University Physics IV · topic UPIV-15.04
Zero resistance, the Meissner effect, and critical parameters
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.
Zero resistance, Meissner and critical fields: predict a finite change
A superconductor is a material that below its critical temperature carries direct current with no detectable dissipation and actively expels magnetic flux from its bulk, screening an applied field to zero within a few penetration depths λ of its surface. Use the supplied topic model Y_out = 2*b0/(exp(d*sqrt(ns)/106.28)+1/exp(d*sqrt(ns)/106.28)), where Y_out is MID-PLANE FIELD (mT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. d is Slab thickness d, measured in nm. Keep ns (Superconducting density n_s) = 4 10²⁸ m⁻³; b0 (Applied field B₀) = 20 mT fixed. For this exercise use only 60.375 nm ≤ d ≤ 359.625 nm 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 = 126.21 nm and 168.105 nm. 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.
MID-PLANE FIELD against d
Predictions from the supplied Zero resistance, Meissner and critical 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): MID-PLANE FIELD / mT
Horizontal axis (x): d: Slab thickness d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Slab thickness d / nm | MID-PLANE FIELD / mT |
|---|---|
| 60.375 | 11.642102318714104 |
| 135.1875 | 3.1228321554458454 |
| 210 | 0.7684995994953936 |
| 284.8125 | 0.18809543481003832 |
| 359.625 | 0.04602259485452007 |
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 = 126.21 nm, substitute into Y_out = 2*b0/(exp(d*sqrt(ns)/106.28)+1/exp(d*sqrt(ns)/106.28)): Y₀ = 3.6885 mT.
- At d = 168.105 nm, the same fixed-parameter model gives Y₁ = 1.68819 mT.
- Signed change: ΔY_out = Y₁ − Y₀ = -2.00031 mT; the input change is Δd = 41.895 nm.
- Average sensitivity is ΔY_out/Δd = -0.0477457 mT per nm. 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 = 3.6885 mT, Y₁ = 1.68819 mT; ΔY_out = -2.00031 mT; average sensitivity = -0.0477457 mT per nm.
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.
Zero resistance, Meissner and critical fields: infer an input with uncertainty
A superconductor is a material that below its critical temperature carries direct current with no detectable dissipation and actively expels magnetic flux from its bulk, screening an applied field to zero within a few penetration depths λ of its surface. Use the supplied topic model Y_out = 2*b0/(exp(d*sqrt(ns)/106.28)+1/exp(d*sqrt(ns)/106.28)), where Y_out is MID-PLANE FIELD (mT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. d is Slab thickness d, measured in nm. Keep ns (Superconducting density n_s) = 4 10²⁸ m⁻³; b0 (Applied field B₀) = 20 mT fixed. For this exercise use only 60.375 nm ≤ d ≤ 359.625 nm 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 = 0.41374661 mT, with a bounded reading uncertainty ±0.055159 mT. 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 0.029925 nm, 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.
MID-PLANE FIELD against d
Predictions from the supplied Zero resistance, Meissner and critical 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): MID-PLANE FIELD / mT
Horizontal axis (x): d: Slab thickness d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Slab thickness d / nm | MID-PLANE FIELD / mT |
|---|---|
| 60.375 | 11.642102318714104 |
| 135.1875 | 3.1228321554458454 |
| 210 | 0.7684995994953936 |
| 284.8125 | 0.18809543481003832 |
| 359.625 | 0.04602259485452007 |
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(60.375) = 11.6421 mT and Y_out(359.625) = 0.0460226 mT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 60.375 + (0.413747 − 11.6421)(359.625 − 60.375)/(0.0460226 − 11.6421) = 350.135 nm. 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 = 242.917 nm; substituting back gives Y_out = 0.41374661 mT.
- The interpolation error in the input is |d_line − d| = 107.218 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.35858761 mT and 0.46890561 mT. Sort the roots even if the response decreases: 236.266 nm ≤ d ≤ 250.522 nm.
- The lower and upper input deviations are 6.65197 nm and 7.60472 nm. 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 ≈ 242.917 nm; compatible interval [236.266, 250.522] nm. Endpoint interpolation gives 350.135 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.