Question

A uniform magnetic field points upward, parallel to the page, and has a magnitude of 7.85 mt. a negatively charged particle (q=−3.32 μc, m=2.05 pg) moves through this field with a speed of 67.3 km/s perpendicular to the magnetic field, as shown. the magnetic force on this particle is a centripetal force and causes the particle to move in a circular path. what is the radius of the particle's circular path?

Answer 1

The Lorentz force exerted on the particle due to the magnetic field provides the centripetal force that keeps the particle in circular motion:
$m \frac{v^2}{r} = qvB$
where
m is the particle mass
v is its velocity
r its orbital radius
q its charge
B the magnetic field intensity
and where we neglected the factor $\sin \theta$ in the Lorentz force formula because the particle is traveling perpendicular to the magnetic field, so $\sin \theta=1$ .

Re-arranging the formula, we get
$r= \frac{mv}{qB}$ (1)
The problem gives us all the data about the particle and the magnetic field:
$m=2.05 pg= 2.05 \cdot 10^{-12} g=2.05 \cdot 10^{-15} kg$
$v=67.3 km/s = 67.3 \cdot 10^3 m/s$
$q=3.32 \mu C = 3.32 \cdot 10^{-6} C$ (we are only interested in the magnitude of the charge)
$B=7.85 mT = 7.85 \cdot 10^{-3}T$

And by plugging these numbers into eq.(1), we find the radius of the orbit
$r= \frac{mv}{qB}= \frac{(2.05 \cdot 10^{-15} kg)(67.3 \cdot 10^3 m/s)}{(3.32 \cdot 10^{-6} C)(7.85 \cdot 10^{-3} T)}=5.29 \cdot 10^{-3} m$