Consider an infinitely long wire carrying a current $I(t)$,with $\frac{dI}{dt} = \lambda = \text{constant}$. Find the current produced in the rectangular loop of wire $ABCD$ if its resistance is $R$ as shown in the figure.

(N/A) Consider a small area element having thickness $dr$ and length $l$ on the closed rectangular loop $ABCD$ at a distance $r$ from the very long current-carrying wire as shown in the figure.
The magnetic field at this strip,due to the very long current-carrying wire,is given by:
$B = \frac{\mu_0 I}{2 \pi r}$ (directed into or out of the plane depending on current direction).
The magnetic flux linked with this strip is:
$d\phi = B \cdot da = B \cdot l \cdot dr = \frac{\mu_0 I l}{2 \pi r} dr$
The net flux linked with the $ABCD$ loop is:
$\phi = \int_{x_0}^{x} \frac{\mu_0 I l}{2 \pi r} dr = \frac{\mu_0 I l}{2 \pi} [\ln r]_{x_0}^{x} = \frac{\mu_0 I l}{2 \pi} \ln \left( \frac{x}{x_0} \right)$
The induced $emf$ is given by Faraday's law:
$\varepsilon = \left| \frac{d\phi}{dt} \right| = \frac{d}{dt} \left[ \frac{\mu_0 I l}{2 \pi} \ln \left( \frac{x}{x_0} \right) \right] = \frac{\mu_0 l}{2 \pi} \ln \left( \frac{x}{x_0} \right) \frac{dI}{dt}$
Since $\frac{dI}{dt} = \lambda$,the induced $emf$ is:
$\varepsilon = \frac{\mu_0 l \lambda}{2 \pi} \ln \left( \frac{x}{x_0} \right)$
The induced current $I_{ind}$ is:
$I_{ind} = \frac{\varepsilon}{R} = \frac{\mu_0 l \lambda}{2 \pi R} \ln \left( \frac{x}{x_0} \right)$