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A charged particle projected in a magnetic field [latex]B = (3hat{i}+4hat{j})times 10^{-2}T[/latex] .The acceleration of the particle is found to be [latex]a = (xhat{i}+2hat{j})m/s^{2}[/latex]. Find the value of x.

Answer: -8/3 m/s2 Solution: \(\begin{array}{l}\vec{F}=q(\vec{v}\times \vec{B})\end{array} \)   \(\begin{array}{l}\vec{m\vec{a}}=q(\vec{v}\times \vec{B})\end{array} \)   \(\begin{array}{l}\vec{a}\propto (\vec{v}\times \vec{B})\end{array} \)   \(\begin{array}{l}\vec{a}\perp \vec{V}\end{array} \)... View Article