University Physics II · topic UPII-06.03
Current density and drift speed
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.
Current density and drift speed: predict a finite change
Current density J = n q v(d) is current per unit cross-sectional area, a vector along the conventional current. Drift velocity v(d) is the slow average velocity the field adds to the carriers' fast random motion. Use the supplied topic model Y_out = i/max(a2,0.05), where Y_out is J, NARROW SECTION (MA m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a2 is Narrow cross-section A₂, measured in mm². Keep i (Current I) = 3 A fixed. For this exercise use only 0.765625 mm² ≤ a2 ≤ 2.73438 mm² 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 a2 = 1.19875 mm² and 1.47437 mm². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
J, NARROW SECTION against a2
Predictions from the supplied Current density and drift speed 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): J, NARROW SECTION / MA m⁻²
Horizontal axis (x): a2: Narrow cross-section A₂ / mm²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a2: Narrow cross-section A₂ / mm² | J, NARROW SECTION / MA m⁻² |
|---|---|
| 0.765625 | 3.9183673469387754 |
| 1.2578125 | 2.3850931677018634 |
| 1.75 | 1.7142857142857142 |
| 2.2421875 | 1.3379790940766552 |
| 2.734375 | 1.0971428571428572 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a2 = 1.19875 mm², substitute into Y_out = i/max(a2,0.05): Y₀ = 2.50261 MA m⁻².
- At a2 = 1.47437 mm², the same fixed-parameter model gives Y₁ = 2.03476 MA m⁻².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.467846 MA m⁻²; the input change is Δa2 = 0.275625 mm².
- Average sensitivity is ΔY_out/Δa2 = -1.6974 MA m⁻² per mm². 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 = 2.50261 MA m⁻², Y₁ = 2.03476 MA m⁻²; ΔY_out = -0.467846 MA m⁻²; average sensitivity = -1.6974 MA m⁻² per mm².
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.
Current density and drift speed: infer an input with uncertainty
Current density J = n q v(d) is current per unit cross-sectional area, a vector along the conventional current. Drift velocity v(d) is the slow average velocity the field adds to the carriers' fast random motion. Use the supplied topic model Y_out = i/max(a2,0.05), where Y_out is J, NARROW SECTION (MA m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a2 is Narrow cross-section A₂, measured in mm². Keep i (Current I) = 3 A fixed. For this exercise use only 0.765625 mm² ≤ a2 ≤ 2.73438 mm² 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 = 1.5255045 MA m⁻², with a bounded reading uncertainty ±0.064254 MA m⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 mm², 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.
J, NARROW SECTION against a2
Predictions from the supplied Current density and drift speed 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): J, NARROW SECTION / MA m⁻²
Horizontal axis (x): a2: Narrow cross-section A₂ / mm²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a2: Narrow cross-section A₂ / mm² | J, NARROW SECTION / MA m⁻² |
|---|---|
| 0.765625 | 3.9183673469387754 |
| 1.2578125 | 2.3850931677018634 |
| 1.75 | 1.7142857142857142 |
| 2.2421875 | 1.3379790940766552 |
| 2.734375 | 1.0971428571428572 |
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(a2) = 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(0.765625) = 3.91837 MA m⁻² and Y_out(2.73438) = 1.09714 MA m⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a2_line = 0.765625 + (1.5255 − 3.91837)(2.73438 − 0.765625)/(1.09714 − 3.91837) = 2.43545 mm². 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 a2 = 1.96656 mm²; substituting back gives Y_out = 1.5255045 MA m⁻².
- The interpolation error in the input is |a2_line − a2| = 0.468887 mm². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4612505 MA m⁻² and 1.5897585 MA m⁻². Sort the roots even if the response decreases: 1.88708 mm² ≤ a2 ≤ 2.05304 mm².
- The lower and upper input deviations are 0.0794835 mm² and 0.0864735 mm². 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 a2 ≈ 1.96656 mm²; compatible interval [1.88708, 2.05304] mm². Endpoint interpolation gives 2.43545 mm².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.