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

\[\vec{B}  = \frac{\mu_0 i}{2\pi a}\]

Now,

Induced emf in the rod:-

\[de = Bvda\]

\[= \frac{\mu_0 i}{2\pi a} \times v \times da\]

Integrating from `x-l/2` to `x+l/2,` we get

\[e =  \int\limits_{x - \frac{l}{2}}^{x + \frac{l}{2}} de\]

\[     =  \int\limits_{x - \frac{l}{2}}^{x + \frac{l}{2}} \frac{\mu_0 i}{2\pi a} vda\]

\[     = \frac{\mu_0 iv}{2\pi}\left[ \ln\left( x + \frac{l}{2} \right) - \ln\left( x - \frac{l}{2} \right) \right]\]

\[       = \frac{\mu_0 iv}{2\pi}\ln\left[ \frac{x + \frac{l}{2}}{x - \frac{l}{2}} \right]\]