# How may the magnetic resistance for a magnet moving through a copper pipe be calculated?

When a magnet moves past non-magnetic, electrically conducting material such as copper, the changing magnetic field induces electric eddy currents in the copper, which create their own magnetic field that repels the magnet's movement past the copper. A magnet dropped through a copper pipe will descend slower than free-fall. If the magnet is strong enough the descent may be very gradual. How may the magnetic resistance of a magnet moving through a copper pipe be calculated?

[Bonus] What would happen if the copper pipe were cut lenghtwise so as to make a nearly closed C instead of an O? Would the eddy currents still form and repel the magnet?

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This question seemed relevant: physics.stackexchange.com/questions/3344/… – Dale Oct 19 '11 at 6:16

Ok, i consider a magnet, falling inside a long, vertical cylindrical hole(with radius $R$), drilled in a non-magnetic conductor. For simplicity, i assume that the magnetic field $\overrightarrow B$ of the magnet is so strong that we can ignore induced magnetic fields by eddy currents. This assumes also that terminal velocity of the magnet is moderate(i.e. the magnet is light in weight). Friction forces with the exception caused by the eddy currents are supposed to be negligible. I am using CGS system.

External magnetic field produced by the magnet:

$$\overrightarrow B=3\frac{\overrightarrow{r'}(\overrightarrow P_m\cdot \overrightarrow{r'})}{r'^5}-\frac{\overrightarrow P_m }{r'^3}$$ Here $\overrightarrow P_m$ is magnetic dipole moment of the magnet, $\overrightarrow{r'}$ is radius vector of a point where $\overrightarrow B$ is calculated.

Let the axis of the cylindrical hole be the z-axis, pointing down. We place the center of the magnetic dipole at z=0 and assume that $\overrightarrow P_m$ is always parallel to the z-axis. Let $v_t$ be terminal velocity of the falling magnet. Then we can find $v_t$ from the formula:

$$W=F\cdot v_t=mg\cdot v_t$$ where $W$ is power, generated by eddy currents in the conductor, $mg$ is the weight of the magnet.
So we need to find $W$.

Consider a thin wall cylinder(concentric with the hole), with radius $r>R$, wall thickness $dr$, in the conductor. In this cylinder we consider a ring, with the cross-sectional area $dz\cdot dr$.
At this ring let $B_n$ be a coponent of $\overrightarrow B$, perpendicular to the wall of the cylinder. Then the magnitude of the electromotive force(emf) induced in the ring is:

$$E=\frac{v_t}{c}B_n2\pi r$$ Conductance of the ring:

$$dg=\frac{\gamma drdz}{2\pi r}$$ where $\gamma$ is the electrical conductivity of the conductor. Thus, the current induced in the ring:

$$I=Edg=\frac{v_t}{c}B_n\gamma drdz$$ Power, released in the ring:

$$dW=\frac{I^2}{dg}=E^2dg=\frac{v_t^2}{c^2}\gamma B_n^22\pi r drdz$$ The only thing that has yet to be found is $B_n$. It comes from the expression of $\overrightarrow B$ above. Because this is a pure math i skip this step and write down the final formula for $dW$:

$$dW=18\pi\frac{v_t^2}{c^2}\gamma P_m^2\frac{z^2r^3drdz}{(z^2+r^2)^5}$$ Now, to get the whole power released in the conductor we need to integrate throughout the conductor:
$$W=18\pi\frac{v_t^2}{c^2}\gamma P_m^2 \int_{-\infty }^{+\infty } \int_{R}^{+\infty } \frac{z^2r^3drdz}{(z^2+r^2)^5}=18\pi\frac{v_t^2}{c^2}\gamma P_m^2\frac{5\pi}{384R^3}$$ So, terminal velocity of the falling magnet:

$$v_t=\frac{64}{15\pi^2}\frac{mgc^2}{\gamma P_m^2}R^3$$

If you want to consider only a pipe then integrate $dW$ only by $z$.

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