Cambridge International AS & A Level · 20 · structured
Derive Hall voltage and use a Hall probe
Derive Hall voltage and use a Hall probe · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 10
- Topics
- 2
- Answer
- Complete
A rectangular conducting strip carries current I along x. Its width w is along y and thickness t is along z. A uniform magnetic field B is along z. Let n be mobile carrier number density, q the magnitude of one carrier's charge and v_d the drift-speed magnitude. The Hall voltage is measured across width w.
- (a)
Explain the charge separation and derive V_H=BI/(ntq), beginning with the balance of electric and magnetic forces. Show why thickness t, not width w, remains in the denominator.
5 marks - (b)
Calculate the Hall-voltage magnitude for B=0.300 T, I=0.0200 A, n=1.00×10²² m⁻³, t=1.00×10⁻³ m and q=1.60×10⁻¹⁹ C.
2 marks - (c)
Describe zero correction and orientation of a calibrated Hall probe. Starting with the sensing-plane normal parallel to B, predict the reading when that normal is rotated through 90° and 180° relative to B; distinguish rotation about the normal.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Explain the charge separation and derive V_H=BI/(ntq), beginning with the balance of electric and magnetic forces. Show why thickness t, not width w, remains in the denominator.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
Magnetic force deflects moving carriers sideways, producing charge separation and a transverse electric field.
- 2
At equilibrium electric and magnetic force magnitudes balance: qE_H=qv_dB, so E_H=v_dB.
- 3
V_H=E_Hw=v_dBw.
- 4
Current I=nqv_dA=nqv_dwt, so v_d=I/(nqwt).
- 5
Substitute into V_H: width cancels and V_H=BI/(ntq); polarity depends on carrier sign and field/current directions.
(b)
Calculate the Hall-voltage magnitude for B=0.300 T, I=0.0200 A, n=1.00×10²² m⁻³, t=1.00×10⁻³ m and q=1.60×10⁻¹⁹ C.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
V_H=0.300×0.0200/[(1.00×10²²)(1.00×10⁻³)(1.60×10⁻¹⁹)].
- 2
The denominator is 1.60, so V_H=0.00375 V=3.75 mV.
(c)
Describe zero correction and orientation of a calibrated Hall probe. Starting with the sensing-plane normal parallel to B, predict the reading when that normal is rotated through 90° and 180° relative to B; distinguish rotation about the normal.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
Use a fixed probe current and its voltage-to-field calibration; subtract any zero offset measured without the applied field.
- 2
Place the sensing element at the required point with the field perpendicular to its sensing plane for maximum |V_H|.
- 3
Rotating the sensing-plane normal from parallel to B through 90° makes the normal component zero; through 180° reverses polarity. Rotation about the normal alone does not change that component.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- (a) Magnetic force deflects moving carriers sideways, producing charge separation and a transverse electric field.
- (a) At equilibrium electric and magnetic force magnitudes balance: qE_H=qv_dB, so E_H=v_dB.
- (a) V_H=E_Hw=v_dBw.
- (a) Current I=nqv_dA=nqv_dwt, so v_d=I/(nqwt).
- (a) Substitute into V_H: width cancels and V_H=BI/(ntq); polarity depends on carrier sign and field/current directions.
- (b) V_H=0.300×0.0200/[(1.00×10²²)(1.00×10⁻³)(1.60×10⁻¹⁹)].
- (b) The denominator is 1.60, so V_H=0.00375 V=3.75 mV.
- (c) Use a fixed probe current and its voltage-to-field calibration; subtract any zero offset measured without the applied field.
- (c) Place the sensing element at the required point with the field perpendicular to its sensing plane for maximum |V_H|.
- (c) Rotating the sensing-plane normal from parallel to B through 90° makes the normal component zero; through 180° reverses polarity. Rotation about the normal alone does not change that component.
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