Question 18M.2.SL.TZ1.5
| Date | May 2018 | Marks available | [Maximum mark: 5] | Reference code | 18M.2.SL.TZ1.5 |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Explain, State | Question number | 5 | Adapted from | N/A |
An electron moves in circular motion in a uniform magnetic field.
The velocity of the electron at point P is 6.8 × 105 m s–1 in the direction shown.
The magnitude of the magnetic field is 8.5 T.
State the direction of the magnetic field.
[1]
out of the page plane / ⊙
Do not accept just “up” or “outwards”.
[1 mark]
Calculate, in N, the magnitude of the magnetic force acting on the electron.
[1]
1.60 × 10–19 × 6.8 × 105 × 8.5 = 9.2 × 10–13 «N»
[1 mark]
Explain why the electron moves at constant speed.
[1]
the magnetic force does not do work on the electron hence does not change the electron’s kinetic energy
OR
the magnetic force/acceleration is at right angles to velocity
[1 mark]
Explain why the electron moves on a circular path.
[2]
the velocity of the electron is at right angles to the magnetic field
(therefore) there is a centripetal acceleration / force acting on the charge
OWTTE
[2 marks]
