Consider a strip of length $l$ and width dr at a distance $\mathrm{r}$ from the surface of infinite long current carrying wire. The magnetic field at strip due to current carrying wire is
Field $B(r)=\frac{\mu_0 I}{2 \pi r}$
Where, $\mathrm{B}(\mathrm{r})$ is perpendicular to the paper.
So, total flux through the strip is
$$
\phi=\frac{\mu_0 \mathrm{I}}{2 \pi} l \int_{x_0}^{\mathrm{x}} \frac{\mathrm{dr}}{\mathrm{r}}=\frac{\mu_0 \mathrm{I}}{2 \pi} \ln \left[\frac{\mathrm{x}}{\mathrm{x}_0}\right] \ldots \text { (i) }
$$
The induced emf induced can be obtained by differentiating eq. (i) and then applying Ohm's law
$$
\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}, \Rightarrow \mathrm{IR}=\frac{\mathrm{d} \phi}{\mathrm{dt}} \quad\left(\therefore \frac{\varepsilon}{\mathrm{R}}=\mathrm{I}\right)
$$
We have, induced current
$$
\begin{aligned}
\mathrm{I}=\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}} &=\frac{1}{\mathrm{R}} \frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\mu_0 \mathrm{Il}}{2 \pi} \ln \left(\frac{\mathrm{x}}{\mathrm{x}_0}\right)\right] \\
&=\frac{\mu_0 l}{2 \pi \mathrm{R}} \ln \left(\frac{\mathrm{x}}{\mathrm{x}_0}\right) \cdot\left[\frac{\mathrm{dI}}{\mathrm{dt}}\right]
\end{aligned}
$$

$$
\mathrm{I}=\frac{\mu_0 l}{2 \pi} \frac{\lambda}{\mathrm{R}} \ln \frac{\mathrm{x}}{\mathrm{x}_0}\left[\because \frac{\mathrm{dI}}{\mathrm{dt}}=\lambda \text { (given) }\right]
$$