Question: A doughnut-shaped solenoid winding has dimensions $R=1$ meter, diameter of the loop $=10 \mathrm{~cm}$, and the number of windings $=1000$. If a current of $10$ amperes runs through the wire, what is the magnitude of the force on one loop?
Answer: The constitutive equation for electric fields in a dielectric medium moving with velocity $v$ in a magnetic field $B$ is

$$
\mathbf{D}=k \varepsilon_{0} \mathbf{E}+\varepsilon_{0}(k-1) \mathbf{v} \times \mathbf{B},
$$

where $k$ is its relative dielectric constant. For a point distance $r$ from the axis of rotation, $\mathbf{v}=\boldsymbol{\omega} \times \mathbf{r}$ and $\mathbf{v} \times \mathbf{B}=(\boldsymbol{\omega} \cdot \mathbf{B}) \mathbf{r}-(\mathbf{r} \cdot \mathbf{B}) \boldsymbol{\omega}=\omega B \mathbf{r}$ as $\mathbf{r}$ is perpendicular to $B$. As there are no free charges, Gauss' flux theorem $\oint \mathbf{D} \cdot d \mathbf{S}=0$ gives $\mathbf{D}=0$. Then from $\mathbf{D}=\varepsilon_{0} \mathbf{L}+\mathbf{P}$ we get

$$
\mathbf{P}=-\varepsilon_{0} \mathbf{E}=\varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B \mathbf{r} .
$$

Hence the volume bound charge density is

$$
\rho^{\prime}=-\nabla \cdot \mathbf{P}=-\frac{1}{r} \frac{\partial}{\partial r}\left(r P_{r}\right)=-2 \varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B
$$

and the surface bound charge density is

$$
\sigma^{\prime}=P_{r}=\varepsilon_{0}\left(1-\frac{1}{k}\right) \omega B a,
$$

as $r=a$ for the cylinder's surface.By symmetry and using Ampère's circuital law, we obtain the magnetic induction in a doughnut-shaped solenoid:

$$
B=\frac{\mu_{0} N I}{2 \pi r},
$$

where $r$ is the distance from the center of the doughnut. Consider a small section of length $d l$ of the solenoid. This section contains $\frac{N}{2 \pi R} d l$ turns of the winding, where $R$ is the radius of the doughnut. Take as current element a segment of this section which subtends an angle $d \theta$ at the axis of the solenoid:

$$
\Delta I=\frac{N I d l}{2 \pi R} \rho d \theta,
$$

where $\theta$ is the angle made by the radius from the axis to the segment and the line from the axis to center of the doughnut and $\rho$ is the radius of a loop of winding. The magnetic force on the current element is in the radial direction and has magnitude

$$
\begin{aligned}
d F &=\Delta I \cdot \frac{B}{2}=\frac{N I \rho}{4 \pi R} B d \theta d l \\
&=\frac{\mu_{0} N^{2} I^{2} \rho}{8 \pi^{2} R r} d \theta d l,
\end{aligned}
$$

where $B / 2$ is used, instead of $B$, because the magnetic field established by the current element itself has to be taken out from the total field. Note that $d F$ is perpendicular to the surface of the solenoid and only its component $d F \cdot \cos \theta$ along the line from the axis to the center of the doughnut is not canceled out with another element at $2 \pi-\theta$. As

$$
r=R+\rho \cos \theta,
$$

we have the total force on the doughnut

$$
\begin{aligned}
F &=\int \cos \theta d F \\
&=\frac{\mu_{0} N^{2} I^{2}}{8 \pi^{2} R} \int_{0}^{2 \pi R} d l \int_{0}^{2 \pi} \frac{\rho \cos \theta}{R+\rho \cos \theta} d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{4 \pi} \int_{0}^{2 \pi}\left(1-\frac{R}{R+\rho \cos \theta}\right) d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{4 \pi} \int_{0}^{2 \pi}\left[1-\left(1+\frac{\rho}{R} \cos \theta\right)^{-1}\right] d \theta \\
&=\frac{\mu_{0} N^{2} I^{2}}{2}\left\{1-\left[1-\left(\frac{\rho}{R}\right)^{2}\right]^{-\frac{1}{2}}\right\} \\
&=\frac{4 \pi \times 10^{-7} \times 1000^{2} \times 10^{2}}{2}\left[1-\frac{1}{\sqrt{1-0.05^{2}}}\right] \\
&=-0.079 \mathrm{~N}
\end{aligned}
$$

Hence, the force on one loop is

$$
\frac{F}{N}=-\frac{0.079}{1000}=-7.9 \times 10^{-5} \mathrm{~N}
$$

and points to the center of the doughnut.