What is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T?
A) 9.9 mm
B) 1.1 cm
C) 1.3 cm
D) 1.4 cm
E) 7.6 mm

Question

Quizplus · Accepted Answer

The radius of curvature $r$ for a charged particle moving in a magnetic field is given by the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is the velocity of the particle, $q$ is the charge of the particle, and $B$ is the magnetic field strength. For a proton, the kinetic energy $KE = \frac{1}{2}mv^2$, so we can solve for $v = \sqrt{\frac{2KE}{m}}$. Given that the kinetic energy $KE = 3.0 \, \text{keV} = 3.0 \times 10^3 \, \text{eV} = 3.0 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J}$ (since $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$), the mass of a proton $m = 1.67 \times 10^{-27} \, \text{kg}$, and the charge of a proton $q = 1.6 \times 10^{-19} \, \text{C}$, we can find $v$ and subsequently $r$.Solving for $v$, we get $v = \sqrt{\frac{2 \times 3.0 \times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$ m/s. Plugging this value into the formula for $r$, along with $q = 1.6 \times 10^{-19} \, \text{C}$ and $B = 0.80 \, \text{T}$, gives us the radius of curvature.After calculating, the radius $r$ comes out to be approximately $9.9 \, \text{mm}$, which corresponds to choice A.

What Is the Radius of Curvature of the Path of a 3.0-KeV