Q122 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
A charge of $+1 \mathrm{C}$ is moving with velocity $\vec{V}=(2 \mathrm{i}+2 \mathrm{j}-\mathrm{k}) \mathrm{ms}^{-1}$ through a region in which electric field $\vec{E}=(\mathrm{i}+\mathrm{j}-3 \mathrm{k}) \quad \mathrm{NC}^{-1}$ and magnetic field $\vec{B}=(\mathrm{i}-2 \mathrm{j}+3 \mathrm{k}) \mathrm{T}$ are present. The force experienced by the charge is
- A$(5 i+6 j-7 k) N$
- B$(-5 i-6 j-7 k) N$
- C$(-5 i+6 j+9 k) N$
- Correct answerD$(5 i-6 j-9 k) N$
Solution
$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$, where $q = +1$ C, $\vec{v} = 2\hat{i}+2\hat{j}-\hat{k}$, $\vec{E} = \hat{i}+\hat{j}-3\hat{k}$, $\vec{B} = \hat{i}-2\hat{j}+3\hat{k}$
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 1 & -2 & 3 \end{vmatrix}$
$= \hat{i}(6-2) - \hat{j}(6+1) + \hat{k}(-4-2) = 4\hat{i} - 7\hat{j} - 6\hat{k}$
$\vec{F} = (\hat{i}+\hat{j}-3\hat{k}) + (4\hat{i}-7\hat{j}-6\hat{k})$
$\vec{F} = 5\hat{i} - 6\hat{j} - 9\hat{k}$ N
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 1 & -2 & 3 \end{vmatrix}$
$= \hat{i}(6-2) - \hat{j}(6+1) + \hat{k}(-4-2) = 4\hat{i} - 7\hat{j} - 6\hat{k}$
$\vec{F} = (\hat{i}+\hat{j}-3\hat{k}) + (4\hat{i}-7\hat{j}-6\hat{k})$
$\vec{F} = 5\hat{i} - 6\hat{j} - 9\hat{k}$ N
Hence, the correct answer is option: 'D'.