Question

a)Find the minimum magnetic field needed to exert a 5.7 fN force on an electron moving at 23Mm/s .b)Find the field strength required if the field were at 45 degrees to the electron's velocity.

Answer 1

a) $1.55\cdot 10^{-3} T$

The magnetic force exerted on a charged particle in motion is given by:

$F=qvB sin \theta$

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

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

The expression can be rewritten as

$B=\frac{F}{qv sin \theta}$

We see that the minimum magnetic field needed is the one for which $sin \theta=1$, so with $\theta=90^{\circ}$. In this problem, we have:

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

$f=5.7 fN=5.7\cdot 10^{-15}N$ is the force

$v=23 Mm/s = 23\cdot 10^6 m/s$ is the electron's velocity

Substituting, we find

$B=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 90^{\circ}}=1.55\cdot 10^{-3} T$

b) $2.19\cdot 10^{-3} T$

In this case, the field is at 45 degrees to the electron's velocity, so we have

$\theta=45^{\circ}$

Therefore, the field strength required to obtain a force of

$f=5.7 fN=5.7\cdot 10^{-15}N$ is the force

will be equal to

$B=\frac{F}{qv sin \theta}=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 45^{\circ}}=2.19\cdot 10^{-3} T$