Answer
$2.12\times10^{-17}\; T$
Work Step by Step
Given: $R=3\;cm$
$\Phi_E=(0.600\;V.m/s)(\frac{r}{R})t$
or, $\frac{d\Phi_{E}}{dt}=(0.600\;V.m/s)(\frac{r}{R})$
Here, $r=5\;cm$, but the electric flux will be enclosed by a maximum radius of $R=3\;cm.$ Therefore, we have to put $r=3\;cm$ in the left side of above equation.
$\therefore\;\frac{d\Phi_{E}}{dt}|_{r=5\;cm}=(0.600\;V.m/s)(\frac{3\;cm}{3\;cm})=0.6\;V.m/s$
Again,
$\oint \vec{B}\cdot d\vec{s}=B\times2\pi r$
$\therefore\;\oint \vec{B}\cdot d\vec{s}|_{r=5\;cm=0.05\;m}=B\times2\pi \times0.05=0.1\pi B$
According to Maxwell's law of induction,
$\oint \vec{B}\cdot d\vec{s}=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$
$\implies B\times2\pi r=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$
Putting the known values at $r=5\;cm$, we obtain
$0.1\pi B=4\pi\times10^{-7}\times8.85\times10^{-12}\times0.6$
or, $B=\frac{4\pi\times10^{-7}\times8.85\times10^{-12}\times0.6}{0.1\pi}\;T$
or, $B=2.12\times10^{-17}\; T$
$\therefore\;$ The magnitude of the induced magnetic field at radial distance $5.00\;cm$ is $2.12\times10^{-17}\; T$
