Question

An electron moves through a uniform magnetic field given by $\overrightarrow{B}=B_x\hat{i}+\left(3B_x\right)\hat{j}$.At a particular instant,the electron has the velocity $\overrightarrow{v}=(2.0\hat{i}+4.0\hat{j})m/s$ and the magnetic force acting on it is $(6.4\times {10}^{-19}N)\hat{k}$.Find $x$ such that $x=-B_x$.

Answer 1

The force on a moving charged particle with velocity $\overrightarrow{v}$ placed in magnetic field $\overrightarrow{B}$ is given by

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$⟹6.4\times {10}^{-19}\hat{k}=(-1.6\times {10}^{-19})\left(\right[2\hat{i}+4\hat{j}]\times [B_x\hat{i}+3B_x\hat{j}\left]\right)$

$⟹-4T=6B_x-4B_x$

$⟹B_x=-2T$