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

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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, $\theta = 90 ^{\circ} \rightarrow \sin \theta = \sin 90 ^{\circ} = 1$

$\Rightarrow F = qvB = -1.6 \times 10^{-19} C \times 3.0 \times 10^{6} m/s \times 4.0 \times 10^{-2} T = -1.9 \times 10^{-14} N$

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