Subfield: Electromagnetism


Question: Consider a rectangular loop made of superconducting material with length $\ell=200 \mathrm{~cm}$ and width $w=2 \mathrm{~cm}$. The radius of this particular wire is $r=0.5 \mathrm{~mm}$. This superconducting rectangular loop initially has a current $I_{1}=5 \mathrm{~A}$ in the counterclockwise direction as shown in the figure below. This rectangular loop is situated a distance $d=1 \mathrm{~cm}$ above an infinitely long wire that initially contains no current. Suppose that the current in the infinitely long wire is increased to some current $I_{2}$ such that there is an attractive force $F$ between the rectangular loop and the long wire. Find the maximum possible value of $F$. Write your answer in newtons.
Hint: You may neglect the magnetic field produced by the vertical segments in the rectangular loop.

![](https://cdn.mathpix.com/cropped/2023_12_21_125ad4b445367283c5eag-1.jpg?height=319&width=1214&top_left_y=1689&top_left_x=453)

Solution:
The key idea is that the superconducting loop must have constant flux. If it did not, by Faraday's Law, an emf
$$
\mathcal{E}=-\frac{d \Phi}{d t}
$$

would be generated in the loop. Since superconducting materials have no resistance, this would imply an infinite current, hence a contradiction.

We will first compute the flux through the rectangular loop when there is a current $I_{1}$. Since $w \ll \ell$, we can assume that the vertical segments produce negligible amounts of magnetic field. We can furthermore approximate the field produced by one of the horizontal wires a distance $r$ away as



$\frac{\mu_{0} I}{2 \pi r}$ (this is valid for an infinitely long wire, and therefore is also valid in the regime where $w \ll \ell$ ).

Thus, the total flux through the rectangular loop when there is a current $I_{1}$ is

$$
\Phi_{1}=\int_{r}^{w} B\left(\ell d r^{\prime}\right)=\int_{r}^{w-r}\left(\frac{\mu_{0} I_{1}}{2 \pi r^{\prime}}+\frac{\mu_{0} I_{1}}{2 \pi\left(w-r^{\prime}\right)}\right) \ell d r^{\prime}=\frac{\mu_{0} I_{1} \ell}{\pi} \ln \left(\frac{w}{r}\right)
$$

Note that the self inductance of the loop is $L=\frac{\Phi}{I_{1}}=\frac{\mu_{0} \ell}{\pi} \ln \left(\frac{w}{r}\right)$.

Now, we will determine the flux through the rectangular loop due to the long current-carrying wire. This is

$$
\Phi_{2}=\int_{d}^{d+w} \frac{\mu_{0} I_{2}}{2 \pi r}(\ell d r)=\frac{\mu_{0} I_{2} \ell}{2 \pi} \ln \left(\frac{d+w}{d}\right)
$$

The mutual inductance is $M=\frac{\Phi_{2}}{I_{2}}=\frac{\mu_{0} \ell}{2 \pi} \ln \left(\frac{d+w}{d}\right)$. In to maintain the same flux in the loop, the current will change to $I_{3}$ where

$$
L I_{1}=M I_{2}+L I_{3}
$$

or

$$
I_{3}=I_{1}-\frac{M}{L} I_{2}
$$

Now, we compute the force between the rectangular loop and the long, current-carrying wire. The forces on the vertical sides cancel out because the current in the loop is in opposite directions on these sides. From the horizontal sides, we have the force is

$$
\begin{aligned}
F=\sum\left(I_{3} \vec{\ell} \times \vec{B}\right) & =I_{3} \ell\left(\frac{\mu_{0} I_{2}}{2 \pi d}-\frac{\mu_{0} I_{2}}{2 \pi(d+w)}\right) \\
& =\frac{\mu_{0} \ell w}{2 \pi d(d+w)}\left[I_{2}\left(I_{1}-\frac{M}{L} I_{2}\right)\right] .
\end{aligned}
$$

This quadratic in $I_{2}$ is maximized when $I_{2}=\frac{L}{2 M} I_{1}$ in which case the force becomes

$$
F=\frac{\mu_{0} \ell w}{2 \pi d(d+w)} \frac{L I_{1}^{2}}{4 M}=\frac{\mu_{0} \ell w I_{1}^{2}}{4 \pi d(d+w)} \frac{\ln \left(\frac{w}{r}\right)}{\ln \left(\frac{d+w}{d}\right)}
$$

Note: If the size of the wires is considered when computing flux, a slightly different answer is obtained. In the contest, all answers between $1.11 \times 10^{-3}$ and $1.18 \times 10^{-3}$ were accepted.

Final answer: $1.12 \times 10^{-3}$ N