University Physics II · topic UPII-09.01
The Biot-Savart law
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.
The Biot-Savart law: predict a finite change
The Biot-Savart law gives the magnetic field contributed by a short piece of a current: its size falls as the inverse square of the distance and as sinθ, and its direction is the cross product of the element with the direction to the field point. Use the supplied topic model Y_out = 2000*cur*sind(th)/(d*d), where Y_out is dB AT P (nT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. d is Distance r to the field point, measured in cm. Keep cur (Current I) = 10 A; th (Angle θ from the element) = 60 ° fixed. For this exercise use only 3.31875 cm ≤ d ≤ 5.68125 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 = 3.8385 cm and 4.16925 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.
dB AT P against d
Predictions from the supplied The Biot-Savart law 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): dB AT P / nT
Horizontal axis (x): d: Distance r to the field point / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Distance r to the field point / cm | dB AT P / nT |
|---|---|
| 3.31875 | 1572.5756637890793 |
| 3.9093750000000003 | 1133.3028074426343 |
| 4.5 | 855.3337321327789 |
| 5.090625 | 668.3722579469228 |
| 5.68125 | 536.6273782619139 |
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 = 3.8385 cm, substitute into Y_out = 2000*cur*sind(th)/(d*d): Y₀ = 1175.54 nT.
- At d = 4.16925 cm, the same fixed-parameter model gives Y₁ = 996.425 nT.
- Signed change: ΔY_out = Y₁ − Y₀ = -179.115 nT; the input change is Δd = 0.33075 cm.
- Average sensitivity is ΔY_out/Δd = -541.542 nT 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 = 1175.54 nT, Y₁ = 996.425 nT; ΔY_out = -179.115 nT; average sensitivity = -541.542 nT 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.
The Biot-Savart law: infer an input with uncertainty
The Biot-Savart law gives the magnetic field contributed by a short piece of a current: its size falls as the inverse square of the distance and as sinθ, and its direction is the cross product of the element with the direction to the field point. Use the supplied topic model Y_out = 2000*cur*sind(th)/(d*d), where Y_out is dB AT P (nT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. d is Distance r to the field point, measured in cm. Keep cur (Current I) = 10 A; th (Angle θ from the element) = 60 ° fixed. For this exercise use only 3.31875 cm ≤ d ≤ 5.68125 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 = 764.48599 nT, with a bounded reading uncertainty ±34.179 nT. 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 0.00023625 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.
dB AT P against d
Predictions from the supplied The Biot-Savart law 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): dB AT P / nT
Horizontal axis (x): d: Distance r to the field point / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Distance r to the field point / cm | dB AT P / nT |
|---|---|
| 3.31875 | 1572.5756637890793 |
| 3.9093750000000003 | 1133.3028074426343 |
| 4.5 | 855.3337321327789 |
| 5.090625 | 668.3722579469228 |
| 5.68125 | 536.6273782619139 |
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(3.31875) = 1572.58 nT and Y_out(5.68125) = 536.627 nT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 3.31875 + (764.486 − 1572.58)(5.68125 − 3.31875)/(536.627 − 1572.58) = 5.16161 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 = 4.75988 cm; substituting back gives Y_out = 764.48599 nT.
- The interpolation error in the input is |d_line − d| = 0.401739 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 730.30699 nT and 798.66499 nT. Sort the roots even if the response decreases: 4.65691 cm ≤ d ≤ 4.86998 cm.
- The lower and upper input deviations are 0.102963 cm and 0.11011 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 ≈ 4.75988 cm; compatible interval [4.65691, 4.86998] cm. Endpoint interpolation gives 5.16161 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.