Question

An electron is travelling with velocity $\vec{v}=3\hat{i}+5\hat{j}m{s}^{-1}$ in a magnetic field $\vec{B}=6\hat{i}+4\hat{j}$ tesla. What is the magnitude and direction of the force $\vec{F}$ acting on the electron?

• A. 18 e N along + ve z -axis
• B. 18 e N along - ve z -axis
• C. 36 e N along - ve z-axis
• D. 54 e N along +ve z -axis

Answer 1

Answer: (A)

Given: $\vec{v}=3\hat{i}+5\hat{j},\vec{B}=6\hat{i}+4\hat{j}$
Force on a particle of charge $q$, moving with velocity $\vec{v}$ is placed in a magnetic field $\vec{B}$ is given by $\vec{F}=q(\vec{v}\times \vec{B})$
$\vec{v}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 0\\ 6& 4& 0\end{array}\right|=\hat{i}[0-0]-\hat{j}[0-0]+\hat{k}[12-30]=-18\hat{k}$
For an electron $q=(-e)$
$\therefore \vec{F}=(-e)[-18\hat{k}]=18e\hat{k}$ along $+$ ve $z$ axis
$|\vec{F}|=18eN$