A point charge q is located at the center of a toroidal coil of rectangular cross section, inner radius a, outer radius a$+W$, and height h, which carries a total of N tightly-wound turns and current I.

(a) Find the electromagnetic momentum p of this configuration, assuming that w and h are both much less than a (so you can ignore the variation of the fields over the cross section).

(b) Now the current in the toroid is turned off, quickly enough that the point charge does not move appreciably as the magnetic field drops to zero. Show that the impulse imparted to q is equal to the momentum originally stored in the electromagnetic fields. [Hint: You might want to refer to Prob. 7.19.]

Write the expression for the electromagnetic momentum of the toroidal configuration.

$p=\int gd\tau$…… (1)

Here, g is the momentum density and $\int d\tau$is the total cross-sectional area of the toroid.

(a)

Write the expression for the momentum density.

$$\begin{gathered} g={\mu }_0{\epsilon }_{0}S \\ g={\epsilon }_0(E\times B) \end{gathered}$$…… (2)

Here, E is the electric field on the toroid due to the charge inside the toroid, and B is the magnetic field inside the toroidal coil.

Write the expression for the electric field on the toroid due to the charge inside the toroid.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\hat{s}$

Here, q is the charge inside the toroid and is the electric permeability in free space.

Write the expression for the magnetic field inside the toroidal coil.

$B=\frac{{\mu }_{0}Nl}{2\pi a}\hat{\phi }$

Here, N is the number of turns, I is the current, and a is the inner radius of the coil.

Substitute the value of E and B in equation (2).

$$\begin{gathered} g={\epsilon }_0\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\hat{s}\times \frac{{\mu }_{0}Nl}{2\pi a}\hat{\varphi }\right) \\ g={\epsilon }_0\left(\frac{q}{4\pi {\epsilon }_0{a}^2}\right)\left(\frac{{\mu }_{0}Nl}{2\pi a}\right)\left(\hat{s}\times \hat{\varphi }\right) \\ g=\frac{{\mu }_0}{8{\pi }^2}\frac{qNl}{a^3}\hat{z} \end{gathered}$$

Write the expression for the total cross-sectional area of the toroid.

$\int d\tau =2\pi \alpha wh$

Here, a is the radius, w is the width and h is the height.

Substitute the value of g and$\int d\tau$ in equation (1).

$$\begin{gathered} p=\left(\int \frac{{\mu }_0}{8{\pi }^2}\frac{qNl}{a^3}\right)d\tau \\ p=\left(\int \frac{{\mu }_0}{8{\pi }^2}\frac{qNl}{a^3}\right)\left(2\pi awh\right) \\ p=\int \frac{{\mu }_0}{4{\pi }^{}}\frac{qNlwh}{a^3}\hat{z} \end{gathered}$$

Therefore, the required electromagnetic momentum of the toroidal configuration is $p=\frac{{\mu }_0}{4\pi }\frac{qNlwh}{a^2}\hat{z}$.

(b)

Write the expression for the impulse derived to the charge q.

$l=\int Fdt$…… (3)

Here, F is the electric force.

Write the expression for the electric force.

F=qE …… (4)

Write the expression for the electric field at a point z above the center of the toroid.

$E=-\left(\frac{{\mu }_0}{4\pi }\right)\frac{Nhwka}{{\left(a^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}\hat{z}$…… (5)

At the center of the toroid, the value of z will be zero, and the value of k will be,

\

$k=\frac{dl}{dt}$

Substitute the value of kand in equation (5).

localid="1657534722299" $E=\left(-\frac{{\mu }_0}{4\pi }\right)\frac{Nhwa}{{\left(a^2+{\left(0\right)}^2\right)}^{{\displaystyle \frac{3}{2}}}}\left(\frac{dl}{dt}\right)\hat{z}$

Substitute the value of E in equation (4).

$$\begin{gathered} E=q\left(-\frac{{\mu }_0}{4{\pi }^2}\right)Nhwa\left(\frac{dl}{dt}\right)\hat{z} \\ F=\left(-\frac{{\mu }_0}{4{\pi }^2}\right)Nhwa\left(\frac{dl}{dt}\right)\hat{z} \end{gathered}$$

Substitute the value of F in equation (3).

$$\begin{gathered} l=\int \left(\left(-\frac{{\mu }_{0}q}{4\pi a^2}\right)Nhw\frac{dl}{dt}\hat{z}\right)dt \\ l=\int \left(-\frac{{\mu }_{0}q}{4\pi a^2}\right)Nhw\hat{z}\frac{dl}{dt}dt \\ l=\int \left(-\frac{{\mu }_{0}q}{4\pi a^2}\right)Nhw\hat{z}\frac{dl}{dt}\int dl \\ l=\frac{{\mu }_0}{4\pi }\frac{qNlwh}{a^2}\hat{z} \end{gathered}$$

On comparing the above obtained result with part (a), the impulse imparted to charge q will be equal to the momentum originally stored in the electromagnetic fields.

Therefore, the impulse imparted to charge q is equal to the momentum originally stored in the electromagnetic fields is localid="1657536086084" $l=\frac{{\mu }_0}{4\pi }\frac{qNlwh}{a^2}\hat{z}$.