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 $\vec{F}=q(\vec{v} \times \vec{B})$
$\vec{v} \times \vec{B}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 0 \\ 6 & 4 & 0\end{vmatrix}=\hat{i}[0-0]-\hat{j}[0-0]+\hat{k}[12-30]=-18 \hat{k}$
For an electron $q=(-e)$
$\therefore \vec{F}=(-e)[-18 \hat{k}]=18 e \hat{k}$ along $+$ ve $z$ axis
$|\vec{F}|=18 e\, N$