A proton with a kinetic energy of 0.20 keV follows a circular path in a region where the magnetic field is uniform and has a magnitude of 60 mT. What is the radius of this path?
A) 4.1 cm
B) 2.9 cm
C) 3.4 cm
D) 5.1 cm
E) 2.4 cm

Question

Quizplus · Accepted Answer

The radius of the circular path is given by:
$r=\frac{mv}{qB}$

where $m$ is the mass of the proton, $v$ is its velocity, $q$ is its charge, and $B$ is the magnitude of the magnetic field.

The velocity of the proton can be found using its kinetic energy:
$K=\frac{1}{2}mv^2$
$v=\sqrt{\frac{2K}{m}}$

Substituting these values into the equation for the radius, we get:
$r=\frac{m}{qB}\sqrt{\frac{2K}{m}}$
$r=\frac{\sqrt{2mK}}{qB}$

Plugging in the given values for $m$, $K$, $q$, and $B$, we get:
$r=\frac{\sqrt{2(1.67	imes 10^{-27}\,kg)(0.20\,keV)	imes 1.6	imes 10^{-19}\,C}}{(1.6	imes 10^{-19}\,C)(0.06\,T)}\approx 3.4\,cm$

Therefore, the answer is C: 3.4 cm.

A Proton with a Kinetic Energy of 0