Question

The magnetic field within a long, straight solenoid with a circular cross section and radius R is increasing at a rate of dB/dt. (a) What is the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with centre on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance r1 from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance r2 from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance r from the axis from r = 0 to r = 2R. (e) What is the magnitude of the induced emf in a circular turn of radius R/2 that has its centre on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is R? (g) What is the induced emf if the radius in part (e) is 2R?

Short answer

(a) the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with centre on the solenoid axis is $\frac{d\varphi B}{dt}=\mathrm{\pi }{r}^2\frac{dB}{dt}$

(b) the magnitude of the induced electric field inside the solenoid, at a distance r1 from its axis is$E=\frac{r_1}{2}\frac{dB}{dt}$

(c) the magnitude of the induced electric field outside the solenoid, at a distance r2 from the axis is$E=\frac{R^2}{2r_2}\frac{dB}{dt}$

(d) Graph the magnitude of the induced electric field as a function of the distance r from the axis from$r=0tor=2R$ is given below:

(e) the magnitude of the induced emf in a circular turn of radiusthat has its centre on the solenoid axis is$\epsilon =\frac{{\mathrm{\pi R}}^2}{4}\frac{dB}{dt}$

(f) is the magnitude of the induced emf if the radius in part e is$\epsilon =\pi R^{2}2\frac{dB}{dt}$

(g) the induced emf if the radius in part (e) is 2R is$\epsilon =\pi R^2\frac{dB}{dt}$

Step-by-step solution

rate of change of flux through a circle of radius r_1

The rate of change of flux is given by:

$$\begin{gathered} \frac{d\varphi B}{dt}=A\frac{dB}{dt} \\ =\pi r_1^2\frac{dB}{dt}\frac{d\varphi B}{dt} \\ =\pi r_1^2\frac{dB}{dt} \end{gathered}$$

Now, according to the faradays law we get:

$$\begin{gathered} \epsilon =\int \overrightarrow{E}\cdot \overline{dl} \\ =-\frac{d\varphi B}{dt} \end{gathered}$$

Now, the magnetic flux is:$$\begin{gathered} \varphi =AB \\ ={\mathrm{\pi r}}_1^2\mathrm{B} \end{gathered}$$

Now, integrating LHS gives:

$\begin{array}{r}\overrightarrow{E}\cdot \overrightarrow{dl}\\ =E\left(2\pi r_1\right)\end{array}$

Therefore:

$E2\pi r_1=\frac{d\left(B\pi r_1^2\right)}{dt}=\pi r_1^2\frac{dB}{dt}$

Or

$E=\frac{r_1}{2}\frac{dB}{dt}$

The direction of the induced electric field is clockwise.

Calculating for the flux for r_2

Since all flux is within r>R then the magnetic field within r_2>R is:

$\varphi =AB=\pi R^{2}B$

Integrating LHS we get:

$$\begin{gathered} \oint \overrightarrow{E}\cdot \left(\overrightarrow{dl}\right)=E\left(2\pi r_2\right) \\ ,E\left(2\pi r_2\right)=\pi R^2\frac{dB}{dt} \end{gathered}$$

So

$E=\frac{R^2}{2r_2}\frac{dB}{dt}$

Graph

To draw the graph of the magnitude of the induced electric field as a function as a function of the distance r from the axis $r=0tor=2r$. Is shown below:

Calculating the rest emfs

To calculate the induced emf in the circular turn radius$r=R/2$. We can use the equation used above with$r1\to R/2$

$$\begin{gathered} \epsilon =\pi (R/2{)}^2\frac{dB}{dt} \\ =\frac{\pi R^2}{4}\frac{dB}{dt} \\ \epsilon =\frac{\pi R^2}{4}\frac{dB}{dt} \end{gathered}$$

Now, the magnitude of the induced emf in a circular turn of radius$r=R.$ we use:

$\epsilon =\frac{\pi R^2}{4}\frac{dB}{dt}$

Finally, we need to calculate the magnitude of the induced emf in a circular turn of radius$r=2R$. All the flux is within$r<R$, therefore:

$$\begin{gathered} \epsilon =\frac{d\varphi B}{dt} \\ =\pi R^2\frac{dB}{dt} \end{gathered}$$

Therefore, the induced emf is$\pi R^2\frac{dB}{dt}$.