An electron moves in a region where the magnetic field is uniform, has a magnitude of 60 μT, and points in the positive x direction. At t = 0 the electron has a velocity that has an x component of 30 km/s, a y component of 40 km/s, and a z component of zero. What is the radius of the resulting helical path?
A) 4.7 mm
B) 18 mm
C) 3.8 mm
D) 2.8 mm
E) 5.7 mm

Question

Quizplus · Accepted Answer

The radius $r$ of the helical path of an electron moving in a magnetic field is given by the formula $r = \frac{mv_{\perp}}{|q|B}$, where $m$ is the mass of the electron ($9.11 \times 10^{-31}$ kg), $v_{\perp}$ is the component of the velocity perpendicular to the magnetic field (in this case, the y-component, 40 km/s or $4 \times 10^4$ m/s), $q$ is the charge of the electron ($1.6 \times 10^{-19}$ C), and $B$ is the magnetic field (60 μT or $60 \times 10^{-6}$ T). Plugging in the values:$$r = \frac{9.11 \times 10^{-31} \cdot 4 \times 10^4}{1.6 \times 10^{-19} \cdot 60 \times 10^{-6}}$$$$r = \frac{3.644 \times 10^{-26}}{9.6 \times 10^{-25}}$$$$r \approx 3.8 \times 10^{-3} \text{ m}$$$$r \approx 3.8 \text{ mm}$$

An Electron Moves in a Region Where the Magnetic Field