University Physics II · topic UPII-08.01
Magnetic fields and field representation
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.
Magnetic Fields & Field Representation: predict a finite change
The magnetic field B at a point is the vector whose direction is the one axis along which a moving charge feels no magnetic force, the sense being that taken by a compass needle's north end, and whose magnitude is the largest force per unit charge per unit speed found there. Use the supplied topic model Y_out = ba*sind(phi), where Y_out is EAST COMPONENT (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. phi is Bearing of B_a from north, measured in °. Keep bh (Horizontal Earth field B_h) = 20 μT; ba (Applied field B_a) = 20 μT fixed. For this exercise use only 94.725 ° ≤ phi ≤ 160.875 ° 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 phi = 109.278 ° and 118.539 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δphi across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EAST COMPONENT against phi
Predictions from the supplied Magnetic Fields & Field Representation 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): EAST COMPONENT / μT
Horizontal axis (x): phi: Bearing of B_a from north / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Bearing of B_a from north / ° | EAST COMPONENT / μT |
|---|---|
| 94.725 | 19.932030790546634 |
| 111.26249999999999 | 18.638575507790026 |
| 127.8 | 15.803100247513811 |
| 144.3375 | 11.660191579667014 |
| 160.875 | 6.552603591233874 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of phi separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At phi = 109.278 °, substitute into Y_out = ba*sind(phi): Y₀ = 18.8786 μT.
- At phi = 118.539 °, the same fixed-parameter model gives Y₁ = 17.5698 μT.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.30871 μT; the input change is Δphi = 9.261 °.
- Average sensitivity is ΔY_out/Δphi = -0.141314 μT per °. 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 = 18.8786 μT, Y₁ = 17.5698 μT; ΔY_out = -1.30871 μT; average sensitivity = -0.141314 μT per °.
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.
Magnetic Fields & Field Representation: infer an input with uncertainty
The magnetic field B at a point is the vector whose direction is the one axis along which a moving charge feels no magnetic force, the sense being that taken by a compass needle's north end, and whose magnitude is the largest force per unit charge per unit speed found there. Use the supplied topic model Y_out = ba*sind(phi), where Y_out is EAST COMPONENT (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. phi is Bearing of B_a from north, measured in °. Keep bh (Horizontal Earth field B_h) = 20 μT; ba (Applied field B_a) = 20 μT fixed. For this exercise use only 94.725 ° ≤ phi ≤ 160.875 ° 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 = 14.123241 μT, with a bounded reading uncertainty ±0.87132 μT. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.006615 °, 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.
EAST COMPONENT against phi
Predictions from the supplied Magnetic Fields & Field Representation 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): EAST COMPONENT / μT
Horizontal axis (x): phi: Bearing of B_a from north / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Bearing of B_a from north / ° | EAST COMPONENT / μT |
|---|---|
| 94.725 | 19.932030790546634 |
| 111.26249999999999 | 18.638575507790026 |
| 127.8 | 15.803100247513811 |
| 144.3375 | 11.660191579667014 |
| 160.875 | 6.552603591233874 |
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(phi) = 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(94.725) = 19.932 μT and Y_out(160.875) = 6.5526 μT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 94.725 + (14.1232 − 19.932)(160.875 − 94.725)/(6.5526 − 19.932) = 123.445 °. 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 phi = 135.076 °; substituting back gives Y_out = 14.123241 μT.
- The interpolation error in the input is |phi_line − phi| = 11.6319 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 13.251921 μT and 14.994561 μT. Sort the roots even if the response decreases: 131.433 ° ≤ phi ≤ 138.502 °.
- The lower and upper input deviations are 3.64333 ° and 3.42534 °. 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 phi ≈ 135.076 °; compatible interval [131.433, 138.502] °. Endpoint interpolation gives 123.445 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.