University Physics II · unit UPII-09
Sources of Magnetic Fields
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
Field of a long straight wire: infer an input with uncertainty
The field of a long straight wire is azimuthal: at every point outside the conductor it is tangent to the circle through that point centred on the wire, with magnitude μ₀I/2πr, where r is the perpendicular distance from the axis. Use the supplied topic model Y_out = 20*i/d, where Y_out is FIELD AT THE PROBE (µT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Probe distance d, measured in cm. Keep i (Current I) = 12 A fixed. For this exercise use only 3.6375 cm ≤ d ≤ 8.3625 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 = 36.811227 µT, with a bounded reading uncertainty ±1.2168 µT. 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 0.0004725 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.
FIELD AT THE PROBE against d
Predictions from the supplied Field of a long straight wire 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): FIELD AT THE PROBE / µT
Horizontal axis (x): d: Probe distance d / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Probe distance d / cm | FIELD AT THE PROBE / µT |
|---|---|
| 3.6375 | 65.97938144329896 |
| 4.8187500000000005 | 49.805447470817114 |
| 6 | 40 |
| 7.18125 | 33.42036553524804 |
| 8.3625 | 28.699551569506724 |
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(3.6375) = 65.9794 µT and Y_out(8.3625) = 28.6996 µT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 3.6375 + (36.8112 − 65.9794)(8.3625 − 3.6375)/(28.6996 − 65.9794) = 7.33439 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 = 6.51975 cm; substituting back gives Y_out = 36.811227 µT.
- The interpolation error in the input is |d_line − d| = 0.814643 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 35.594427 µT and 38.028027 µT. Sort the roots even if the response decreases: 6.31113 cm ≤ d ≤ 6.74263 cm.
- The lower and upper input deviations are 0.208615 cm and 0.222878 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 ≈ 6.51975 cm; compatible interval [6.31113, 6.74263] cm. Endpoint interpolation gives 7.33439 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Fields of loops, arcs and coils: infer an input with uncertainty
The on-axis field of a circular coil is the Biot-Savart integral evaluated with the ring's symmetry: B = μ₀NIR²/2(R² + z²)^(3/2), pointing along the axis in the sense the right hand gives. Use the supplied topic model Y_out = 1885*r^2/(r^2+z^2)^1.5, where Y_out is FIELD AT THE PROBE (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Probe distance z on the axis, measured in cm. Keep r (Coil radius R) = 6 cm fixed. For this exercise use only 3.44375 cm ≤ z ≤ 21.5562 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 = 17.584684 μT, with a bounded reading uncertainty ±1.7291 μT. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00181125 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.
FIELD AT THE PROBE against z
Predictions from the supplied Fields of loops, arcs and coils 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): FIELD AT THE PROBE / μT
Horizontal axis (x): z: Probe distance z on the axis / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Probe distance z on the axis / cm | FIELD AT THE PROBE / μT |
|---|---|
| 3.44375 | 204.9570159456949 |
| 7.971874999999999 | 68.31983085838122 |
| 12.499999999999998 | 25.45741677182713 |
| 17.028125 | 11.531168152754821 |
| 21.55625 | 6.057368723296098 |
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(z) = 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(3.44375) = 204.957 μT and Y_out(21.5562) = 6.05737 μT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives z_line = 3.44375 + (17.5847 − 204.957)(21.5562 − 3.44375)/(6.05737 − 204.957) = 20.5065 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 z = 14.4924 cm; substituting back gives Y_out = 17.584684 μT.
- The interpolation error in the input is |z_line − z| = 6.01416 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 15.855584 μT and 19.313784 μT. Sort the roots even if the response decreases: 13.9684 cm ≤ z ≤ 15.0866 cm.
- The lower and upper input deviations are 0.523964 cm and 0.594224 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 z ≈ 14.4924 cm; compatible interval [13.9684, 15.0866] cm. Endpoint interpolation gives 20.5065 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Solenoids & Toroids: infer an input with uncertainty
A solenoid is a wire wound in a close helix, whose interior magnetic field is axial and uniform with magnitude B = μ₀nI, set by the turns per metre n and the current I and by nothing else. Use the supplied topic model Y_out = len/(2*sqrt(len^2+r^2)), where Y_out is AT THE MOUTH (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. len is Winding length ℓ, measured in cm. Keep r (Winding radius R) = 2 cm; z (Probe position z) = 0 cm fixed. For this exercise use only 6.7625 cm ≤ len ≤ 27.2375 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 = 0.49732367, with a bounded reading uncertainty ±0.00020007. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown len by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0020475 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.
AT THE MOUTH against len
Predictions from the supplied Solenoids & Toroids 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): AT THE MOUTH
Horizontal axis (x): len: Winding length ℓ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| len: Winding length ℓ / cm | AT THE MOUTH |
|---|---|
| 6.762499999999999 | 0.4794705228083441 |
| 11.881249999999998 | 0.4930631242761421 |
| 17 | 0.4965753021614381 |
| 22.11875 | 0.49796845855522665 |
| 27.237499999999997 | 0.498657501991145 |
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(len) = 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(6.7625) = 0.479471 and Y_out(27.2375) = 0.498658. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives len_line = 6.7625 + (0.497324 − 0.479471)(27.2375 − 6.7625)/(0.498658 − 0.479471) = 25.8141 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 len = 19.2522 cm; substituting back gives Y_out = 0.49732367.
- The interpolation error in the input is |len_line − len| = 6.56189 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.4971236 and 0.49752374. Sort the roots even if the response decreases: 18.565 cm ≤ len ≤ 20.0209 cm.
- The lower and upper input deviations are 0.687226 cm and 0.768683 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 len ≈ 19.2522 cm; compatible interval [18.565, 20.0209] cm. Endpoint interpolation gives 25.8141 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.