Question

if an electron in an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T west, what are the direction and the magnitude of the velocity?

Answer 1

Answer:

$1.51\cdot 10^6 m/s$ north

Explanation:

When a charged particle moves in a magnetic field, the particle experiences a force given by the formula:

$F=qvB sin \theta$

where

q is the magnitude of the charge

v is its velocity

B is the magnetic field

$\theta$ is the angle between the directions of v and B

In this problem,

$q=1.6\cdot 10^{-19}C$ (charge of the electron)

$B=8.3\cdot 10^{-2} T$ (strength of magnetic field)

$F=2.0\cdot 10^{-14} N$ (force)

$\theta=90^{\circ}$

Therefore, the velocity is

$v=\frac{F}{qB sin \theta}=\frac{2.0\cdot 10^{-14}}{(1.6\cdot 10^{-19})(8.3\cdot 10^{-2})(sin 90^{\circ})}=1.51\cdot 10^6 m/s$

The direction of the force is perpendicular to both the direction of the velocity and the magnetic field, and it can be found using the right-hand rule:

. Thumb: direction of the force (downward) --> however the charge is negative, so this direction must be reversed: upward

- Middle finger: direction of the field (west)

- Index finger: direction of velocity --> north

So, the electron is travelling north.