Question

An electron moves north at a velocity of 4.5 × 104 m/s and has a magnetic force of 7.2 × 10-18 n exerted on it. if the magnetic field points upward, what is the magnitude of the magnetic field? use 1.60 × 10–19 c for the magnitude of the charge of an electron

Answer 1

The magnetic force on a moving charge is given by:

$F=qvB \sin \theta$

where q is the charge, v is the speed of the charge, B is the magnetic field intensity and $\theta$ is the angle between the directions of B and v.

In our problem, the charge is an electron ($q=1.6 \cdot 10^{-19}C$), the velocity of the charge is $v=4.5 \cdot 10^4 m/s$, the magnetic force is $F=7.2 \cdot 10^{-18} N$ and the angle between the direction of v and B is $\theta=90^{\circ}$, so $\sin \theta=1$ and we can ignore the sine in the formula. Therefore, if we rearrange the equation and we put these numbers in, we can find the intensity of the magnetic field:

$B=\frac{F}{qv}=\frac{7.2 \cdot 10^{-18} N}{(1.6 \cdot 10^{-19}C)(4.5 \cdot 10^{-4} m/s)}=0.001 T$