Physics (AS & A Level) · Feb / March 2016 · Paper 42 · Question 9
9702
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The question as text
A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s–1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10–2 T, as shown in Fig. 9.1. particle charge +q mass m speed 1.6 × 105 m s–1 uniform magnetic field out of the plane of the paper flux density 9.7 × 10–2 T path of particle Fig. 9.1 The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields.
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9(a)[2]
direction of force due to electric field opposite to force due to magnetic field electric field is up the page [2]
9(b)[3]
force due to electric field = force due to magnetic field or Eq = Bqv E = Bv = 9.7 × 10–2 × 1.6 × 105 = 1.6 (1.55) × 104 V m–1 [3]
9(c)[3]
q / m = v / Br = 1.6 × 105 / (9.7 ×10–2 × 4.0 × 10–2) = 4.1 (4.12) × 107 C kg–1 [3]
9(d)(i)[2]
m = (3 × 1.60 × 10–19) / (4.12 × 107) m = 1.16 × 10–26 / 1.66 × 10–27 = 7(.0) u (allow 7.1 u) [2]
9(d)(ii)[1]
3 protons, 4 neutrons [1]
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