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香港数学试题


(2003CEQ6) (a) Two essential components: *a coil *a magnet *a commutator *a pair of brushes *a soft iron core (b) When the blades are turned, the coil inside the motor will turn in the magnetic field of the magnet. An induced e. m. f. will be set up in the coil. The induced current flows through the bulb and lights up the bulb. (1993CEQ5) (a) (i) Sprinkle iron filings on the board and tap the board gently. The magnetic field pattern is shown by the pattern of the filings. (ii) 圖形 (b) (i)When s is pressed, current flows through the coil and the soft iron core is magnetized. It attracts the spring towards the left. The hammer strikes the left metal plate to produce the first note. (ii) Replace one metal plate with anther made of a different metal *Vary the length (or other dimensions) of one of the tow metal plates *Stick a lump of plasticine to one plate (iii)Statement 1 is correct because copper is not a magnetic material. Statement 2 is incorrect. If the polarities of the battery are reversed, the soft iron core will still be magnetized.

1994Q7(C) (C) (i) When a current flows in the direction ABCD, forces acting on AD and BC as shown below; 圖形 The coil rotates and moves pointer. The rotation of the coil is opposed by the hairspring. The coil will come to a rest and the current is shown by the deflection of the pointer. (ii) Any Two of the following; *Increasing the number of turns of the coil *Increasing the strength of the magnetic field *Using weaker hairsprings *Increasing the area of the coil

*Using a light-beam galvanometer (iii)The galvanometer cannot measure 50 Hz alternating currents because *the pointer will always point at zero reading. *the pointer will vibrate about the zero reading quickly and is unable to give a steady reading. OR The galvanometer can measure alternating currents only if a diode is built into it. The diode allows current to flow in only one direction through the meter.
[1986ALQ5] 5. (a) (i) The result shows that the induced e.m.f. in the search coil is proportional to 1/r.B ??1/rfor a long, current carrying, straight wire. (ii) When the distance of the search coil from thestraight wire is comparable to the finite length of the wire,the induced e.m.f. is less than it should be fora infinitely long wire. Therefore, those data points lie above the fitted straight line. (b) (i) B = ?0I/2?r = ?0I0 sin (2?ft)/2?r emf = NAdB/dt = (?0NAI0f/r) cos (2?ft) V = 2?0NAI0f/r (ii) Slope of line = 1800 V-1m-1 = 1/(2?0NAI0f) N =1/(2 ×1.26×10^?6 ×?3.14×?10^?4 ×14.1×?50) turns ~ 1000 turns 1 (c) His argument is wrong. The earth’s magnetic field is more or less a steady, intensity fieldwhich would not give rise to any measurable inducede.m.f. in the search coil.

[1998AL IB Q6]

2000Q7

2000Q8:

2001ALQ4(a)

1999ALQ3

2003CEQ6

2002CEQ6

1995CEQ5(b)

1993Q5

1994Q7(C)

2003Q1

2003Q4

1993ALQ10

96-IA-Q1

03-1B-Q7(AL)

2000-I-Q1

1989ALQ10
10. (a) (i) I = nevA 50 ??10-3 = 1029 ??1.6 ??10-19 ?? (0.02 ??0.1 ??10-3) ??v v = 1.6 ??10-6 ms-1 (ii) direction marked correctly F= Bev

= 1.5 ??(1.6 ??10-19) ??(1.6 ??10-6) N = 3.8 ??10-25 N 1 (b) - force moves electron to far end - negative charges build up at far end - E field set up until it is large enough to prevent further movement - steadyp.d. develops (c) (i) semiconductor has lower carrier concentration larger Hall voltage observed (ii) - X, Y not exactly opposite - smallp.d. exists due to longitudinal electric field set up by battery - adjust R until voltmeter gives nil deflection (d) Bev = eVH/d VH = Bvd= B(I/neA)d = BI/net = (0.2)(10-3)(1020)-1(1.6 ??10-19)-1(0.1 ??10-3)-1 V = 0.125 V (e) Hall probe usedin magnetic field measurements

1998Q7

1996ALQ8

1995ALQ4


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