University Physics II · topic UPII-10.02
Faraday's law
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Faraday's law: predict a finite change
Faraday's law states that the emf induced around a closed loop equals minus the rate at which the magnetic flux through that loop changes, and that a coil of N identical turns gives N times as much. Use the supplied topic model Y_out = 2*pi*f*phi*cos(2*pi*f*tm/1000), where Y_out is SLOPE dΦ/dt (mWb s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Frequency f, measured in Hz. Keep phi (Flux amplitude Φ₀ per turn) = 2 mWb; tm (Instant t on the graph) = 12 ms fixed. For this exercise use only 12.25 Hz ≤ f ≤ 27.3438 Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at f = 15.5706 Hz and 17.6837 Hz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SLOPE dΦ/dt against f
Predictions from the supplied Faraday's law model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SLOPE dΦ/dt / mWb s⁻¹
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | SLOPE dΦ/dt / mWb s⁻¹ |
|---|---|
| 12.25 | 92.81379192132255 |
| 16.0234375 | 71.43326148786548 |
| 19.796875 | 19.421257195326593 |
| 23.5703125 | -60.69053379119238 |
| 27.34375 | -161.97743245892636 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = 15.5706 Hz, substitute into Y_out = 2*pi*f*phi*cos(2*pi*f*tm/1000): Y₀ = 75.6187 mWb s⁻¹.
- At f = 17.6837 Hz, the same fixed-parameter model gives Y₁ = 52.2768 mWb s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -23.3419 mWb s⁻¹; the input change is Δf = 2.11313 Hz.
- Average sensitivity is ΔY_out/Δf = -11.0462 mWb s⁻¹ per Hz. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 75.6187 mWb s⁻¹, Y₁ = 52.2768 mWb s⁻¹; ΔY_out = -23.3419 mWb s⁻¹; average sensitivity = -11.0462 mWb s⁻¹ per Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Faraday's law: infer an input with uncertainty
Faraday's law states that the emf induced around a closed loop equals minus the rate at which the magnetic flux through that loop changes, and that a coil of N identical turns gives N times as much. Use the supplied topic model Y_out = 2*pi*f*phi*cos(2*pi*f*tm/1000), where Y_out is SLOPE dΦ/dt (mWb s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Frequency f, measured in Hz. Keep phi (Flux amplitude Φ₀ per turn) = 2 mWb; tm (Instant t on the graph) = 12 ms fixed. For this exercise use only 12.25 Hz ≤ f ≤ 27.3438 Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -12.678466 mWb s⁻¹, with a bounded reading uncertainty ±15.824 mWb s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00150938 Hz, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
SLOPE dΦ/dt against f
Predictions from the supplied Faraday's law model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SLOPE dΦ/dt / mWb s⁻¹
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | SLOPE dΦ/dt / mWb s⁻¹ |
|---|---|
| 12.25 | 92.81379192132255 |
| 16.0234375 | 71.43326148786548 |
| 19.796875 | 19.421257195326593 |
| 23.5703125 | -60.69053379119238 |
| 27.34375 | -161.97743245892636 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(f) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(12.25) = 92.8138 mWb s⁻¹ and Y_out(27.3438) = -161.977 mWb s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 12.25 + (-12.6785 − 92.8138)(27.3438 − 12.25)/(-161.977 − 92.8138) = 18.4993 Hz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives f = 21.4572 Hz; substituting back gives Y_out = -12.678466 mWb s⁻¹.
- The interpolation error in the input is |f_line − f| = 2.95786 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -28.502466 mWb s⁻¹ and 3.145534 mWb s⁻¹. Sort the roots even if the response decreases: 20.6727 Hz ≤ f ≤ 22.1913 Hz.
- The lower and upper input deviations are 0.784451 Hz and 0.734109 Hz. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 21.4572 Hz; compatible interval [20.6727, 22.1913] Hz. Endpoint interpolation gives 18.4993 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.