University Physics II · topic UPII-10.06
Eddy 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.
Eddy Currents and Magnetic Braking: predict a finite change
Eddy currents are the closed loops of current that a changing magnetic flux drives inside a bulk conductor. They dissipate energy as heat and produce a force opposing whatever change created them. Use the supplied topic model Y_out = 2*0.0749*v*L/(1+(0.0749*v*L)^2), where Y_out is DRAG ÷ PEAK DRAG (×). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. L is Field region size L, measured in mm. Keep v (Speed v) = 0.5 m s⁻¹ fixed. For this exercise use only 13.1875 mm ≤ L ≤ 25.7875 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 L = 15.9595 mm and 17.7235 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/ΔL across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DRAG ÷ PEAK DRAG against L
Predictions from the supplied Eddy Currents and Magnetic Braking 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): DRAG ÷ PEAK DRAG / ×
Horizontal axis (x): L: Field region size L / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| L: Field region size L / mm | DRAG ÷ PEAK DRAG / × |
|---|---|
| 13.1875 | 0.7940640427983193 |
| 16.3375 | 0.8903707544976913 |
| 19.4875 | 0.952366263978113 |
| 22.637500000000003 | 0.9865175117798372 |
| 25.7875 | 0.99939274330398 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of L separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At L = 15.9595 mm, substitute into Y_out = 2*0.0749*v*L/(1+(0.0749*v*L)^2): Y₀ = 0.880743 ×.
- At L = 17.7235 mm, the same fixed-parameter model gives Y₁ = 0.921511 ×.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0407685 ×; the input change is ΔL = 1.764 mm.
- Average sensitivity is ΔY_out/ΔL = 0.0231114 × per mm. Its positive sign means that this response increases 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 = 0.880743 ×, Y₁ = 0.921511 ×; ΔY_out = 0.0407685 ×; average sensitivity = 0.0231114 × 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.
Eddy Currents and Magnetic Braking: infer an input with uncertainty
Eddy currents are the closed loops of current that a changing magnetic flux drives inside a bulk conductor. They dissipate energy as heat and produce a force opposing whatever change created them. Use the supplied topic model Y_out = 2*0.0749*v*L/(1+(0.0749*v*L)^2), where Y_out is DRAG ÷ PEAK DRAG (×). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. L is Field region size L, measured in mm. Keep v (Speed v) = 0.5 m s⁻¹ fixed. For this exercise use only 13.1875 mm ≤ L ≤ 25.7875 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 = 0.97042384 ×, with a bounded reading uncertainty ±0.0043453 ×. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown L by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00126 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.
DRAG ÷ PEAK DRAG against L
Predictions from the supplied Eddy Currents and Magnetic Braking 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): DRAG ÷ PEAK DRAG / ×
Horizontal axis (x): L: Field region size L / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| L: Field region size L / mm | DRAG ÷ PEAK DRAG / × |
|---|---|
| 13.1875 | 0.7940640427983193 |
| 16.3375 | 0.8903707544976913 |
| 19.4875 | 0.952366263978113 |
| 22.637500000000003 | 0.9865175117798372 |
| 25.7875 | 0.99939274330398 |
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(L) = 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(13.1875) = 0.794064 × and Y_out(25.7875) = 0.999393 ×. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives L_line = 13.1875 + (0.970424 − 0.794064)(25.7875 − 13.1875)/(0.999393 − 0.794064) = 24.0098 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 L = 20.8735 mm; substituting back gives Y_out = 0.97042384 ×.
- The interpolation error in the input is |L_line − L| = 3.13632 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.96607854 × and 0.97476914 ×. Sort the roots even if the response decreases: 20.5019 mm ≤ L ≤ 21.2788 mm.
- The lower and upper input deviations are 0.371595 mm and 0.405294 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 L ≈ 20.8735 mm; compatible interval [20.5019, 21.2788] mm. Endpoint interpolation gives 24.0098 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.