Question

A particle of charge \(q\) and mass \(m\) starts moving from the origin under the action of an electric field \(E=E_{0}\) cap i and \(B=B_{0}\) cap i with velocity vector \(v=v_{0}\) cap \(j\) The speed of the particle will become \(2 \mathrm{v}_{0}\) after a time.
(A) \(\mathrm{t}=\frac{2 \mathrm{mv}_{0}}{\mathrm{qE}}\)
(B) \(\mathrm{t}=\frac{2 \mathrm{~Bq}}{\mathrm{mv}_{0}}\)
(C) \(\mathrm{t}=\frac{3 \mathrm{~Bq}}{\mathrm{mv}_{0}}\)
(D) \(t=\frac{\sqrt{3} m v_{0}}{q E}\)

Answer 1

Final velocity of the particle
$$
\begin{aligned}
&=v=\sqrt{v_{0}^{2}+\left(\frac{q E t}{m}\right)^{2}}=2 v_{0} \\
&v_{0}^{2}+\left(\frac{q E t}{m}\right)^{2}=4 v_{0}^{2} \\
&\left(\frac{q E t}{m}\right)^{2}=3 v_{0}^{2} \quad \Rightarrow t=\frac{\sqrt{3} m v_{0}}{q E}
\end{aligned}
$$

Answer 2

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Answer 3

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