Question

An electron moving at 4.40 ✕ 103 m/s in a 1.45 T magnetic field experiences a magnetic force of 1.40 ✕ 10−16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°.

Answer 1

Step-by-step explanation:

Charge on electron, e = - 1.6 x 10^-19 C

Magnetic field, B = 1.45 T

Force, F = 1.4 x 10^-16 N

The formula for the magnetic force is given by

F = q x v x B x Sinθ

Where, θ is the angle between the velocity vector and the magnetic field vector.

1.4 x 10^-16 = - 1.6 x 10^-19 x 4.4 x 10^3 x 1.45 x Sinθ

Sin θ = - 0.1371

θ = 172°

Answer 2

Step-by-step explanation:

Given that,

Speed of an electron,
`v=4.4* 10^3\ m/s`

Magnetic field, B = 1.45 T

Magnetic force,
`F=1.4* 10^(-16)\ N`

The magnetic force is given by in terms of charge, speed and magnetic field on which the charged particle is kept :

`F=qvB\sin\theta\\\\\sin\theta=(F)/(qvB)\\\\\sin\theta=(1.4* 10^(-16))/(1.6* 10^(-19)* 4.4* 10^3* 1.45)\\\\\sin\theta=0.137\\\\\theta=7.88^(\circ)`

or

`\theta=180-7.88=172.12^(\circ)`

Hence, this is the required solution.