University Physics V · topic UPV-10.08
Magnetic moments, the g factor, and Larmor precession
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 moments and Larmor precession: predict a finite change
Larmor precession is the rigid rotation of a spin's expectation value ⟨S⟩ about the applied field, at angular rate ω = |γ|B, generated by the Zeeman Hamiltonian Ĥ = −γ B⋅Ŝ. Use the supplied topic model Y_out = 28.025*b, where Y_out is LARMOR f = |γ|B/2π (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Field B along z, measured in mT. Keep th (Tip angle θ from B) = 60 °; t (Elapsed time t) = 12 ns fixed. For this exercise use only 0.39125 mT ≤ b ≤ 1.80875 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. Calculate Y_out at b = 0.7031 mT and 0.90155 mT. 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.
LARMOR f = |γ|B/2π against b
Predictions from the supplied Magnetic moments and Larmor precession 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): LARMOR f = |γ|B/2π / MHz
Horizontal axis (x): b: Field B along z / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Field B along z / mT | LARMOR f = |γ|B/2π / MHz |
|---|---|
| 0.39125 | 10.96478125 |
| 0.745625 | 20.896140624999997 |
| 1.1 | 30.8275 |
| 1.4543749999999998 | 40.75885937499999 |
| 1.8087499999999999 | 50.69021874999999 |
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 = 0.7031 mT, substitute into Y_out = 28.025*b: Y₀ = 19.7044 MHz.
- At b = 0.90155 mT, the same fixed-parameter model gives Y₁ = 25.2659 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = 5.56156 MHz; the input change is Δb = 0.19845 mT.
- Average sensitivity is ΔY_out/Δb = 28.025 MHz per mT. 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 = 19.7044 MHz, Y₁ = 25.2659 MHz; ΔY_out = 5.56156 MHz; average sensitivity = 28.025 MHz per mT.
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 moments and Larmor precession: infer an input with uncertainty
Larmor precession is the rigid rotation of a spin's expectation value ⟨S⟩ about the applied field, at angular rate ω = |γ|B, generated by the Zeeman Hamiltonian Ĥ = −γ B⋅Ŝ. Use the supplied topic model Y_out = 28.025*b, where Y_out is LARMOR f = |γ|B/2π (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Field B along z, measured in mT. Keep th (Tip angle θ from B) = 60 °; t (Elapsed time t) = 12 ns fixed. For this exercise use only 0.39125 mT ≤ b ≤ 1.80875 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 = 35.197298 MHz, with a bounded reading uncertainty ±2.3239 MHz. 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 0.00014175 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.
LARMOR f = |γ|B/2π against b
Predictions from the supplied Magnetic moments and Larmor precession 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): LARMOR f = |γ|B/2π / MHz
Horizontal axis (x): b: Field B along z / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Field B along z / mT | LARMOR f = |γ|B/2π / MHz |
|---|---|
| 0.39125 | 10.96478125 |
| 0.745625 | 20.896140624999997 |
| 1.1 | 30.8275 |
| 1.4543749999999998 | 40.75885937499999 |
| 1.8087499999999999 | 50.69021874999999 |
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(0.39125) = 10.9648 MHz and Y_out(1.80875) = 50.6902 MHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.39125 + (35.1973 − 10.9648)(1.80875 − 0.39125)/(50.6902 − 10.9648) = 1.25592 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 = 1.25592 mT; substituting back gives Y_out = 35.197298 MHz.
- The interpolation error in the input is |b_line − b| = 2.22045e-16 mT. This is within the requested numerical tolerance.
- Invert both limits Y_out = 32.873398 MHz and 37.521198 MHz. Sort the roots even if the response decreases: 1.173 mT ≤ b ≤ 1.33885 mT.
- The lower and upper input deviations are 0.0829224 mT and 0.0829224 mT. 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 ≈ 1.25592 mT; compatible interval [1.173, 1.33885] mT. Endpoint interpolation gives 1.25592 mT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.