An electron moves with a speed of $8.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ along the $+ x$-axis. It enters a region where there is a magnetic field of $2.5 \mathrm {~T}$, directed at an angle of $60 ^ { \circ }$ to the $+ x$-axis and lying in the $x y$-plane.
Calculate the magnitude of the magnetic force on the electron. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$
A) $3.2 \times 10 ^ { - 12 } \mathrm {~N}$
B) $0 \mathrm {~N}$
C) $3.2 \times 10 ^ { - 10 } \mathrm {~N}$
D) $2.8 \times 10 ^ { - 12 } \mathrm {~N}$
E) $2.8 \times 10 ^ { - 10 } \mathrm {~N}$

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by $ F = qvB \sin \theta $. Here, $ q = 1.60 \times 10^{-19} \, \mathrm{C} $, $ v = 8.0 \times 10^6 \, \mathrm{m/s} $, $ B = 2.5 \, \mathrm{T} $, and $ \theta = 60^\circ $. Thus, $ F = (1.60 \times 10^{-19})(8.0 \times 10^6)(2.5) \sin 60^\circ = 2.8 \times 10^{-12} \, \mathrm{N} $.

An Electron Moves with a Speed Of 8.0×106 m/s8.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }8.0×106 m/s

An Electron Moves with a Speed Of $8.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$