Question

An electron experiences a magnetic force of magnitude 4.60 xx 10^-15 N, when moving at an angle of 60^@ with respect to a magnetic field of magnitude 3.50 xx 10^-3 T. Find the speed of the electron.

Answer 1

The speed of the electron is

Explanation:

Given :

Force $F = 4.60 \times 10^{-15}$ N

Magnetic field $B = 3.50 \times 10^{-3}$ T

According to force experience in a magnetic field,

   $F = qvB \sin \theta$

Where $\theta =$ angle between velocity and magnetic field.

In our case angle between velocity and magnetic field is 60°

So speed of electron is,

    $v = \frac{F}{qB \sin 60}$                                       ( $q = 1.6 \times 10^{-19}$ )

    $v = \frac{4.6 \times 10^{-15} }{1.6 \times 10^{-19 \times 3.5 \times 10^{-3} \times 0.866} }$

    $v = 9.48 \times 10^{6}$ $\frac{m}{s}$