University Physics II · unit UPII-12
Maxwell's Equations and Electromagnetic Waves
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.
Gauss's law for electricity: infer an input with uncertainty
Gauss's law says the electric flux out through any closed surface equals the net charge enclosed divided by ε₀, whatever the surface's shape and wherever inside it the charge sits. Use the supplied topic model Y_out = 89880*q*min(rg,rb)/(max(rg,rb)^2*rb), where Y_out is FIELD AT r (N C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. rg is Gaussian sphere radius r, measured in cm. Keep q (Ball charge Q) = 8 nC; rb (Ball radius R) = 5 cm fixed. For this exercise use only 5.4275 cm ≤ rg ≤ 13.5125 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 = 6700.2042 N C⁻¹, with a bounded reading uncertainty ±414.32 N C⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rg by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0008085 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 r against rg
Predictions from the supplied Gauss's law for electricity 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 r / N C⁻¹
Horizontal axis (x): rg: Gaussian sphere radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rg: Gaussian sphere radius r / cm | FIELD AT r / N C⁻¹ |
|---|---|
| 5.427499999999999 | 24409.19104285141 |
| 7.4487499999999995 | 12959.440553690134 |
| 9.469999999999999 | 8017.760749501847 |
| 11.49125 | 5445.2585359376235 |
| 13.5125 | 3938.0537259073344 |
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(rg) = 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(5.4275) = 24409.2 N C⁻¹ and Y_out(13.5125) = 3938.05 N C⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rg_line = 5.4275 + (6700.2 − 24409.2)(13.5125 − 5.4275)/(3938.05 − 24409.2) = 12.4216 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 rg = 10.3593 cm; substituting back gives Y_out = 6700.2042 N C⁻¹.
- The interpolation error in the input is |rg_line − rg| = 2.06225 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6285.8842 N C⁻¹ and 7114.5242 N C⁻¹. Sort the roots even if the response decreases: 10.0532 cm ≤ rg ≤ 10.6953 cm.
- The lower and upper input deviations are 0.306167 cm and 0.335959 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 rg ≈ 10.3593 cm; compatible interval [10.0532, 10.6953] cm. Endpoint interpolation gives 12.4216 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Gauss's law for magnetism: infer an input with uncertainty
Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero: ∮B⃗⋅dA⃗ = 0. There is no magnetic charge to act as a source, so field lines close on themselves rather than beginning or ending. Use the supplied topic model Y_out = f*B*A*100, where Y_out is OUT THROUGH THE SIDES (μWb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Share leaking through the side wall, dimensionless. Keep B (Field in the iron B) = 0.8 T; A (Bar cross-section A) = 2 cm² fixed. For this exercise use only 0.095625 ≤ f ≤ 0.804375 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 = 84.474 μWb, with a bounded reading uncertainty ±6.6339 μWb. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000070875, 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.
OUT THROUGH THE SIDES against f
Predictions from the supplied Gauss's law for magnetism 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): OUT THROUGH THE SIDES / μWb
Horizontal axis (x): f: Share leaking through the side wall
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Share leaking through the side wall | OUT THROUGH THE SIDES / μWb |
|---|---|
| 0.095625 | 15.300000000000002 |
| 0.2728125 | 43.650000000000006 |
| 0.45 | 72.00000000000001 |
| 0.6271874999999999 | 100.34999999999998 |
| 0.804375 | 128.7 |
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(f) = 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(0.095625) = 15.3 μWb and Y_out(0.804375) = 128.7 μWb. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.095625 + (84.474 − 15.3)(0.804375 − 0.095625)/(128.7 − 15.3) = 0.527963. 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 f = 0.527962; substituting back gives Y_out = 84.474 μWb.
- The interpolation error in the input is |f_line − f| = 1.11022e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 77.8401 μWb and 91.1079 μWb. Sort the roots even if the response decreases: 0.486501 ≤ f ≤ 0.569424.
- The lower and upper input deviations are 0.0414619 and 0.0414619. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 0.527962; compatible interval [0.486501, 0.569424]. Endpoint interpolation gives 0.527963.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Faraday's law as field circulation: infer an input with uncertainty
Faraday's law states that the circulation of the electric field round any closed loop fixed in space equals minus the rate of change of the magnetic flux through any surface that loop bounds. Use the supplied topic model Y_out = 5*k*min(rp, rs*rs/max(rp,0.05)), where Y_out is E AT PROBE (mV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. rp is Probe radius r, measured in cm. Keep rs (Solenoid radius R) = 4 cm; k (Field ramp rate dB/dt) = 2 T s⁻¹ fixed. For this exercise use only 4.59125 cm ≤ rp ≤ 16.1938 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 = 13.711808 mV m⁻¹, with a bounded reading uncertainty ±0.57472 mV m⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00116025 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.
E AT PROBE against rp
Predictions from the supplied Faraday's law as field circulation 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): E AT PROBE / mV m⁻¹
Horizontal axis (x): rp: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rp: Probe radius r / cm | E AT PROBE / mV m⁻¹ |
|---|---|
| 4.5912500000000005 | 34.84889735910699 |
| 7.491875 | 21.356469508634355 |
| 10.392500000000002 | 15.395718065912916 |
| 13.293125000000002 | 12.03629695801401 |
| 16.19375 | 9.880355075260516 |
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(rp) = 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(4.59125) = 34.8489 mV m⁻¹ and Y_out(16.1938) = 9.88036 mV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rp_line = 4.59125 + (13.7118 − 34.8489)(16.1938 − 4.59125)/(9.88036 − 34.8489) = 14.4133 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 rp = 11.6688 cm; substituting back gives Y_out = 13.711808 mV m⁻¹.
- The interpolation error in the input is |rp_line − rp| = 2.74456 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 13.137088 mV m⁻¹ and 14.286528 mV m⁻¹. Sort the roots even if the response decreases: 11.1994 cm ≤ rp ≤ 12.1793 cm.
- The lower and upper input deviations are 0.469413 cm and 0.510484 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 rp ≈ 11.6688 cm; compatible interval [11.1994, 12.1793] cm. Endpoint interpolation gives 14.4133 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Ampere-Maxwell law: infer an input with uncertainty
The Ampere-Maxwell law says the circulation of B around a closed loop equals μ₀ times the sum of the conduction current through any surface bounded by that loop and ε₀ times the rate of change of electric flux through that same surface. Use the supplied topic model Y_out = 20*cur*min(rl,rp)/(rp*max(rl,rp)), where Y_out is B ON THE LOOP (μT). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. rl is Amperean loop radius r, measured in cm. Keep cur (Charging current I) = 1.2 A; rp (Plate radius R) = 4 cm fixed. For this exercise use only 4.1625 cm ≤ rl ≤ 9.9375 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 = 3.1228652 μT, with a bounded reading uncertainty ±0.10617 μT. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0005775 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.
B ON THE LOOP against rl
Predictions from the supplied The Ampere-Maxwell 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): B ON THE LOOP / μT
Horizontal axis (x): rl: Amperean loop radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Amperean loop radius r / cm | B ON THE LOOP / μT |
|---|---|
| 4.1625 | 5.765765765765766 |
| 5.606249999999999 | 4.280936454849499 |
| 7.05 | 3.404255319148936 |
| 8.49375 | 2.82560706401766 |
| 9.9375 | 2.4150943396226414 |
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(rl) = 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(4.1625) = 5.76577 μT and Y_out(9.9375) = 2.41509 μT. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rl_line = 4.1625 + (3.12287 − 5.76577)(9.9375 − 4.1625)/(2.41509 − 5.76577) = 8.71763 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 rl = 7.68525 cm; substituting back gives Y_out = 3.1228652 μT.
- The interpolation error in the input is |rl_line − rl| = 1.03238 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.0166952 μT and 3.2290352 μT. Sort the roots even if the response decreases: 7.43256 cm ≤ rl ≤ 7.95573 cm.
- The lower and upper input deviations are 0.252689 cm and 0.270476 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 rl ≈ 7.68525 cm; compatible interval [7.43256, 7.95573] cm. Endpoint interpolation gives 8.71763 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.