An electron is travelling in a vacuum at a speed of 3.4 × 107 m s-1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in Fig. 8.1.

 Question 17

An electron is travelling in a vacuum at a speed of 3.4 × 10⁷ m s^-1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in Fig. 8.1.

Fig. 8.1

 

The initial direction of the electron is at an angle of 30° to the direction of the magnetic field.

 

(a) When the electron enters the magnetic field, the component of its velocity vN normal to the direction of the magnetic field causes the electron to begin to follow a circular path.

 

Calculate:

(i) vN [1]

(ii) the radius of this circular path. [3]

 

 

(b) State the magnitude of the force, if any, on the electron in the magnetic field due to the component of its velocity along the direction of the field. [1]

 

 

(c) Use information from (a) and (b) to describe the resultant path of the electron in the magnetic field. [1]

[Total: 6]

 

 

Reference: Past Exam Paper – June 2019 Paper 42 Q8

 

Solution:

(a)

(i)

vN = 3.4×10⁷ × sin 30°

vN = 1.7 × 10⁷ m s^-1

 

 

(ii)

{The magnetic force provides the centripetal force.

Magnetic force = Bqv}

 

mv² / r = Bqv               or r = mv / Bq

 

{Making r the subject of formula,

r = mv / Bq }

 

r = (9.11×10^-31 × 1.7×10⁷) / (3.2×10^-3 × 1.60×10^-19)

r = 0.030 m                                                                

 

 

(b) zero

{It is only the component of the velocity which is perpendicular to the field that causes a force.}

 

 

(c) helix/coil

 

{The motion can be described in two components:

- circular motion in one plane (due to the component perpendicular to the field)

- linear motion in the third direction}