Question

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $E=E_{0}iandB=B_{0}i$ with a velocity $V=V_{0}i$.The speed of the article will become$\frac{\sqrt{5}}{2}{v}_0$ after a time

• A. $\frac{m{v}_0}{qE}$
• B. $\frac{m{v}_0}{2qE}$
• C. $\frac{\sqrt{3}m{v}_0}{1qE}$
• D. $\frac{\sqrt{5}-m{v}_0}{1qE}$

Answer 1

The correct option is B $\frac{m{v}_0}{2qE}$

Here , $E$ and $B$ are acting along x-axis and $v$ is along y axis , ie , perpendicular to the bothe $E$ abd $B$ . Therefore , the path particle ia a helix with increasing speed.Speed of particle at time $t$ is $v=\sqrt{v_x^2+v_y^2}$
Here,$v_y=v_0;v_x=\frac{qE}{m}t$.........(i)
and $v=\frac{\sqrt{5}}{2}{v}_0$
Hence , we have , $t=\frac{m{v}_0}{2qE}$