University Physics II · topic UPII-09.06
Forces between parallel currents
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.
Forces Between Parallel Currents: predict a finite change
Two long parallel currents exert equal and opposite forces on each other of magnitude μ₀I₁I₂/(2πd) per unit length, attracting when the currents run the same way and repelling when they oppose. Use the supplied topic model Y_out = 200*abs(i1)/d, where Y_out is B FROM WIRE 1 AT WIRE 2 (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. d is Separation d, measured in mm. Keep i1 (Current I₁, + = left to right) = 16 A; i2 (Current I₂, + = left to right) = 12 A fixed. For this exercise use only 39.5625 mm ≤ d ≤ 110.438 mm 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 d = 55.155 mm and 65.0775 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
B FROM WIRE 1 AT WIRE 2 against d
Predictions from the supplied Forces Between Parallel Currents 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 FROM WIRE 1 AT WIRE 2 / μT
Horizontal axis (x): d: Separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation d / mm | B FROM WIRE 1 AT WIRE 2 / μT |
|---|---|
| 39.5625 | 80.88467614533965 |
| 57.28125 | 55.86470267321331 |
| 75 | 42.666666666666664 |
| 92.71875 | 34.512976070104486 |
| 110.4375 | 28.975664968873797 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 55.155 mm, substitute into Y_out = 200*abs(i1)/d: Y₀ = 58.0183 μT.
- At d = 65.0775 mm, the same fixed-parameter model gives Y₁ = 49.1721 μT.
- Signed change: ΔY_out = Y₁ − Y₀ = -8.84617 μT; the input change is Δd = 9.9225 mm.
- Average sensitivity is ΔY_out/Δd = -0.891526 μT per mm. 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 = 58.0183 μT, Y₁ = 49.1721 μT; ΔY_out = -8.84617 μT; average sensitivity = -0.891526 μT per mm.
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.
Forces Between Parallel Currents: infer an input with uncertainty
Two long parallel currents exert equal and opposite forces on each other of magnitude μ₀I₁I₂/(2πd) per unit length, attracting when the currents run the same way and repelling when they oppose. Use the supplied topic model Y_out = 200*abs(i1)/d, where Y_out is B FROM WIRE 1 AT WIRE 2 (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. d is Separation d, measured in mm. Keep i1 (Current I₁, + = left to right) = 16 A; i2 (Current I₂, + = left to right) = 12 A fixed. For this exercise use only 39.5625 mm ≤ d ≤ 110.438 mm 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 = 38.649093 μT, with a bounded reading uncertainty ±1.451 μT. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0070875 mm, 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 FROM WIRE 1 AT WIRE 2 against d
Predictions from the supplied Forces Between Parallel Currents 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 FROM WIRE 1 AT WIRE 2 / μT
Horizontal axis (x): d: Separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation d / mm | B FROM WIRE 1 AT WIRE 2 / μT |
|---|---|
| 39.5625 | 80.88467614533965 |
| 57.28125 | 55.86470267321331 |
| 75 | 42.666666666666664 |
| 92.71875 | 34.512976070104486 |
| 110.4375 | 28.975664968873797 |
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(d) = 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(39.5625) = 80.8847 μT and Y_out(110.438) = 28.9757 μT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 39.5625 + (38.6491 − 80.8847)(110.438 − 39.5625)/(28.9757 − 80.8847) = 97.2297 mm. 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 d = 82.7963 mm; substituting back gives Y_out = 38.649093 μT.
- The interpolation error in the input is |d_line − d| = 14.4334 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 37.198093 μT and 40.100093 μT. Sort the roots even if the response decreases: 79.8003 mm ≤ d ≤ 86.0259 mm.
- The lower and upper input deviations are 2.99594 mm and 3.22966 mm. 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 d ≈ 82.7963 mm; compatible interval [79.8003, 86.0259] mm. Endpoint interpolation gives 97.2297 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.