1. (a) (i) equally spaced horizontal parallel lines
from plate to plate (1)
arrows towards cathode (1) 2
(ii) ½
mv2 = qV; v = √(2eV/m) = √(2
× 1.6 × 10–19 × 7000/9.1 × 10–31)
so (1)
v = 4.96 × 107 (m s–1)
(1) 2
(b) (i) arrow perpendicular to path towards centre
of arc (1) 1
(ii) out of paper/upwards;using Fleming’s LH
rule (for conventional
current) (2) 2
(iii) mv2/r;
= Bqv; r = mv/Bq ;= 9.4 × 10–2
(m) 4
(c) change
magnitude of current in coils to change field; (1)
change field to change deflection; (1)
reverse field/current to change deflection from up to down (1) max 2 2
[13]
2. At
least 3 field lines inside solenoid parallel to axis; (1)
Lines equally spaced over some of length of solenoid. (1)
Arrows on lines pointing left to right. (1) 3
[3]
4. (i) I = V/R = 12/50 (1)
= 0.24 A (1) 2
(ii) Power
in primary = power in secondary / IpVp = IsVs (1)
Ip = 0.24 × 12 / 230 = 0.0125 A (1) 2
[4]
5.
(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2
(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1)
(2) 2
(d) (i) semicircle to
right of hole (1) ecf(a); (a) and d(i) to be consistent 1
(ii) mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19);
(1)
r = 55 mm;so distance = 2r = 0.11 m (2) 5
[13]
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