Magnet Through a Coil: Peak EMF Versus Integrated Flux
A student drops a cylindrical bar magnet through a coil of 300 turns and area 1.2 × 10⁻³ m². The coil is connected to a data logger that records the induced emf. The magnet passes through the coil and the graph shows a positive peak of 0.38 V followed by a negative peak of similar size. The time between the two zero crossings is 0.045 s. The student uses ε = N B A / Δt and obtains B = (0.38 × 0.045)/(300 × 1.2 × 10⁻³) = 0.048 T. The student says this is the flux density of the magnet and that the negative peak is 'the same emf but opposite because the magnet has turned around'.
1. Explain why the induced emf has a negative peak after the positive peak, and why the student's statement that the magnet has turned around is incorrect.
Read the worked answer
The coil sees a magnetic flux that first increases as the magnet approaches and then decreases as it leaves. According to Faraday's law, ε = −N dΦ/dt. While the flux is increasing, dΦ/dt is positive (in the chosen sense) and the induced emf has one sign; while the flux is decreasing, dΦ/dt is negative and the emf reverses. Lenz's law explains the direction: the induced current produces a magnetic field that opposes the change. On approach, the coil's field repels the approaching pole; on leaving, it attracts the receding pole, so the current must reverse. The magnet does not turn around; it keeps the same orientation throughout the fall. The negative peak is therefore not 'the same emf but opposite because the magnet has turned around', but the emf produced by the decreasing flux as the magnet leaves. The two peaks are similar in size because the flux changes from zero to a maximum and back to zero over similar times, but their shapes and heights depend on the magnet's speed and position.
2. Explain why ε_peak = N B A / Δt is not a valid way to find B from the graph. Describe the graphical measurement that should be used, and state what it gives.
Read the worked answer
The formula ε_peak = N B A / Δt treats the emf as constant over the whole interval and treats Δt as the time for the flux to change from 0 to BA. Neither is true. The emf is proportional to the instantaneous rate of change of flux, so it peaks when the magnet is near the coil and is small elsewhere. The time between zero crossings is not the time for the flux to go from 0 to BA; it is the time between the flux maximum and the later zero-flux condition (or between two flux extrema), and the peak emf occurs at a different instant. The correct graphical method is to integrate the emf: ∫ε dt = −N ΔΦ. The area under one peak (between its zero crossings) equals N times the flux change from zero to maximum, N B A, if the flux returns to zero at the end of that peak. Measure that area in V s, then B = area/(N A). This gives an effective average flux density over the coil area, not the magnet's surface field. The peak value alone cannot give B without a model of the pulse shape.
3. The data logger's voltage scale has a zero error of +0.02 V. Explain how this would affect the area under the positive peak and the negative peak, and whether it would affect the calculated B.
Read the worked answer
A zero error of +0.02 V adds a constant offset to every recorded voltage. The positive peak is shifted upward by 0.02 V, so the area under the positive peak increases by approximately 0.02 × (width of the positive peak). The negative peak is shifted upward too, i.e. it becomes less negative, so the magnitude of its area decreases by approximately 0.02 × (width of the negative peak). If the student uses only the positive peak area, the calculated flux change and B are overestimated. If the student uses the difference between the magnitudes of the two areas, the zero-error contributions partly cancel when the two peaks have equal widths; the residual error depends on the difference in widths. The zero error also shifts the apparent zero crossings, which changes the integration limits slightly. A good procedure is to subtract the baseline (the reading with no magnet) from all voltages before integrating, or to measure the area between the trace and the baseline. This removes the constant offset and gives a more reliable B.