Equation 30.26 is an expression for the induced electric field CALC inside a solenoid$(r<R)$ . Find an expression for the induced electric field outside a solenoid$(r>R)$ in which the magnetic field is changing at the rate$dB/dt$ .

A solenoid of radius R, whose length is larger than its radius, and carries a current of I, is considered.

R Radius of the solenoid

The magnetic field of the solenoid is uniform and is perpendicular to the plane and denoted by the ' x' in the figure.

The induced electric field , $\overrightarrow{E}$is depicted by the circles around the solenoid.$\overrightarrow{E}$ is maximum inside the solenoid and decrease with the distance outside the solenoid.

The radius of the solenoid$=R$

The distance of the point outside the solenoid$=r$

The induced Emf in a solenoid is given by the Faraday's Law

$\oint \overrightarrow{E}\cdot \overrightarrow{dr}=-\frac{d{\varphi }_B}{dt}$

Here$\overrightarrow{E}$ is the electric field, $\frac{d{\varphi }_B}{dt}$is the rate change of the flux.

The emf induced outside the solenoid at distance ' r ' from the center of the solenoid.

$r>R$

The magnetic flux ${\varphi }_B=B\cdot A$

Where A is the area of the circle of radius r

Then, from the faraday's law

$\oint \overrightarrow{E}\cdot \overrightarrow{dr}=-\frac{d(B\cdot A)}{dt}$

$\overrightarrow{E}\oint \overrightarrow{dr}=-A\frac{dB}{dt}\overrightarrow{E}\left(2\pi r\right)=-A\frac{dB}{dt}$$(\therefore$The length of the closed cure$\oint \overrightarrow{dr}=2\pi r)$

$\overrightarrow{E}\left(2\pi r\right)=-\pi R^2\frac{dB}{dt}\overrightarrow{E}\times 2r=-R^2\frac{dB}{dt}\overrightarrow{E}=\frac{R^2}{2r}\left|\frac{dB}{dt}\right|$

Conclusion:

The induced electric field at a distance $(r>R)$outside a solenoid $A$of radius is r given by

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

The expression shows that the induced electric field is due to the changing magnetic field and it is decreased as outside the solenoid. r