# A beam of electrons travels at $3 \times 10^{6} m/s$ through a uniform magnetic field of $4 \times 10^{-2} T$ at right angles to the field, as shown in the figure. How strong is the force acting on each electron?

$\begin{array}{1 1} (A) -1.9 \times 10^{-14} N \\(B) -1.6 \times 10^{-19} N \\ (C) -12 \times 10^{-14} N \\ (D) -4.8 \times 10^{-14} N\end{array}$

Given, $v = 4.0 \times 10^{6} m/s$, $B = 4.0 \times 10^{-2} T$
The force on the beam $F$ as illustrated in the figure is opposite the force given by the third right hand rule due to the electron's negative charge $\rightarrow$ We know that the charge of an electron is $1.6 \times 10^{-19} C$, in this case, we have reverse the sign.
Therefore Force acting on each electron $F$ is determined by $F = qvB \sin \theta$, where $\theta$ is the angle between the charge velocity and the magnetic field.
In this case, $\theta = 90 ^{\circ} \rightarrow \sin \theta = \sin 90 ^{\circ} = 1$
$\Rightarrow F = qvB = -1.6 \times 10^{-19} C \times 3.0 \times 10^{6} m/s \times 4.0 \times 10^{-2} T = -1.9 \times 10^{-14} N$