University Physics II · topic UPII-08.08
The Hall effect
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.
The Hall effect: predict a finite change
The Hall effect is the transverse voltage across a current-carrying conductor in a magnetic field, appearing once the charge driven to one edge builds an electric field that cancels the magnetic force on the carriers. Use the supplied topic model Y_out = 1.5625*i/tt, where Y_out is DRIFT SPEED vd (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. tt is Plate thickness t, measured in mm. Keep i (Probe current I) = 5 mA; B (Field B, along the thickness) = 0.5 T fixed. For this exercise use only 0.103375 mm ≤ tt ≤ 0.276625 mm 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 tt = 0.14149 mm and 0.165745 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DRIFT SPEED vd against tt
Predictions from the supplied The Hall effect 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): DRIFT SPEED vd / m s⁻¹
Horizontal axis (x): tt: Plate thickness t / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tt: Plate thickness t / mm | DRIFT SPEED vd / m s⁻¹ |
|---|---|
| 0.103375 | 75.57436517533253 |
| 0.14668749999999997 | 53.25948018747338 |
| 0.18999999999999997 | 41.11842105263158 |
| 0.23331249999999998 | 33.4851326011251 |
| 0.27662499999999995 | 28.242205151378226 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of tt separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At tt = 0.14149 mm, substitute into Y_out = 1.5625*i/tt: Y₀ = 55.2159 m s⁻¹.
- At tt = 0.165745 mm, the same fixed-parameter model gives Y₁ = 47.1357 m s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -8.08026 m s⁻¹; the input change is Δtt = 0.024255 mm.
- Average sensitivity is ΔY_out/Δtt = -333.138 m s⁻¹ per mm. Its negative sign means that this response decreases 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 = 55.2159 m s⁻¹, Y₁ = 47.1357 m s⁻¹; ΔY_out = -8.08026 m s⁻¹; average sensitivity = -333.138 m s⁻¹ per mm.
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.
The Hall effect: infer an input with uncertainty
The Hall effect is the transverse voltage across a current-carrying conductor in a magnetic field, appearing once the charge driven to one edge builds an electric field that cancels the magnetic force on the carriers. Use the supplied topic model Y_out = 1.5625*i/tt, where Y_out is DRIFT SPEED vd (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. tt is Plate thickness t, measured in mm. Keep i (Probe current I) = 5 mA; B (Field B, along the thickness) = 0.5 T fixed. For this exercise use only 0.103375 mm ≤ tt ≤ 0.276625 mm 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 = 37.370102 m s⁻¹, with a bounded reading uncertainty ±1.3692 m 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 tt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000017325 mm, 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.
DRIFT SPEED vd against tt
Predictions from the supplied The Hall effect 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): DRIFT SPEED vd / m s⁻¹
Horizontal axis (x): tt: Plate thickness t / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tt: Plate thickness t / mm | DRIFT SPEED vd / m s⁻¹ |
|---|---|
| 0.103375 | 75.57436517533253 |
| 0.14668749999999997 | 53.25948018747338 |
| 0.18999999999999997 | 41.11842105263158 |
| 0.23331249999999998 | 33.4851326011251 |
| 0.27662499999999995 | 28.242205151378226 |
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(tt) = 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(0.103375) = 75.5744 m s⁻¹ and Y_out(0.276625) = 28.2422 m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tt_line = 0.103375 + (37.3701 − 75.5744)(0.276625 − 0.103375)/(28.2422 − 75.5744) = 0.243214 mm. 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 tt = 0.209057 mm; substituting back gives Y_out = 37.370102 m s⁻¹.
- The interpolation error in the input is |tt_line − tt| = 0.0341566 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 36.000902 m s⁻¹ and 38.739302 m s⁻¹. Sort the roots even if the response decreases: 0.201669 mm ≤ tt ≤ 0.217008 mm.
- The lower and upper input deviations are 0.00738892 mm and 0.00795095 mm. 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 tt ≈ 0.209057 mm; compatible interval [0.201669, 0.217008] mm. Endpoint interpolation gives 0.243214 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.