Predict induced voltage from flux change

Predict voltage from a changing graph

The coil is stationary. Its area and normal stay fixed while an external source changes the applied field. For the displayed interval, Bz(t) = 1 + s t, where field Bz is in teslas, time t in seconds, and s is the signed ramp rate in teslas per second. This equation includes units: the initial term is 1 T and s × t also has units T.

Begin at θ = 0° and s = +1 T/s. Predict the field and linkage at 0 and 2 s, then estimate linkage slope as change divided by elapsed time. The field grows from +1 to +3 T; flux per turn grows from 0.020 to 0.060 Wb. Linkage grows from 0.400 to 1.200 Wb turns. Its slope is (1.200 − 0.400)/2 = +0.400 Wb turns/s.

Faraday’s law turns slope into emf

For the chosen circulation reference, emf = −d(NΦ)/dt. The derivative means the instantaneous slope of the flux-linkage graph. Because this graph is a straight line, the endpoint change divided by time gives that slope exactly within the model. No calculus technique is needed to evaluate this case.

Emf is measured in volts. Despite the historical name “electromotive force,” it is not a mechanical force in newtons. It represents work per unit charge around the circuit. A time-varying magnetic field produces a circulating electric field, so loop emf should not be treated as a single path-independent electrostatic potential difference between two points. The diagram’s voltage scale reports the loop emf with the stated sign reference.

For the default slope, emf = −0.400 V. The minus sign determines a circulation relative to the normal; the next lesson interprets it physically. With fixed area, turns and orientation, emf = −N A cos θ × s. Read each factor: number of equal turns, area per turn, orientation factor and rate of field change.

Capture rates +0.5, +1 and +2 T/s at 0°. Emf is −0.200, −0.400 and −0.800 V. A faster change produces a greater voltage magnitude. Change to −1 T/s and predict the voltage sign before checking. At a fixed positive rate, changing the angle from 0° to 60° halves both linkage slope and emf magnitude.

Do not confuse an intercept with a slope

At zero ramp rate, the field is +1 T and linkage can be nonzero, but the line is horizontal and emf is zero. At s = −1 T/s and θ = 0°, the field goes from +1 to −1 T and passes through zero at t = 1 s. The linkage also passes through zero then, but its slope remains −0.400 Wb turns/s, so emf stays +0.400 V throughout the modeled ramp. An instantaneous zero field does not imply zero induced emf.

The time plots represent 0–2 s of a linear ramp. They do not specify how the source starts or stops it. Do not infer voltage spikes or an instantaneous circuit response at the interval boundaries; switching and inductive transients are not modeled. Changing a slider chooses a different complete ramp rather than changing the source halfway through a run.

Distinguish voltage from current and heat

The browser assumes a closed circuit with total resistance R = 10 Ω and negligible inductive transient effects. Under that approximation, I = emf/R. At the default −0.400 V, current is −0.040 A = −40 mA. The negative sign describes circulation, not a negative amount of heating.

The resistance dissipates P = I²R = 0.016 W = 16 mW. With constant power over 2 s, accumulated heat is P × 2 = 0.032 J = 32 mJ. The graph and ledger are analytic; nothing runs automatically in real time. The total resistance lumps the dissipative circuit into one value. Temperature rise needs thermal properties and is not calculated.

If resistance doubled while the same imposed emf were maintained, current and resistive power would halve. With an open conducting path, there would be no sustained loop conduction current even though an induced electric field could exist. Resistance and open/closed state are not controls here. A real coil also has inductance and a self-generated field: the resistance-only response is an approximation, not a universal switch-on law.

Transfer to another ramp

A stationary 100-turn coil has area 0.0050 m² per turn and normal along +z. The uniform applied field rises from 0.20 to 0.50 T in 0.10 s at a constant rate. Find emf, then the resistance-only current and power for a 30 Ω closed circuit.

Check the Faraday transfer

The rate is (0.50 − 0.20)/0.10 = +3.0 T/s. Emf is −100 × 0.0050 × 3.0 = −1.5 V. Current is −1.5/30 = −0.050 A. Power is (0.050)² × 30 = 0.075 W, and heat over the 0.10 s ramp is 0.0075 J under the same resistance-only approximation. These parameters differ from the browser constants.

Report a slope with units, an emf with a sign convention, and the assumption needed to infer current. Use one capture to refute the claim that voltage is proportional to the instantaneous field value.

Reference for the linkage-rate relationship: OpenStax: Faraday’s law. Calculations, comparisons and transfer tasks are original.