91影视

$$\begin{gathered} R=3.00\mathrm{cm}=0.003\mathrm{m} \\ \mathrm{蠒}=\left(0.600\mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{\mathrm{r}}{\mathrm{R}}\mathrm{t} \end{gathered}$$

For a non-uniform electric field, we use equation 32-3 for finding the magnetic field inside the circle and outside the circle. If the magnetic induction varies in magnitude and direction at different points in a region, the magnetic field is said to be non-uniform. The magnetic field due to a bar magnet is non-uniform.

The formula is as follows:

$鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{s}={渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}$

Where, $\stackrel{鈫�}{B}$is the magnetic field, $蠒$ is the flux.

From the formula, for magnetic field inside the circle, i.e., $r=002$as

$鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{s}={渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}$

localid="1663150013317" $\begin{array}{rcl}鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{s}& =& {渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}\\ B{鈭�}_0^{2蟺}rd胃& =& {渭}_0{蔚}_0\frac{d蠒}{dt}\\ B2蟺r& =& {渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}\end{array}$

From the given, localid="1663150045833" $\frac{蠒}{t}=\left(0.600\mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{r}{R}$.

So, by putting the value,

$\begin{array}{rcl}B& =& \frac{{渭}_0{\mathcal{E}}_0}{2蟺r}\left(0.600\right)\frac{r}{R}\\ B& =& \frac{{渭}_0{\mathcal{E}}_0}{2蟺}\left(0.600\right)\frac{1}{R}\\ B& =& \frac{\left(2脳{10}^{-7}\right)\left(8.85脳{10}^{-12}\right)\left(0.600\right)}{0.03}\\ B& =& 3.54脳{10}^{-17}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a radial distance 2.00 cm is$B=3.54脳{10}^{-17}\mathrm{T}.$

As $r>R,$i.e.,$r=0.05\mathrm{m},$then the$\frac{r}{R}$is taken as unity, so the equation is

$\begin{array}{rcl}鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{s}& =& {渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}\\ B{鈭�}_0^{2蟺}rd胃& =& {渭}_0{蔚}_0\frac{d蠒}{dt}\\ B2蟺r& =& {渭}_0{\mathcal{E}}_0\frac{d蠒}{dt}\end{array}$

From the given, $\frac{蠒}{t}=\left(0.600\mathrm{V}.\frac{\mathrm{m}}{\mathrm{s}}\right)\frac{r}{R}$.

So, by putting the value,

$\begin{array}{rcl}B& =& \frac{{渭}_0{\mathcal{E}}_0}{2蟺r}\left(0.600\right)\\ B& =& \frac{{渭}_0{\mathcal{E}}_0}{2蟺r}\left(0.600\right)\\ B& =& \frac{\left(4蟺脳{10}^{-7}\right)\left(8.85脳{10}^{-12}\right)\left(0.600\right)}{2蟺r}\\ B& =& \frac{\left(2脳{10}^{-7}\right)\left(8.85脳{10}^{-12}\right)\left(0.600\right)}{0.05}\\ B& =& 2.13脳{10}^{-17}\mathrm{T}\\ & & \end{array}$
Therefore, the magnitude of an induced magnetic field at a radial distance $5.00\mathrm{cm}$is $B=2.13脳{10}^{-17}\mathrm{T}.$

By using the concept of magnetic field for the non-uniform electric field, the magnetic field inside and outside the circle can be found.