University Physics II · topic UPII-12.03
Faraday's law as field circulation
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 as field circulation: predict a finite change
Faraday's law states that the circulation of the electric field round any closed loop fixed in space equals minus the rate of change of the magnetic flux through any surface that loop bounds. Use the supplied topic model Y_out = 5*k*min(rp, rs*rs/max(rp,0.05)), where Y_out is E AT PROBE (mV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. rp is Probe radius r, measured in cm. Keep rs (Solenoid radius R) = 4 cm; k (Field ramp rate dB/dt) = 2 T s⁻¹ fixed. For this exercise use only 4.59125 cm ≤ rp ≤ 16.1938 cm 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 rp = 7.1438 cm and 8.76815 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E AT PROBE against rp
Predictions from the supplied Faraday's law as field circulation 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): E AT PROBE / mV m⁻¹
Horizontal axis (x): rp: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rp: Probe radius r / cm | E AT PROBE / mV m⁻¹ |
|---|---|
| 4.5912500000000005 | 34.84889735910699 |
| 7.491875 | 21.356469508634355 |
| 10.392500000000002 | 15.395718065912916 |
| 13.293125000000002 | 12.03629695801401 |
| 16.19375 | 9.880355075260516 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rp separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rp = 7.1438 cm, substitute into Y_out = 5*k*min(rp, rs*rs/max(rp,0.05)): Y₀ = 22.397 mV m⁻¹.
- At rp = 8.76815 cm, the same fixed-parameter model gives Y₁ = 18.2479 mV m⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.14918 mV m⁻¹; the input change is Δrp = 1.62435 cm.
- Average sensitivity is ΔY_out/Δrp = -2.55436 mV m⁻¹ per cm. 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 = 22.397 mV m⁻¹, Y₁ = 18.2479 mV m⁻¹; ΔY_out = -4.14918 mV m⁻¹; average sensitivity = -2.55436 mV m⁻¹ per cm.
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 as field circulation: infer an input with uncertainty
Faraday's law states that the circulation of the electric field round any closed loop fixed in space equals minus the rate of change of the magnetic flux through any surface that loop bounds. Use the supplied topic model Y_out = 5*k*min(rp, rs*rs/max(rp,0.05)), where Y_out is E AT PROBE (mV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. rp is Probe radius r, measured in cm. Keep rs (Solenoid radius R) = 4 cm; k (Field ramp rate dB/dt) = 2 T s⁻¹ fixed. For this exercise use only 4.59125 cm ≤ rp ≤ 16.1938 cm 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 = 13.711808 mV m⁻¹, with a bounded reading uncertainty ±0.57472 mV m⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00116025 cm, 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.
E AT PROBE against rp
Predictions from the supplied Faraday's law as field circulation 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): E AT PROBE / mV m⁻¹
Horizontal axis (x): rp: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rp: Probe radius r / cm | E AT PROBE / mV m⁻¹ |
|---|---|
| 4.5912500000000005 | 34.84889735910699 |
| 7.491875 | 21.356469508634355 |
| 10.392500000000002 | 15.395718065912916 |
| 13.293125000000002 | 12.03629695801401 |
| 16.19375 | 9.880355075260516 |
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(rp) = 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(4.59125) = 34.8489 mV m⁻¹ and Y_out(16.1938) = 9.88036 mV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rp_line = 4.59125 + (13.7118 − 34.8489)(16.1938 − 4.59125)/(9.88036 − 34.8489) = 14.4133 cm. 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 rp = 11.6688 cm; substituting back gives Y_out = 13.711808 mV m⁻¹.
- The interpolation error in the input is |rp_line − rp| = 2.74456 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 13.137088 mV m⁻¹ and 14.286528 mV m⁻¹. Sort the roots even if the response decreases: 11.1994 cm ≤ rp ≤ 12.1793 cm.
- The lower and upper input deviations are 0.469413 cm and 0.510484 cm. 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 rp ≈ 11.6688 cm; compatible interval [11.1994, 12.1793] cm. Endpoint interpolation gives 14.4133 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.