University Physics II · topic UPII-09.02
Field of a long straight wire
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.
Field of a long straight wire: predict a finite change
The field of a long straight wire is azimuthal: at every point outside the conductor it is tangent to the circle through that point centred on the wire, with magnitude μ₀I/2πr, where r is the perpendicular distance from the axis. Use the supplied topic model Y_out = 20*i/d, where Y_out is FIELD AT THE PROBE (µT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Probe distance d, measured in cm. Keep i (Current I) = 12 A fixed. For this exercise use only 3.6375 cm ≤ d ≤ 8.3625 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 d = 4.677 cm and 5.3385 cm. 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.
FIELD AT THE PROBE against d
Predictions from the supplied Field of a long straight wire 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): FIELD AT THE PROBE / µT
Horizontal axis (x): d: Probe distance d / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Probe distance d / cm | FIELD AT THE PROBE / µT |
|---|---|
| 3.6375 | 65.97938144329896 |
| 4.8187500000000005 | 49.805447470817114 |
| 6 | 40 |
| 7.18125 | 33.42036553524804 |
| 8.3625 | 28.699551569506724 |
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 = 4.677 cm, substitute into Y_out = 20*i/d: Y₀ = 51.3149 µT.
- At d = 5.3385 cm, the same fixed-parameter model gives Y₁ = 44.9564 µT.
- Signed change: ΔY_out = Y₁ − Y₀ = -6.3585 µT; the input change is Δd = 0.6615 cm.
- Average sensitivity is ΔY_out/Δd = -9.61224 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 = 51.3149 µT, Y₁ = 44.9564 µT; ΔY_out = -6.3585 µT; average sensitivity = -9.61224 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.
Field of a long straight wire: infer an input with uncertainty
The field of a long straight wire is azimuthal: at every point outside the conductor it is tangent to the circle through that point centred on the wire, with magnitude μ₀I/2πr, where r is the perpendicular distance from the axis. Use the supplied topic model Y_out = 20*i/d, where Y_out is FIELD AT THE PROBE (µT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Probe distance d, measured in cm. Keep i (Current I) = 12 A fixed. For this exercise use only 3.6375 cm ≤ d ≤ 8.3625 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 = 36.811227 µT, with a bounded reading uncertainty ±1.2168 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 0.0004725 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.
FIELD AT THE PROBE against d
Predictions from the supplied Field of a long straight wire 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): FIELD AT THE PROBE / µT
Horizontal axis (x): d: Probe distance d / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Probe distance d / cm | FIELD AT THE PROBE / µT |
|---|---|
| 3.6375 | 65.97938144329896 |
| 4.8187500000000005 | 49.805447470817114 |
| 6 | 40 |
| 7.18125 | 33.42036553524804 |
| 8.3625 | 28.699551569506724 |
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(3.6375) = 65.9794 µT and Y_out(8.3625) = 28.6996 µT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 3.6375 + (36.8112 − 65.9794)(8.3625 − 3.6375)/(28.6996 − 65.9794) = 7.33439 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 d = 6.51975 cm; substituting back gives Y_out = 36.811227 µT.
- The interpolation error in the input is |d_line − d| = 0.814643 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 35.594427 µT and 38.028027 µT. Sort the roots even if the response decreases: 6.31113 cm ≤ d ≤ 6.74263 cm.
- The lower and upper input deviations are 0.208615 cm and 0.222878 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 d ≈ 6.51975 cm; compatible interval [6.31113, 6.74263] cm. Endpoint interpolation gives 7.33439 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.