Exercise 5.3.1: Induced electric field of a solenoid

By symmetry $E=E_{\varphi }(\rho )a_{\varphi }$. On a circle of radius $\rho$, with $dS=dS{a}_z$ (out of the page) and $dl$ counter-clockwise,

$|E_{\varphi }(2\mathrm{cm})|=0.01\times 100=1.0V/m$ and $|E_{\varphi }(10\mathrm{cm})|=0.0025/(0.2)\times 100=1.25V/m$. The field is largest at the rim ($2.5V/m$) and decays as $1/\rho$ outside.

$e=-d\Phi /dt=-\pi R^{2}dB/dt=-\pi \times 0.0025\times 100=-0.785\mathrm{V}$, that is $|e|=0.785\mathrm{V}$, the same as $E_{\varphi }2\pi \rho =1.25\times 2\pi \times 0.1$.

Units: $\rho dB/dt$ is $\mathrm{m}\cdot T/s=V/m$. There is an electric field in a region where $B$ is zero: Faraday's law is about the flux enclosed, not the field at the wire. The induced $E$ is not conservative ($\nabla \times E\ne 0$), so no potential can be assigned to it. This is the field that drives the current in every transformer secondary and in every induction-motor rotor.