A long thin copper wire of the radius 2mm, carries a time-varying current I=t ampere (uniformly distributed), then the induced electric field on its surface is equal to $k\times {10}^{-7}V{m}^{-1}.$ Find the value of k. Take the induced field along the axis of the wire to be zero.

Let the radius of the wire be R. Consider a rectangle abcd in the wire with the side ab along the axis. The magnetic induction at a distance r from the axis (r < R) is $B=\frac{{\mu }_{0}lr}{2\pi R^2}$ The flux through the elementary shaded area within abcd is Bldr, where $\mathcal{l}$ = ab = cd. The flux $\Phi$ through abcd is $\Phi =\underset{0}{\overset{R}{\int }}\frac{{\mu }_{0}Ir}{2\pi R^2}\mathcal{l}dr=\frac{{\mu }_{0}I\mathcal{l}}{4\pi }.$

The induced emf along the curve abcd is  $\left|\epsilon \right|=\frac{d\varphi }{dt}=\frac{{\mu }_0\mathcal{l}}{4\pi }\frac{dI}{dt}=\frac{{\mu }_0\mathcal{l}}{4\pi } \dots \dots \dots \left(i\right)$

The wire being thin, cd >> da. If $E_s$ is the induced electric field at the surface and $E_0$ that along the axis, then $\left|\epsilon \right|=(E_s-E_0)\mathcal{l}.$ Since $E_0=0,$ we obtain from (i) $E_s=\frac{{\mu }_0}{4\pi }={10}^{-7}V{m}^{-1}$

Clearly,  $E_s$ is independent of the radius of the wire.