Q109 / 180 · AIIMS 2003 · Physics
An electron is travelling along the $x$-direction. It encounters a magnetic field in the $y$-direction. Its subsequent motion will be
- Astraight line along the $x$-direction
- Correct answerBa circle in the $x z$ plane
- Ca circle in the $y z$ plane
- Da circle in the $x y$ plane.
Solution
We know that the force on a charged particle $e$ due to a magnetic field is given by
$$ \begin{array}{ll} & \vec{F}=e(\vec{v} \times \vec{B}) \\ \text { Here } \vec{v}=v_x=\hat{i} \text { and } \vec{B}=B_y \hat{j} \\ \therefore \quad & \vec{F}=e v_x B_y(\hat{i} \times \hat{j})=e v_x B_y \hat{k} \end{array} $$ Therefore the subsequent motion of the charged particle will be a circle in the $x z$ plane.
$$ \begin{array}{ll} & \vec{F}=e(\vec{v} \times \vec{B}) \\ \text { Here } \vec{v}=v_x=\hat{i} \text { and } \vec{B}=B_y \hat{j} \\ \therefore \quad & \vec{F}=e v_x B_y(\hat{i} \times \hat{j})=e v_x B_y \hat{k} \end{array} $$ Therefore the subsequent motion of the charged particle will be a circle in the $x z$ plane.
Hence, the correct answer is option: 'B'.