University Physics II · unit UPII-08
Magnetic Fields and Magnetic Forces
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.
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.
Magnetic Force on a Moving Charge: infer an input with uncertainty
The magnetic force on a moving charge is F = qv×B: its size is |q| v B sin θ, and its direction is perpendicular to both the velocity and the field, with the sign of the charge choosing between the two perpendicular senses. Use the supplied topic model Y_out = sind(theta), where Y_out is ANGLE FACTOR sin θ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Angle θ between v and B, measured in °. Keep v (Speed v) = 3 10⁶ m s⁻¹; q (Charge q) = 1 e fixed. For this exercise use only 94.725 ° ≤ theta ≤ 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 = 0.70616204, with a bounded reading uncertainty ±0.043566. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta 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.
ANGLE FACTOR sin θ against theta
Predictions from the supplied Magnetic Force on a Moving Charge 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): ANGLE FACTOR sin θ
Horizontal axis (x): theta: Angle θ between v and B / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Angle θ between v and B / ° | ANGLE FACTOR sin θ |
|---|---|
| 94.725 | 0.9966015395273318 |
| 111.26249999999999 | 0.9319287753895013 |
| 127.8 | 0.7901550123756905 |
| 144.3375 | 0.5830095789833507 |
| 160.875 | 0.3276301795616937 |
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(theta) = 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) = 0.996602 and Y_out(160.875) = 0.32763. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 94.725 + (0.706162 − 0.996602)(160.875 − 94.725)/(0.32763 − 0.996602) = 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 theta = 135.076 °; substituting back gives Y_out = 0.70616204.
- The interpolation error in the input is |theta_line − theta| = 11.6319 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.66259604 and 0.74972804. Sort the roots even if the response decreases: 131.433 ° ≤ theta ≤ 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 theta ≈ 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.
Charged-particle orbits in uniform fields: infer an input with uncertainty
In a uniform magnetic field a charged particle travels at constant speed along a helix whose axis lies along B, circling at the cyclotron frequency while coasting steadily in the field direction. Use the supplied topic model Y_out = 5.686*vperp/b, where Y_out is ORBIT RADIUS r (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Field strength B, measured in mT. Keep vperp (Perpendicular speed v⊥) = 2 ×10⁶ m s⁻¹; vpar (Parallel speed v∥) = 1.2 ×10⁶ m s⁻¹ fixed. For this exercise use only 2.31875 mT ≤ b ≤ 4.68125 mT 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 = 3.0245686 mm, with a bounded reading uncertainty ±0.089295 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b 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 mT, 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.
ORBIT RADIUS r against b
Predictions from the supplied Charged-particle orbits in uniform fields 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): ORBIT RADIUS r / mm
Horizontal axis (x): b: Field strength B / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Field strength B / mT | ORBIT RADIUS r / mm |
|---|---|
| 2.31875 | 4.904366576819407 |
| 2.9093750000000003 | 3.908743286788399 |
| 3.5 | 3.249142857142857 |
| 4.090625 | 2.780015278838808 |
| 4.68125 | 2.4292656875834444 |
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(b) = 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(2.31875) = 4.90437 mm and Y_out(4.68125) = 2.42927 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 2.31875 + (3.02457 − 4.90437)(4.68125 − 2.31875)/(2.42927 − 4.90437) = 4.11303 mT. 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 b = 3.75988 mT; substituting back gives Y_out = 3.0245686 mm.
- The interpolation error in the input is |b_line − b| = 0.353154 mT. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.9352736 mm and 3.1138636 mm. Sort the roots even if the response decreases: 3.65205 mT ≤ b ≤ 3.87426 mT.
- The lower and upper input deviations are 0.10782 mT and 0.11438 mT. 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 b ≈ 3.75988 mT; compatible interval [3.65205, 3.87426] mT. Endpoint interpolation gives 4.11303 mT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Velocity Selectors: infer an input with uncertainty
A velocity selector is a region of mutually perpendicular electric and magnetic fields in which the two forces on a charged particle cancel, so only particles whose velocity is E/B directed along E × B cross it in a straight line. Use the supplied topic model Y_out = 0.46*(b*v-10*e)/(v*v), where Y_out is OFFSET AT SLIT (+ IS UP) (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Entry speed v, measured in × 10⁶ m s⁻¹. Keep e (Electric field E) = 3 × 10⁴ V m⁻¹; b (Magnetic field B) = 20 mT fixed. For this exercise use only 1.26375 × 10⁶ m s⁻¹ ≤ v ≤ 1.73625 × 10⁶ m s⁻¹ 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 = 0.19852394 mm, with a bounded reading uncertainty ±0.078371 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00004725 × 10⁶ m s⁻¹, 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.
OFFSET AT SLIT (+ IS UP) against v
Predictions from the supplied Velocity Selectors 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): OFFSET AT SLIT (+ IS UP) / mm
Horizontal axis (x): v: Entry speed v / × 10⁶ m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Entry speed v / × 10⁶ m s⁻¹ | OFFSET AT SLIT (+ IS UP) / mm |
|---|---|
| 1.26375 | -1.3609347621269894 |
| 1.381875 | -0.5691046433062272 |
| 1.5 | 0 |
| 1.618125 | 0.4150548693621776 |
| 1.73625 | 0.7209997714221741 |
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(v) = 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(1.26375) = -1.36093 mm and Y_out(1.73625) = 0.721 mm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v_line = 1.26375 + (0.198524 − -1.36093)(1.73625 − 1.26375)/(0.721 − -1.36093) = 1.61767 × 10⁶ m s⁻¹. 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 v = 1.55198 × 10⁶ m s⁻¹; substituting back gives Y_out = 0.19852394 mm.
- The interpolation error in the input is |v_line − v| = 0.0656979 × 10⁶ m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12015294 mm and 0.27689494 mm. Sort the roots even if the response decreases: 1.5306 × 10⁶ m s⁻¹ ≤ v ≤ 1.57462 × 10⁶ m s⁻¹.
- The lower and upper input deviations are 0.0213788 × 10⁶ m s⁻¹ and 0.0226495 × 10⁶ m s⁻¹. 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 v ≈ 1.55198 × 10⁶ m s⁻¹; compatible interval [1.5306, 1.57462] × 10⁶ m s⁻¹. Endpoint interpolation gives 1.61767 × 10⁶ m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.