Question

Calculate the motional EMF induced in the rod of length L.

Answer 1

$\frac{\mu_0 I v}{2\pi} \ln \left(\frac{b}{a}\right)$

Answer 2

The magnetic field due to the current wire is non-uniform, varying as $B(r) = \frac{\mu_0 I}{2\pi r}$ and directed into the page. The rod moves upwards with velocity $v$. The motional EMF is calculated by integrating the effective electric field $\vec{E}_{eff} = \vec{v} \times \vec{B}$ along the length of the rod. $\vec{v} \times \vec{B}$ is directed towards point P. Integrating this field component along the rod from P to Q yields $\mathcal{E}_{PQ} = -\frac{\mu_0 I v}{2\pi} \ln \left(\frac{b}{a}\right)$. The magnitude of this induced EMF is $\frac{\mu_0 I v}{2\pi} \ln \left(\frac{b}{a}\right)$.