University Physics II · topic UPII-10.07
Induced electric fields
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.
Induced electric fields: predict a finite change
An induced electric field is the electric field produced by a changing magnetic flux rather than by charges: its lines form closed loops, and the work it does on a charge carried once round such a loop is not zero. Use the supplied topic model Y_out = 5*rate*min(rprobe, rsol^2/max(rprobe,0.05)), where Y_out is E AT THE 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. rprobe is Probe radius r, measured in cm. Keep rate (dB/dt inside the coil) = 1 T s⁻¹; rsol (Solenoid radius R) = 4 cm fixed. For this exercise use only 4.2525 cm ≤ rprobe ≤ 8.99062 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 rprobe = 5.29489 cm and 5.95823 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrprobe across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E AT THE PROBE against rprobe
Predictions from the supplied Induced electric fields 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 THE PROBE / mV m⁻¹
Horizontal axis (x): rprobe: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rprobe: Probe radius r / cm | E AT THE PROBE / mV m⁻¹ |
|---|---|
| 4.2525 | 18.812463256907698 |
| 5.43703125 | 14.713912118860822 |
| 6.6215625 | 12.081740525744491 |
| 7.80609375 | 10.248403691026642 |
| 8.990625 | 8.898157803267292 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rprobe separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rprobe = 5.29489 cm, substitute into Y_out = 5*rate*min(rprobe, rsol^2/max(rprobe,0.05)): Y₀ = 15.1089 mV m⁻¹.
- At rprobe = 5.95823 cm, the same fixed-parameter model gives Y₁ = 13.4268 mV m⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.6821 mV m⁻¹; the input change is Δrprobe = 0.663338 cm.
- Average sensitivity is ΔY_out/Δrprobe = -2.53581 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 = 15.1089 mV m⁻¹, Y₁ = 13.4268 mV m⁻¹; ΔY_out = -1.6821 mV m⁻¹; average sensitivity = -2.53581 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.
Induced electric fields: infer an input with uncertainty
An induced electric field is the electric field produced by a changing magnetic flux rather than by charges: its lines form closed loops, and the work it does on a charge carried once round such a loop is not zero. Use the supplied topic model Y_out = 5*rate*min(rprobe, rsol^2/max(rprobe,0.05)), where Y_out is E AT THE 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. rprobe is Probe radius r, measured in cm. Keep rate (dB/dt inside the coil) = 1 T s⁻¹; rsol (Solenoid radius R) = 4 cm fixed. For this exercise use only 4.2525 cm ≤ rprobe ≤ 8.99062 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 = 11.200158 mV m⁻¹, with a bounded reading uncertainty ±0.3453 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 rprobe by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000473812 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 THE PROBE against rprobe
Predictions from the supplied Induced electric fields 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 THE PROBE / mV m⁻¹
Horizontal axis (x): rprobe: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rprobe: Probe radius r / cm | E AT THE PROBE / mV m⁻¹ |
|---|---|
| 4.2525 | 18.812463256907698 |
| 5.43703125 | 14.713912118860822 |
| 6.6215625 | 12.081740525744491 |
| 7.80609375 | 10.248403691026642 |
| 8.990625 | 8.898157803267292 |
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(rprobe) = 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(4.2525) = 18.8125 mV m⁻¹ and Y_out(8.99062) = 8.89816 mV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rprobe_line = 4.2525 + (11.2002 − 18.8125)(8.99062 − 4.2525)/(8.89816 − 18.8125) = 7.89048 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 rprobe = 7.14276 cm; substituting back gives Y_out = 11.200158 mV m⁻¹.
- The interpolation error in the input is |rprobe_line − rprobe| = 0.747725 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.854858 mV m⁻¹ and 11.545458 mV m⁻¹. Sort the roots even if the response decreases: 6.92913 cm ≤ rprobe ≤ 7.36997 cm.
- The lower and upper input deviations are 0.213625 cm and 0.227216 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 rprobe ≈ 7.14276 cm; compatible interval [6.92913, 7.36997] cm. Endpoint interpolation gives 7.89048 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.