Question

Figure shows a straight, long wire carrying a current i and a rod of length l coplanar with the wire and perpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distance of the wire from the centre of the rod is x. Find the motional emf induced in the rod.

Answer 1

Here, the magnetic field $\overrightarrow{B}$ due to the long wire varies along the length of the rod. We will consider a small element of the rod of length da at a distance a from the wire. The magnetic field at a distance a is given by

$\overrightarrow{B}=\frac{{\mathrm{\mu }}_{0}i}{2\pi a}$
Now,
Induced emf in the rod:
$$\begin{gathered} de=Bvda \\ =\frac{{\mathrm{\mu }}_{0}i}{2\pi a}\times v\times da \end{gathered}$$
Integrating from $x-\frac{l}{2}$ to $x+\frac{l}{2}$, we get
$$\begin{gathered} e=\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int }}de \\ =\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int \frac{{\mu }_{0}i}{2\pi a}}}vda \\ =\frac{{\mu }_{0}iv}{2\mathrm{\pi }}\left[\ln \left(x+\frac{l}{2}\right)-\ln \left(x-\frac{l}{2}\right)\right] \\ =\frac{{\mu }_{0}iv}{2\mathrm{\pi }}\ln \left[\frac{x+{\displaystyle \frac{l}{2}}}{x-{\displaystyle \frac{l}{2}}}\right] \end{gathered}$$