91Ó°ÊÓ

$\begin{array}{rcl}R& =& 3.00\mathrm{cm}=0.03\mathrm{m}\\ \mathrm{r}& =& 0.02\mathrm{m}\\ \mathrm{E}& =& \left(0.500\frac{\mathrm{V}}{\mathrm{m}.\mathrm{s}}\right)\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)\mathrm{t}\\ & & \end{array}$

For a non-uniform electric field, first, find the electric flux for the region inside and outside the circular the region. Then calculate the magnetic field by using the Maxwell equation for a non-uniform electric field.

Formulae are as follows:

$âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}={\mathrm{μ}}_0{\mathcal{E}}_0\frac{d\mathrm{Ï•}}{dt}$

Where,$\stackrel{→}{B}$is the magnetic field, $\mathrm{ϕ}$is the flux.

By using the formula, find the magnetic field inside the circle as,

$âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}={\mathrm{μ}}_0{\mathcal{E}}_0\frac{d{\mathrm{Ï•}}_E}{dt}…………………………………………………………………………\left(1\right)$

Where, $\mathrm{Ï•}$electric flux for a non-uniform field can be defined as,

localid="1663156293618" $\begin{array}{rcl}{\mathrm{ϕ}}_E& =& ∫EdA\\ {\mathrm{ϕ}}_E& =& {∫}_0^{r}EdA\\ {\mathrm{ϕ}}_E& =& {∫}_0^{r}E\left(2\mathrm{π}rdr\right)\end{array}$

By the given value,

localid="1663156368547" $\begin{array}{rcl}{\mathrm{ϕ}}_E& =& {∫}_0^r\left(0.500\right)\left(1-\frac{r}{R}\right)t2\mathrm{π}rdr\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t2\mathrm{π}{∫}_0^r\left(1-\frac{r}{R}\right)rdr\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t.2\mathrm{π}\left\{{{\left[\frac{r^2}{2}\right]}_0}^r-{\left[\frac{r^3}{3R}\right]}_0^r\right\}\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t.2\mathrm{π}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ & & \end{array}$

Simplify further.

localid="1663156414306" $\begin{array}{rcl}{\mathrm{Ï•}}_E& =& t.\mathrm{Ï€}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ \frac{{\mathrm{Ï•}}_E}{t}& =& \mathrm{Ï€}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\end{array}$

Then equation (1) can be written as,

localid="1663156492531" $\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}& =& {\mathrm{μ}}_0{\mathcal{E}}_0\mathrm{Ï€}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B\left(2\mathrm{Ï€}r\right)& =& {\mathrm{μ}}_0{\mathcal{E}}_0\mathrm{Ï€}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B& =& \frac{{\mathrm{μ}}_0{\mathcal{E}}_0}{2r}\left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B& =& \frac{{\mathrm{μ}}_0{\mathcal{E}}_0}{2}\left(\frac{r}{2}-\frac{r^2}{3R}\right)\\ & & \end{array}$

By substituting the given value,

localid="1663156562796" $\begin{array}{rcl}\mathrm{B}& =& \frac{\left(4\mathrm{π}×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(\frac{\mathrm{r}}{2}-\frac{{\mathrm{r}}^2}{3\mathrm{R}}\right)\\ \mathrm{B}& =& \frac{\left(4\mathrm{π}×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(\frac{0.02}{2}-\frac{{\left(0.02\right)}^2}{3×0.03}\right)\\ \mathrm{B}& =& 5.56×{10}^{-18}×\left(5.6×{10}^{-3}\right)\\ \mathrm{B}& =& 3.09×{10}^{-20}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a given radial distance islocalid="1663156604486" $B=3.09×{10}^{-20}\mathrm{T}.$

For the r> R in the above equation, the limit of integration is 0 to R, i.e.,

For $r=0.05\mathrm{m}$,

$\begin{array}{rcl}{\mathrm{ϕ}}_E& =& ∫EdA\\ {\mathrm{ϕ}}_E& =& {∫}_0^{R}EdA\\ {\mathrm{ϕ}}_E& =& {∫}_0^{R}E\left(2\mathrm{π}rdr\right)\end{array}$

By the given value,

localid="1663157413775" $\begin{array}{rcl}{\mathrm{ϕ}}_E& =& {∫}_0^R\left(0.500\right)\left(1-\frac{r}{R}\right)t2\mathrm{π}rdr\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t2\mathrm{π}{∫}_0^R\left(1-\frac{r}{R}\right)rdr\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t.2\mathrm{π}\left\{{{\left[\frac{r^2}{2}\right]}_0}^R-{\left[\frac{r^3}{3R}\right]}_0^R\right\}\\ {\mathrm{ϕ}}_E& =& \left(0.500\right)t.2\mathrm{π}\left(\frac{R^2}{2}-\frac{R^3}{3R}\right)\end{array}$

Simplify further.

localid="1663157442965" $\begin{array}{rcl}{\mathrm{Ï•}}_E& =& t.\mathrm{Ï€}\left(\frac{R^2}{2}-\frac{R^2}{3}\right)\\ \frac{{\mathrm{Ï•}}_E}{t}& =& \mathrm{Ï€}\left(\frac{R^2}{6}\right)\end{array}$

Then equation (1) can be written as,

localid="1663157466171" $\begin{array}{rcl}âˆ\circledR \stackrel{→}{B}·d\stackrel{→}{s}& =& {\mathrm{μ}}_0{\mathcal{E}}_0\mathrm{Ï€}\left(\frac{R^2}{6}\right)\\ B\left(2\mathrm{Ï€}r\right)& =& {\mathrm{μ}}_0{\mathcal{E}}_0\mathrm{Ï€}\left(\frac{R^2}{6}\right)\\ B& =& \frac{{\mathrm{μ}}_0{\mathcal{E}}_0}{2r}\left(\frac{R^2}{6}\right)\end{array}$

By substituting the given value,

localid="1663157500012" $\begin{array}{rcl}B& =& \frac{\left(4\mathrm{π}×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2×0.05}\left(\frac{{0.03}^2}{6}\right)\\ B& =& \frac{\left(4\mathrm{π}×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(1.5×{10}^{-4}\right)\\ B& =& 1.12×{10}^{-16}×\left(1.5×{10}^{-4}\right)\\ B& =& 1.67×{10}^{-20}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a given radial distance is $\begin{array}{rcl}B& =& 1.67×{10}^{-20}\mathrm{T}\end{array}$.

By using the relation between the non-uniform electric field and electric flux, the magnetic field for outside and inside the given circular region can be found.