Question

A particular of charge q and mass m starts moving from the origin under the action of an electric field $\overrightarrow{E}$ = $E_0^i$ and $\overrightarrow{B}$ = $B_0^i$with a velocity $\overrightarrow{V}$ = ${\nu }_0^j$. The speed of the particle will becomes $\frac{\sqrt{5}}{2}{\nu }_0$ after a time

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

Answer 1

The correct option is C $\frac{\sqrt{3}m{\nu }_0}{2qE}$

$v=\sqrt{v_0^2+(\frac{qEt}{m}{)}^2}=\frac{\sqrt{5}}{2}{v}_0$

$v_0^2+(\frac{qET}{m}{)}^2=\frac{5}{4}{v}_0^2.$

$(\frac{qET}{m}{)}^2=\frac{3}{4}{v}_0^2$

T=$\frac{\sqrt{3}m{v}_0}{2qE}$