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The expression for magnetic field due to a solenoid at an inside point is,

$B=\frac{{渭}_{0}NI}{l}$

The expression for induced electric field in the solenoid.

$\begin{array}{c}E=\frac{r}{2}\left|\frac{dB}{dt}\right|\\ =\frac{r}{2}\left|\frac{d}{dt}\left(\frac{{渭}_{0}NI}{\mathcal{l}}\right)\right|\\ =\frac{{渭}_{0}Nr}{2\mathcal{l}}\end{array}$

Here, ${渭}_0$is permeability of free space, $N$is number of turns, $I$is the current in the solenoid, and $\mathcal{l}$is length of the solenoid..

Substituting values in the above expression

$\begin{array}{c}\frac{{渭}_{0}Nr}{2\mathcal{l}}=\frac{\left(4蟺脳{10}^{鈭�7}\text{T}鈰�\text{m}鈰�{\text{A}}^{鈭�1}\right)\left(400\right)(1\text{cm})\left(\frac{{10}^{鈭�2}\text{m}}{1\text{cm}}\right)}{2(20\text{cm})\left(\frac{{10}^{鈭�2}\text{m}}{1\text{cm}}\right)}\\ =1.25脳{10}^{鈭�5}\text{T}鈰�\text{m}鈰�{\text{A}}^{鈭�1}\end{array}$

In the interval $0<t<0.1\text{s}$, current increases from 0 to 5A. The value of $\frac{{渭}_{0}Nr}{2l}$is .

The electric field is

$\begin{array}{c}E=\left(\frac{{渭}_{0}Nr}{2l}\right)\left(\frac{dI}{dt}\right)\\ =\left(1.25脳{10}^{鈭�5}\text{T}鈰�\text{m}鈰�{\text{A}}^{鈭�1}\right)\left(\frac{5\text{A}}{0.1\text{s}}\right)\\ =6.25脳{10}^{鈭�4}\text{N}/\text{m}\end{array}$

In the interval $0.1\text{s}<t<0.2\text{鈥剆}$, current remains constant and hence $dI=0$. So, the electric field strength is zero $E=0$.

In the interval, current decreases from to 0 . the value of is .

The electric field is

.$\begin{array}{c}E=\left(\frac{{渭}_{0}Nr}{2l}\right)\left(\frac{dI}{dt}\right)\\ =\left(1.25脳{10}^{鈭�5}\text{T}鈰�\text{m}鈰�{\text{A}}^{鈭�1}\right)\left(\frac{5\text{A}}{0.2\text{s}}\right)\\ =3.125脳{10}^{鈭�4}\text{N}/\text{m}\end{array}$