University Physics II · topic UPII-10.01
Magnetic flux
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 flux: predict a finite change
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. Calculate Y_out at b = 0.23975 T and 0.294875 T. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 0.23975 T, substitute into Y_out = b*a*cosd(th)*1000: Y₀ = 3.92783 mWb.
- At b = 0.294875 T, the same fixed-parameter model gives Y₁ = 4.83095 mWb.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.903115 mWb; the input change is Δb = 0.055125 T.
- Average sensitivity is ΔY_out/Δb = 16.383 mWb per T. Its positive sign means that this response increases 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 = 3.92783 mWb, Y₁ = 4.83095 mWb; ΔY_out = 0.903115 mWb; average sensitivity = 16.383 mWb per T.
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 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.