D.4 · IB Physics
Lenz's law
The induced current always flows so that its magnetic field opposes the change that caused it. Push a magnet towards a copper ring or pull it away and see which pole the ring becomes, or drop magnets down a copper tube and a plastic tube and compare how they fall.
Predict first
Start from the default ring experiment (N pole approaching, closed ring, 0.50 m s⁻¹). If you double the Magnet speed to 1.00 m s⁻¹, what happens to the peak induced current?
The EMF is the rate of change of flux, ε = (dΦ/dz) × v, and the magnet passes through the same positions, so the peak EMF and I = ε ÷ R are proportional to v. The peak current rises from 63 mA to 126 mA.
Check it in the model: Move the Magnet speed, v slider from 0.50 to 1.00 m s⁻¹ and compare the Peak current readout.
N pole S pole copper
Φ = μ₀mA² ÷ [2(A² + z²)^(3/2)]
ring radius A = 3.0 cm, m = 1.0 A m²
F = −(dΦ/dz)² v ÷ R (opposes the motion)
heat = ∫ I²R dt
v_t = mg ÷ b, τ = m ÷ b
v = v_t(1 − e^(−t/τ))
y = v_t[t − τ(1 − e^(−t/τ))]
internal energy Q = mgh − ½mv²
plastic tube: free fall, y = ½gt²
Try this
- Use the preset "N pole approaches" and note the induced pole of the ring. Change "Pole facing the ring" to South. What changes in the readouts and what stays the same?
- Set the N pole moving away from the ring, then tick "Ring has a cut". Compare the force and the EMF readouts with the closed ring.
- Choose the tubes with m = 50 g and b = 1.00 kg s⁻¹. Raise the mass to 100 g and compare the terminal speed, the fall time and the energy that becomes internal energy.
Exam tip: explain Lenz's law in linked steps: the flux linkage changes, so an EMF is induced; in a closed circuit a current flows; its field opposes the change in flux, so the force opposes the motion. Then finish with energy: the work done against that force becomes internal energy in the conductor.