velocity is $\vec{v} = \vec{\omega} \times \vec{r}$. The induced emf in this element is $d\mathcal{E} = (\vec{v} \times \vec{B}) \cdot d\vec{l}$.

Setting up a coordinate system with A at the origin, B at $(0, \ell)$ and C at $(\ell, \ell)$, and with $\vec{\omega} = \omega \hat{k}$ and $\vec{B} = -B \hat{k}$, the velocity of a point P on BC at $(x, \ell)$ is $\vec{v}_P = \omega x \hat{j} - \omega \ell \hat{i}$. The element $d\vec{l}$ along BC is $dx \hat{i}$.

Calculating $(\vec{v}_P \times \vec{B}) \cdot d\vec{l}$ gives $-\omega x B dx$. Integrating this from B ($x=0$) to C ($x=\ell$) yields the total emf $\mathcal{E}_{BC} = \int_0^\ell -\omega x B dx = -\frac{1}{2} B \omega \ell^2$. The magnitude of the induced emf is $\frac{1}{2} B \omega \ell^2$.