91Ó°ÊÓ

Radius of circle$r=1.5\text{m}$

Electric field strength$E=4.0\text{mV}/\text{m}$

$\frac{4.0\text{mV}}{1}×\frac{{10}^{−3}\text{V}}{1\text{mV}}=4.0×{10}^{−3}\text{V}$

The induced electric field in a region where the radius of circle is less than radius of circular region the charge is confined is given by

$E=\frac{r}{2}\left|\frac{dB}{dt}\right|$

Here $\frac{dB}{dt}$is the rate of changing magnetic field, $r$is the radius of the solenoid.

Therefore, the Rate of changing of magnetic field is given by

$\frac{dB}{dt}=\frac{2E}{r}$

The rate of changing of magnetic field is given by

$\frac{dB}{dt}=\frac{2E}{r}$

Substituting the values of radius , electric field strength (E) in above equation, the rate of change of magnetic field is calculated as

$\begin{array}{l}\frac{dB}{dt}=\frac{2\left(4×{10}^{−3}\text{V}/\text{m}\right)}{(1.5\text{m})}\\ \frac{dB}{dt}=5.3\text{T}/\text{s}\end{array}$

The values of rate of electric field strength and radius given in the question are used to calculate the changing magnetic field using the formula $\frac{dB}{dt}=\frac{2E}{r}$

The rate at which the magnetic field is changing is$5.3\text{T}/\text{s}$