Why a Magnet Dropped Through a Copper Pipe Falls in Slow Motion

A magnet dropped down a copper pipe drifts down slowly, even though copper is not magnetic. The falling magnet induces currents in the pipe that oppose its motion — the same induction that runs generators, chargers and transformers.

This lesson covers Faraday's and Lenz's laws with motional emf, self and mutual inductance, and the transformer.

**Faraday's law states that the induced emf equals the rate of change of magnetic flux linkage, $\epsilon =-N\frac{d\Phi }{dt}$, and Lenz's law, the minus sign, says the induced current opposes the change that causes it.

Magnetic flux** is $\Phi =BA\cos \theta$, where $\theta$ is the angle between B and the normal to the area. It can change through B, A or $\theta$.

Worked example 1. A 200-turn coil of area 0.010 m${}^2$ sits in a 0.50 T field, which falls to zero in 0.10 s:

$$\epsilon =200\times \frac{0.010\times 0.50}{0.10}=10\text{ V}$$

Motional emf. A rod of length l sliding at speed v across a field B sweeps area $lv$ each second, so $\epsilon =Blv$.

Worked example 2. A 0.50 m rod moves at 4.0 m s${}^{-1}$ through 0.20 T, on rails joined by $2.0\Omega$:

$$\epsilon =0.20\times 0.50\times 4.0=0.40\text{ V}I=0.20\text{ A}F=BIl=0.020\text{ N}$$

Keeping it moving takes $Fv=0.080$ W, exactly the $I^{2}R=0.080$ W dissipated.

An everyday example. The copper pipe slows the magnet because eddy currents in the pipe create a field that pushes back on it.

The substance. Lenz's law is energy conservation — if the induced current helped the change, the magnet would speed up and create energy from nothing.

**Self-inductance L is the flux linkage per unit current in a coil, $L=\frac{{\mu }_0{N}^{2}A}{l}$ for a long solenoid, and mutual inductance between two coaxial solenoids is $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$, where A is the inner solenoid's area.

Self-induction.** A changing current in a coil induces $\epsilon =-L\frac{dI}{dt}$ in the same coil, opposing the change. The energy stored is $U=\frac{1}{2}L{I}^2$.

Mutual induction. A changing current in one coil induces ${\epsilon }_2=-M\frac{d{I}_1}{dt}$ in the other.

Worked example 1. A solenoid 0.20 m long with 500 turns and area $4.0\times 10^{-4}$ m${}^2$:

$$L=\frac{4\pi \times 10^{-7}\times 500^2\times 4.0\times 10^{-4}}{0.20}=6.28\times 10^{-4}\text{ H}$$

If its 2.0 A current falls to zero in 0.010 s, $\epsilon =6.28\times 10^{-4}\times 200=0.126$ V, and the energy released is $\frac{1}{2}\times 6.28\times 10^{-4}\times 2.0^2=1.26\times 10^{-3}$ J.

Worked example 2. A 100-turn coil wound over the same solenoid gives

$$M=\frac{4\pi \times 10^{-7}\times 500\times 100\times 4.0\times 10^{-4}}{0.20}=1.26\times 10^{-4}\text{ H}$$

An everyday example. Switching off a motor can produce a spark at the switch, because the coil's self-inductance drives a large emf as the current collapses.

The substance. Inductance depends only on geometry and core material, not on the current — an iron core multiplies it by ${\mu }_r$.

**A transformer uses mutual induction between two coils on a shared iron core to change AC voltage, with $\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$ for an ideal transformer.

Construction. Primary and secondary coils are wound on a laminated soft-iron core, which links almost all the flux of one coil to the other. A step-up transformer has more secondary turns, a step-down one fewer.

Worked example.** A charger steps 230 V down to 12 V with 1150 primary turns and supplies 2.0 A:

$$N_s=1150\times \frac{12}{230}=60I_p=2.0\times \frac{12}{230}=0.104\text{ A}$$

If the input is 25 W for an output of 24 W, the efficiency is $\frac{24}{25}=0.96$, or 96 per cent.

Energy losses and their fixes:

- Copper loss, $I^{2}R$ heating in the windings — thick copper wire
- Eddy currents in the core — thin insulated laminations
- Hysteresis loss from repeated magnetisation — soft iron
- Flux leakage — coils wound one over the other

An everyday example. Power stations step voltage up before transmission, so the same power flows at a smaller current and wastes less as $I^{2}R$ heat.

The substance. A transformer does not work on steady DC, because a constant current gives no changing flux.