Question

An experiment's diary reads as follows; "a charged particle is projected in a magnetic field of $(7.0\hat{i}-3.0\hat{j})\times {10}^{-3}T$. The acceleration of the particles is found to be $(x\hat{i}+7.0\hat{j})\times {10}^{-6}m/s^2$. Find the value of $x$.

Answer 1

We know that- $\overrightarrow{F}=m\overrightarrow{a}=q\overrightarrow{v}\times B$

$⟹\overrightarrow{a}\perp \overrightarrow{B}$

$⟹\overrightarrow{a}.\overrightarrow{B}=0$

$⟹7x-21=0$

$⟹x=3$