A proton travelling at 3.0 × 10$^6$ m s$^{-1}$ enters a uniform magnetic field of flux density 0.25 T. Its velocity is perpendicular to the field.
(a) Explain why the proton moves in a circular path. [2]
(b) Calculate the radius of the path. [2]
(c) Calculate the time taken for the proton to travel through a semicircle. [1]
- (a) the magnetic force ($BQv$) is always perpendicular to the velocity (Fleming's left-hand rule)(1)
- (a) so no work is done / speed is constant and the force (constant magnitude) acts towards a centre, providing the centripetal force(1)
- (b) $BQv = \frac{mv^2}{r}$ so $r = \frac{mv}{BQ}$(1)
- (b) $r = \frac{1.673\times10^{-27}\times3.0\times10^6}{0.25\times1.60\times10^{-19}}$ = 0.13 m (0.125)(1)
- (c) $t = \frac{\pi r}{v}$ (or $\frac{\pi m}{BQ}$) = 1.3 × 10$^{-7}$ s (allow 1.4 × 10$^{-7}$ s from $r$ = 0.13 m; ecf from (b))(1)
- Answer: (b) 0.13 m (c) 1.3 × 10$^{-7}$ s
- Examiner tip: For the 'explain', say the force is perpendicular to the velocity — 'the force is towards the centre' alone just restates what you are explaining.