Question

A very long solenoid of radius a, with n turns per unit length, carries a current $l_s$. Coaxial with the solenoid, at radius$b>>a$ , is a circular ring of wire, with resistance R. When the current in the solenoid is (gradually) decreased, a current$i_r$ is induced in the ring.

a) Calculate role="math" localid="1657515994158" $l_r$, in terms ofrole="math" localid="1657515947581" $\frac{d{l}_s}{dt}$ .

(b) The power role="math" localid="1657515969938" $\left(l_r^{2}R\right)$delivered to the ring must have come from the solenoid. Confirm this by calculating the Poynting vector just outside the solenoid (the electric field is due to the changing flux in the solenoid; the magnetic field is due to the current in the ring). Integrate over the entire surface of the solenoid, and check that you recover the correct total power.

Short answer

(a) The value of$l_r$ in terms of$\frac{d{l}_s}{dt}$ is$l_r=-\frac{1}{R}\left({\mu }_0\pi a^{2}n\right)\frac{d{i}_s}{dt}$ .

(b) The Poynting vector is$S=-\frac{1}{4}{\mu }_0{l}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\stackrel{⏜}{S}$and the total power is .

$P=l_r^2$

Step-by-step solution

Expression for the induced emf in the ring:

Write the expression for the induced emf in the ring.

$\mathit{\epsilon }\mathbf{=}{\mathit{l}}_{\mathbf{r}}\mathit{R}$…… (1)

Here, $l_r$is the current and R is the resistance.

Determine the current lr in terms of dlsdt :

(a)

Write the expression for the induced emf.

$\epsilon =-\frac{d\varphi }{dt}$…… (2)

Here, $\varphi$is the magnetic flux.

Write the expression for the magnetic flux.

$$\begin{gathered} \varphi =BA \\ \varphi =\pi a^{2}B \end{gathered}$$

Here, B is the magnetic field, and a is the radius.

$B={\mu }_{0}n{l}_s$

Here, ${\mu }_0$is the permeability of free space, n is the number of turns and $l_s$is the current.

Substitute all the known values in equation (2).

$$\begin{gathered} \epsilon =-\frac{d}{dt}\left(\pi a^{2}B\right) \\ \epsilon =-\frac{d}{dt}\left[\left(\pi a^2\right)\left({\mu }_{0}n{l}_s\right)\right] \\ \epsilon =-{\mu }_{0}n\pi a^2\left(\frac{d{l}_s}{dt}\right) \end{gathered}$$

Substitute the known values in equation (1).

localid="1657521563292" $$\begin{gathered} -{\mu }_{0}n\pi a^2\left(\frac{d{l}_s}{dt}\right)=l,R \\ {l}_r=\frac{-{\mu }_{0}n\pi a^2{\displaystyle \frac{d{l}_s}{dt}}}{R} \\ {l}_r=-\frac{1}{R}\left({\mu }_0\pi a^{2}n\right)\frac{d{l}_s}{dt} \end{gathered}$$

Therefore, the value of $l_r$in terms of $\frac{d{l}_s}{dt}$is$l_r=-\frac{1}{R}\left({\mu }_0\pi A^{2}n\right)\frac{d{l}_s}{dt}$.

Determine the delivered power by the Poynting vector:

(b)

Write the expression for the Poynting vector.

$S=\frac{1}{{\mu }_0}(E\times B)$…… (3)

Here, E is the electric field, and B is the magnetic field.

Using Gauss law, write the expression for an electric field.

$\oint E.dl=-\frac{d\varphi }{dt}$

Substitute the known values in the above expression.

$$\begin{gathered} E\left(2\pi a\right)=-{\mu }_{0}n\pi a^2\left(\frac{d{l}_s}{dt}\right) \\ E=-\frac{1}{2}{\mu }_{0}an\left(\frac{d{l}_s}{dt}\right)\hat{\varphi } \end{gathered}$$

Write the expression for a magnetic field.

$B=\frac{{\mu }_{0l_r}}{2}\frac{b^2}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{Z}$

Substitute all the known values in equation (3).

$$\begin{gathered} S=\frac{1}{{\mu }_0}-\frac{1}{2}{\mu }_{0}an\left(\frac{d{l}_s}{dt}\right)\hat{\varphi }\times \frac{{\mu }_0{l}_r}{2}\frac{b^2}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}} \\ S=-\frac{1}{4}{\mu }_0{l}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{S} \end{gathered}$$

Write the expression for the delivered power.

$$\begin{gathered} P=\int S.da \\ P={\int }_{-00}^{00}S\left(2\pi a\right)dz \end{gathered}$$

Substitute the known values in the above expression.

$$\begin{gathered} P=\int -\frac{1}{4}{\mu }_0{L}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2+)}^{{\displaystyle \frac{3}{2}}}}\widehat{S\left(2\pi a\right)dz} \\ P=-\frac{1}{2}\pi {\mu }_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}{\int }_{-00}^{00}\frac{1}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}dz \end{gathered}$$

Take the surface integral of the above equation.

localid="1657521729321" $$\begin{gathered} P=-\frac{1}{2}\pi {\mu }_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}{\int }_{-\infty }^{\infty }d\frac{1}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}} \\ P=-\frac{1}{2}\pi {\mu }_0{a}^2{b}^{2}n{l}_r{\left[\frac{d{l}_s}{dt}\frac{z}{b^2\sqrt{z^2}+b^2}\right]}_{-\infty }^{\infty } \\ P=-\frac{1}{2}\pi {\mu }_0{a}^2{b}^{2}n{l}_r\frac{d{l}_s}{dt}\left(\frac{2}{b^2}\right) \end{gathered}$$

On further solving,

$P=-\left(\pi {\mu }_0{a}^{2}n\frac{d{l}_s}{dt}\right)l_r$

Here, $R{l}_r=-{\mu }_0\pi a^{2}n\frac{d{l}_s}{dt}$.

Hence,

$P=L_r^2$

Therefore, the Poynting vector is $S=-\frac{1}{4}{\mu }_0{l}_r\frac{d{l}_s}{dt}\frac{a{b}^{2}n}{{(b^2+z^2)}^{{\displaystyle \frac{3}{2}}}}\hat{S}$, and the total power is $p=-l_r^{2}R$.