Electrons enter a rectangular slice PQRSEFGH of a semiconductor material at right-angles to face PQFE, as shown in Fig. 8.1. [Figure 8.1] A uniform magnetic field of flux density B is directed into the slice, at right-angles to face PQRS.
- (a)The electrons each have charge –q and drift speed v in the slice. State the magnitude and the direction of the force due to the magnetic field on each electron as it enters the slice.[2]
- (b)(i)State the two faces between which the voltage V_H is established. face ..................................................... and face ..................................................... [1][1]
- (b)(ii)Use letters from Fig. 8.1 to identify the distance t.[1]
- (c)(i)Show that the number of charge carriers per unit volume in aluminium is 6.0 × 10²⁸ m⁻³.[2]
- (c)(ii)A sample of aluminium foil has a thickness of 0.090 mm. The current in the foil is 4.6 A. A uniform magnetic field of flux density 0.15 T acts at right-angles to the foil. Use the value in (i) to calculate the voltage V_H that is generated. V_H = ..................................................... V [2][2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [B1] magnitude: (force =) Bqv
- [B1] direction: P→Q or E→F or S→R or H→G
- [B1] EHSP and FGRQ
- [B1] PE or QF or RG or SH
- [C1] any one correct starting point from: (mass of 1 atom =) 27 × 1.66 × 10⁻²⁷ / (amount of substance per unit volume =) 2.7/27 / 27 g (of substance) contains 6.02 × 10²³ atoms / (2.7 g mass contains) 0.1 mol / (1 cm³ volume contains) 0.1 mol / (1 m³ volume contains) 10⁵ mol
- [A1] n = (2.7 × 10³) / (27 × 1.66 × 10⁻²⁷) = 6.0 × 10²⁸ or n = (2.7/27) × 10⁶ × 6.02 × 10²³ = 6.0 × 10²⁸
- [C1] VH = (0.15 × 4.6) / (6.0 × 10²⁸ × 0.090 × 10⁻³ x 1.60 × 10⁻¹⁹)
- [A1] = 8.0 x 10⁻⁷ V
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