University Physics II · unit UPII-10
Electromagnetic Induction
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 flux: infer an input with uncertainty
The magnetic flux through a surface is the component of the magnetic field along the surface normal, added over the whole surface: Φ_B = ∫ B⋅dA, measured in webers. Use the supplied topic model Y_out = b*a*cosd(th)*1000, where Y_out is FLUX Φ = B A cos θ (mWb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. b is Field magnitude B, measured in T. Keep a (Loop area A) = 0.02 m²; th (Angle θ from n̂ to B) = 35 ° fixed. For this exercise use only 0.153125 T ≤ b ≤ 0.546875 T 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 = 6.4436548 mWb, with a bounded reading uncertainty ±0.37737 mWb. 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 0.000039375 T, 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.
FLUX Φ = B A cos θ against b
Predictions from the supplied Magnetic flux 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): FLUX Φ = B A cos θ / mWb
Horizontal axis (x): b: Field magnitude B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Field magnitude B / T | FLUX Φ = B A cos θ / mWb |
|---|---|
| 0.153125 | 2.5086531356350377 |
| 0.2515625 | 4.121358722828989 |
| 0.35 | 5.734064310022942 |
| 0.4484375 | 7.346769897216895 |
| 0.546875 | 8.959475484410849 |
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(0.153125) = 2.50865 mWb and Y_out(0.546875) = 8.95948 mWb. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.153125 + (6.44365 − 2.50865)(0.546875 − 0.153125)/(8.95948 − 2.50865) = 0.393313 T. 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 = 0.393313 T; substituting back gives Y_out = 6.4436548 mWb.
- The interpolation error in the input is |b_line − b| = 5.55112e-17 T. This is within the requested numerical tolerance.
- Invert both limits Y_out = 6.0662848 mWb and 6.8210248 mWb. Sort the roots even if the response decreases: 0.370278 T ≤ b ≤ 0.416347 T.
- The lower and upper input deviations are 0.0230342 T and 0.0230342 T. 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 b ≈ 0.393313 T; compatible interval [0.370278, 0.416347] T. Endpoint interpolation gives 0.393313 T.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Faraday's law: infer an input with uncertainty
Faraday's law states that the emf induced around a closed loop equals minus the rate at which the magnetic flux through that loop changes, and that a coil of N identical turns gives N times as much. Use the supplied topic model Y_out = 2*pi*f*phi*cos(2*pi*f*tm/1000), where Y_out is SLOPE dΦ/dt (mWb s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Frequency f, measured in Hz. Keep phi (Flux amplitude Φ₀ per turn) = 2 mWb; tm (Instant t on the graph) = 12 ms fixed. For this exercise use only 12.25 Hz ≤ f ≤ 27.3438 Hz 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 = -12.678466 mWb s⁻¹, with a bounded reading uncertainty ±15.824 mWb s⁻¹. 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 0.00150938 Hz, 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.
SLOPE dΦ/dt against f
Predictions from the supplied Faraday's 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): SLOPE dΦ/dt / mWb s⁻¹
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | SLOPE dΦ/dt / mWb s⁻¹ |
|---|---|
| 12.25 | 92.81379192132255 |
| 16.0234375 | 71.43326148786548 |
| 19.796875 | 19.421257195326593 |
| 23.5703125 | -60.69053379119238 |
| 27.34375 | -161.97743245892636 |
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(12.25) = 92.8138 mWb s⁻¹ and Y_out(27.3438) = -161.977 mWb s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 12.25 + (-12.6785 − 92.8138)(27.3438 − 12.25)/(-161.977 − 92.8138) = 18.4993 Hz. 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 = 21.4572 Hz; substituting back gives Y_out = -12.678466 mWb s⁻¹.
- The interpolation error in the input is |f_line − f| = 2.95786 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -28.502466 mWb s⁻¹ and 3.145534 mWb s⁻¹. Sort the roots even if the response decreases: 20.6727 Hz ≤ f ≤ 22.1913 Hz.
- The lower and upper input deviations are 0.784451 Hz and 0.734109 Hz. 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 f ≈ 21.4572 Hz; compatible interval [20.6727, 22.1913] Hz. Endpoint interpolation gives 18.4993 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Lenz's Law & Induced Direction: infer an input with uncertainty
Lenz's law: a current induced in a circuit flows in the sense whose own magnetic field opposes the change in magnetic flux through that circuit — the change, not the flux itself. Use the supplied topic model Y_out = abs(rate)*pi*(a/100)^2/r, where Y_out is INDUCED CURRENT (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. r is Loop resistance, measured in Ω. Keep rate (Rate of change dB/dt) = 2 T s⁻¹; a (Loop radius) = 6 cm fixed. For this exercise use only 0.150938 Ω ≤ r ≤ 0.899062 Ω 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 = 0.037246336 A, with a bounded reading uncertainty ±0.0018131 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125 Ω, 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.
INDUCED CURRENT against r
Predictions from the supplied Lenz's Law & Induced Direction 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): INDUCED CURRENT / A
Horizontal axis (x): r: Loop resistance / Ω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Loop resistance / Ω | INDUCED CURRENT / A |
|---|---|
| 0.1509375 | 0.14985982347558763 |
| 0.33796875 | 0.06692768815414595 |
| 0.5249999999999999 | 0.04308469924923145 |
| 0.7120312499999999 | 0.03176752018376513 |
| 0.8990624999999999 | 0.0251589484667045 |
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(r) = 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(0.150938) = 0.14986 A and Y_out(0.899062) = 0.0251589 A. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.150938 + (0.0372463 − 0.14986)(0.899062 − 0.150938)/(0.0251589 − 0.14986) = 0.826546 Ω. 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 r = 0.607294 Ω; substituting back gives Y_out = 0.037246336 A.
- The interpolation error in the input is |r_line − r| = 0.219252 Ω. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.035433236 A and 0.039059436 A. Sort the roots even if the response decreases: 0.579104 Ω ≤ r ≤ 0.638369 Ω.
- The lower and upper input deviations are 0.02819 Ω and 0.0310749 Ω. 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 r ≈ 0.607294 Ω; compatible interval [0.579104, 0.638369] Ω. Endpoint interpolation gives 0.826546 Ω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Motional emf: infer an input with uncertainty
Motional emf is the emf that appears across a conductor moved through a magnetic field, equal to the work per unit charge that the magnetic force does in driving charge from one end of the conductor to the other. Use the supplied topic model Y_out = b*b*l*l*v*v/0.25, where Y_out is POWER P = Fv (W). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Rod speed v, measured in m s⁻¹. Keep b (Field B, out of the page) = 0.35 T; l (Rod length L, the rail gap) = 0.4 m fixed. For this exercise use only 2.6375 m s⁻¹ ≤ v ≤ 7.3625 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 = 2.388663 W, with a bounded reading uncertainty ±0.27649 W. 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 0.0004725 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.
POWER P = Fv against v
Predictions from the supplied Motional emf 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): POWER P = Fv / W
Horizontal axis (x): v: Rod speed v / m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Rod speed v / m s⁻¹ | POWER P = Fv / W |
|---|---|
| 2.6375 | 0.5453822500000001 |
| 3.81875 | 1.1432955625 |
| 5 | 1.96 |
| 6.18125 | 2.9954955625 |
| 7.3625 | 4.24978225 |
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(2.6375) = 0.545382 W and Y_out(7.3625) = 4.24978 W. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v_line = 2.6375 + (2.38866 − 0.545382)(7.3625 − 2.6375)/(4.24978 − 0.545382) = 4.98862 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 = 5.51975 m s⁻¹; substituting back gives Y_out = 2.388663 W.
- The interpolation error in the input is |v_line − v| = 0.531127 m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.112173 W and 2.665153 W. Sort the roots even if the response decreases: 5.19047 m s⁻¹ ≤ v ≤ 5.83046 m s⁻¹.
- The lower and upper input deviations are 0.32928 m s⁻¹ and 0.310713 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 ≈ 5.51975 m s⁻¹; compatible interval [5.19047, 5.83046] m s⁻¹. Endpoint interpolation gives 4.98862 m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.