University Physics II · topic UPII-11.01
Self-inductance
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.
Self-inductance: predict a finite change
Self-inductance is the flux linkage a circuit's own current produces per unit current, L = NΦ_B/I — a constant fixed by the winding's shape and the material inside it, not by the current. Use the supplied topic model Y_out = 0.00024674*n*n/lc, where Y_out is SELF-INDUCTANCE L (mH). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lc is Winding length ℓ, measured in cm. Keep n (Turns N) = 350; cur (Current I) = 2 A fixed. For this exercise use only 16.85 cm ≤ lc ≤ 23.15 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 lc = 18.236 cm and 19.118 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SELF-INDUCTANCE L against lc
Predictions from the supplied Self-inductance 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): SELF-INDUCTANCE L / mH
Horizontal axis (x): lc: Winding length ℓ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lc: Winding length ℓ / cm | SELF-INDUCTANCE L / mH |
|---|---|
| 16.85 | 1.793807121661721 |
| 18.425 | 1.64046947082768 |
| 20 | 1.5112825 |
| 21.575 | 1.4009571263035923 |
| 23.15 | 1.3056436285097193 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lc = 18.236 cm, substitute into Y_out = 0.00024674*n*n/lc: Y₀ = 1.65747 mH.
- At lc = 19.118 cm, the same fixed-parameter model gives Y₁ = 1.581 mH.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0764667 mH; the input change is Δlc = 0.882 cm.
- Average sensitivity is ΔY_out/Δlc = -0.0866969 mH 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 = 1.65747 mH, Y₁ = 1.581 mH; ΔY_out = -0.0764667 mH; average sensitivity = -0.0866969 mH 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.
Self-inductance: infer an input with uncertainty
Self-inductance is the flux linkage a circuit's own current produces per unit current, L = NΦ_B/I — a constant fixed by the winding's shape and the material inside it, not by the current. Use the supplied topic model Y_out = 0.00024674*n*n/lc, where Y_out is SELF-INDUCTANCE L (mH). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lc is Winding length ℓ, measured in cm. Keep n (Turns N) = 350; cur (Current I) = 2 A fixed. For this exercise use only 16.85 cm ≤ lc ≤ 23.15 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 = 1.4606703 mH, with a bounded reading uncertainty ±0.023254 mH. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 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.
SELF-INDUCTANCE L against lc
Predictions from the supplied Self-inductance 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): SELF-INDUCTANCE L / mH
Horizontal axis (x): lc: Winding length ℓ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lc: Winding length ℓ / cm | SELF-INDUCTANCE L / mH |
|---|---|
| 16.85 | 1.793807121661721 |
| 18.425 | 1.64046947082768 |
| 20 | 1.5112825 |
| 21.575 | 1.4009571263035923 |
| 23.15 | 1.3056436285097193 |
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(lc) = 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(16.85) = 1.79381 mH and Y_out(23.15) = 1.30564 mH. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lc_line = 16.85 + (1.46067 − 1.79381)(23.15 − 16.85)/(1.30564 − 1.79381) = 21.1493 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 lc = 20.693 cm; substituting back gives Y_out = 1.4606703 mH.
- The interpolation error in the input is |lc_line − lc| = 0.456302 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4374163 mH and 1.4839243 mH. Sort the roots even if the response decreases: 20.3687 cm ≤ lc ≤ 21.0278 cm.
- The lower and upper input deviations are 0.324272 cm and 0.334764 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 lc ≈ 20.693 cm; compatible interval [20.3687, 21.0278] cm. Endpoint interpolation gives 21.1493 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.