Subfield: Electromagnetism

Part C. Magnetic straws

Consider a cylindrical tube made of a superconducting material. The length of the tube is $l$ and the inner radius is $r$ with $l \gg r$. The centre of the tube coincides with the origin, and its axis coincides with the $z$-axis.

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There is a magnetic flux $\Phi$ through the central cross-section of the tube, $z=0, x^{2}+y^{2}<r^{2}$. A superconductor is a material which expels any magnetic field (the field is zero inside the material).
Context question:
i. Sketch five such magnetic field lines, which pass through the five red dots marked on the axial cross-section of the tube, on the designated diagram on the answer sheet.
Context answer:
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Context question:
ii. Find the tension force $T$ along the $z$-axis in the middle of the tube (i.e. the force by which two halves of the tube, $z>0$ and $z<0$, interact with each other).
Context answer:
\boxed{$T=\frac{\Phi^{2}}{2 \mu_{0} \pi r^{2}}$}


Question: iii. Consider another tube, identical and parallel to the first one.

<img_4385>

The second tube has the same magnetic field but in the opposite direction and its centre is placed at $y=l, x=z=0$ (so that the tubes form opposite sides of a square). Determine the magnetic interaction force $F$ between the two tubes.

Solution:
Let us analyse, what would be the change of the magnetic energy when one of the straws is displaced to a small distance. The magnetic field inside the tubes will remain constant due to the conservation of magnetic flux, but outside, the magnetic field will be changed. The magnetic field outside the straws is defined by the following condition: there is no circulation of $\vec{B}$ (because there are no currents outside the straws); there are no sources of the field lines, other than the endpoints of the straws; each of the endpoints of the straws is a source of streamlines with a fixed magnetic flux $\pm \Phi$. These are exactly the same condition as those which define the electric field of four charges $\pm Q$. We know that if the distance between charges is much larger than the geometrical size of a charge, the charges can be considered as point charges (the electric field near the charges remains almost constant, so that the respective contribution to the change of the overall electric field energy is negligible). Therefore we can conclude that the endpoints of the straws can be considered as magnetic point charges. In order to calculate the force between two magnetic charges (magnetic monopoles), we need to establish the correspondence between magnetic and electric quantities.

For two electric charges $Q$ separated by a distance $a$, the force is $F=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{a^{2}}$, and at the position of one charge, the electric field of the other charge has energy density $w=\frac{1}{32 \pi^{2} \varepsilon_{0}} \frac{Q^{2}}{a^{4}}$; hence we can write $F=8 \pi w a^{2}$. This is a universal expression for the force (for the case when the field lines have the same shape as in the case of two opposite and equal by modulus electric charges) relying only on the energy density, and not related to the nature of the field; so we can apply it to the magnetic



field. Indeed, the force can be calculated as a derivative of the full field energy with respect to a virtual displacement of a field line source (electric or magnetic charge); if the energy densities of two fields are respectively equal at one point, they are equal everywhere, and so are equal the full field energies. As it follows from the Gauss law, for a point source of a fixed magnetic flux $\Phi$ at a distance $a$, the induction $B=\frac{1}{4 \pi} \frac{\Phi}{a^{2}}$. So, the energy density $w=\frac{B^{2}}{2 \mu_{0}}=\frac{1}{32 \pi^{2} \mu_{0}} \frac{\Phi^{2}}{a^{4}}$, hence

$$
F=\frac{1}{4 \pi \mu_{0}} \frac{\Phi^{2}}{a^{2}}
$$

For the two straws, we have four magnetic charges. The longitudinal (along a straw axis) forces cancel out (the diagonally positioned pairs of same-sign-charges push in opposite directions). The normal force is a superposition of the attraction due to the two pairs of opposite charges, $F_{1}=\frac{1}{4 \pi \mu_{0}} \frac{\Phi^{2}}{l^{2}}$, and the repulsive forces of diagonal pairs, $F_{2}=\frac{\sqrt{2}}{8 \pi \mu_{0}} \frac{\Phi^{2}}{2 l^{2}}$. The net attractive force will be

$$
F=2\left(F_{1}-F_{2}\right)=\frac{4-\sqrt{2}}{8 \pi \mu_{0}} \frac{\Phi^{2}}{l^{2}}
$$

Final answer: $F=\frac{4-\sqrt{2}}{8 \pi \mu_{0}} \frac{\Phi^{2}}{l^{2}}$