A long solenoid has a diameter of 12.0 cm.When a current i exists in its windings, a uniform magnetic field of magnitude B = 30.0 mTis produced in its interior. By decreasing i, the field is caused to decrease at the rate of $\mathbf{6}\mathbf{.}\mathbf{50}\frac{\mathbf{mT}}{\mathbf{s}}$.(a) Calculate the magnitude of the induced electric field 2.20 cmfrom the axis of the solenoid.

(b) Calculate the magnitude of the induced electric field 8.20 cmfrom the axis of the solenoid.

$\frac{dB}{dt}=6.50\mathrm{mT}/\mathrm{s}=6.50\times {10}^{-3}\mathrm{T}/\mathrm{s}$

Diameter of solenoid$D=12.0\times {10}^{-2}\mathrm{m}$

Radius of solenoid is$R=6.0\times {10}^{-2}\mathrm{m}$

Use Faraday’s law of electromagnetic induction, which relates line integral of electric field with the rate of change of magnetic flux, and from this, calculate the electric field inside and outside of the solenoid at a radial distance r.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

$$\begin{gathered} d\varphi =\overrightarrow{B}.d\overrightarrow{A} \\ E=\frac{r}{2}\frac{dB}{dt} \end{gathered}$$

Where,$\mathit{\Phi }$is magnetic flux, B is magnetic field, A is area, 𝜀 is emf, is the distance,E is electric field.

Magnitude of electric field insidethe solenoid at distancefrom the axis of solenoid:

By usingthe formula,

$\frac{d\varphi }{dt}=\oint \overrightarrow{E}.d\overrightarrow{s}$

Taking only magnitude,

$$\begin{gathered} E\oint ds=\frac{d}{dt}\left(BA\right)=A.\frac{dB}{dt}=\tau \tau r^2.\frac{dB}{dt} \\ E2\tau \tau r=\frac{d}{dt}\left(BA\right)=A.\frac{dB}{dt}=\tau \tau r^2.\frac{dB}{dt} \\ E=\frac{r}{2}\frac{dB}{dt} \end{gathered}$$

This is the required relation for the field.

$$\begin{gathered} \therefore E=\frac{2.20\times {10}^{-2}}{2}.6.50\times {10}^{-3} \\ E=7.15\times {10}^{-5}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, magnitude of electric field inside the solenoid is $7.15\times {10}^{-5}\mathrm{V}/\mathrm{m}$.

Magnitude of electric field outsidethe solenoid at distancefrom the axis of solenoid:

By usingthe formula,

$\frac{d{\varphi }_B}{dt}=\oint \overrightarrow{E}.d\overrightarrow{s}$

Taking only magnitude,

$$\begin{gathered} E\oint ds=\frac{d}{dt}\left(BA\right)=A.\frac{dB}{dt}=\tau \tau R^2.\frac{dB}{dt} \\ E2\tau \tau r=\frac{d}{dt}\left(BA\right)=A.\frac{dB}{dt}=\tau \tau R^2.\frac{dB}{dt} \\ E=\frac{R^2}{2r}\frac{dB}{dt} \end{gathered}$$

This is the required relation for the field.

$$\begin{gathered} \therefore E=\frac{{(6.0\times {10}^{-2})}^2}{2\times 2.20\times {10}^{-2}}.6.50\times {10}^{-3} \\ E=1.43\times {10}^{-4}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, magnitude of electric field outsidethe solenoid is$1.43\times {10}^{-4}\mathrm{V}/\mathrm{m}$.

Therefore, by using the concept of Faraday’s law and equation, the magnetic field can be determined.