University Physics II · topic UPII-12.04
The Ampere-Maxwell 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.
The Ampere-Maxwell law: predict a finite change
The Ampere-Maxwell law says the circulation of B around a closed loop equals μ₀ times the sum of the conduction current through any surface bounded by that loop and ε₀ times the rate of change of electric flux through that same surface. Use the supplied topic model Y_out = 20*cur*min(rl,rp)/(rp*max(rl,rp)), where Y_out is B ON THE LOOP (μT). 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. rl is Amperean loop radius r, measured in cm. Keep cur (Charging current I) = 1.2 A; rp (Plate radius R) = 4 cm fixed. For this exercise use only 4.1625 cm ≤ rl ≤ 9.9375 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 rl = 5.433 cm and 6.2415 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrl across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
B ON THE LOOP against rl
Predictions from the supplied The Ampere-Maxwell 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): B ON THE LOOP / μT
Horizontal axis (x): rl: Amperean loop radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Amperean loop radius r / cm | B ON THE LOOP / μT |
|---|---|
| 4.1625 | 5.765765765765766 |
| 5.606249999999999 | 4.280936454849499 |
| 7.05 | 3.404255319148936 |
| 8.49375 | 2.82560706401766 |
| 9.9375 | 2.4150943396226414 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rl separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rl = 5.433 cm, substitute into Y_out = 20*cur*min(rl,rp)/(rp*max(rl,rp)): Y₀ = 4.41745 μT.
- At rl = 6.2415 cm, the same fixed-parameter model gives Y₁ = 3.84523 μT.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.572219 μT; the input change is Δrl = 0.8085 cm.
- Average sensitivity is ΔY_out/Δrl = -0.707754 μT 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 = 4.41745 μT, Y₁ = 3.84523 μT; ΔY_out = -0.572219 μT; average sensitivity = -0.707754 μT 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.
The Ampere-Maxwell law: infer an input with uncertainty
The Ampere-Maxwell law says the circulation of B around a closed loop equals μ₀ times the sum of the conduction current through any surface bounded by that loop and ε₀ times the rate of change of electric flux through that same surface. Use the supplied topic model Y_out = 20*cur*min(rl,rp)/(rp*max(rl,rp)), where Y_out is B ON THE LOOP (μT). 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. rl is Amperean loop radius r, measured in cm. Keep cur (Charging current I) = 1.2 A; rp (Plate radius R) = 4 cm fixed. For this exercise use only 4.1625 cm ≤ rl ≤ 9.9375 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 = 3.1228652 μT, with a bounded reading uncertainty ±0.10617 μT. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0005775 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.
B ON THE LOOP against rl
Predictions from the supplied The Ampere-Maxwell 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): B ON THE LOOP / μT
Horizontal axis (x): rl: Amperean loop radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Amperean loop radius r / cm | B ON THE LOOP / μT |
|---|---|
| 4.1625 | 5.765765765765766 |
| 5.606249999999999 | 4.280936454849499 |
| 7.05 | 3.404255319148936 |
| 8.49375 | 2.82560706401766 |
| 9.9375 | 2.4150943396226414 |
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(rl) = 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(4.1625) = 5.76577 μT and Y_out(9.9375) = 2.41509 μT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rl_line = 4.1625 + (3.12287 − 5.76577)(9.9375 − 4.1625)/(2.41509 − 5.76577) = 8.71763 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 rl = 7.68525 cm; substituting back gives Y_out = 3.1228652 μT.
- The interpolation error in the input is |rl_line − rl| = 1.03238 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.0166952 μT and 3.2290352 μT. Sort the roots even if the response decreases: 7.43256 cm ≤ rl ≤ 7.95573 cm.
- The lower and upper input deviations are 0.252689 cm and 0.270476 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 rl ≈ 7.68525 cm; compatible interval [7.43256, 7.95573] cm. Endpoint interpolation gives 8.71763 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.