Question

A particle of charge $q$and mass $m$moves in the uniform fields role="math" localid="1648977372617" $\stackrel{\rightharpoonup }{E}=E_0\hat{k}$and$\stackrel{\rightharpoonup }{B}=B_0\hat{k}$. At $t=0$, the particle has velocity role="math" localid="1648977203095" $\stackrel{\rightharpoonup }{V_0}=V_0\hat{i}$. What is the particle's speed at a later time $t$?

Short answer

The particle's speed at a later time$t$is$V=\sqrt{{v_0}^2+{\left(\frac{q{E}_{0}t}{m}\right)}^2}$

Step-by-step solution

Introduction

The given is $\stackrel{\rightharpoonup }{E}=E_0\hat{k},\stackrel{\rightharpoonup }{B}=B_0\hat{k},\stackrel{\rightharpoonup }{V}=V_0\hat{i},t=0$. The objective is to find the particle's speed at a later time $t$.

Step 1

$$\begin{gathered} \stackrel{\rightharpoonup }{F_E}=q\stackrel{\rightharpoonup }{E}=q{E}_0\hat{k} \\ \stackrel{\rightharpoonup }{F_M}=q\stackrel{\rightharpoonup }{V}\times \stackrel{\rightharpoonup }{B}=q{V}_0{B}_0(-j) \end{gathered}$$

In the$x-y$plane, for circular motion:

Step 2

$$\begin{gathered} \stackrel{\rightharpoonup }{F_M}=q{V}_0{B}_0\hat{r}\iff F_E=q{E}_0\hat{k} \\ \sum _{\stackrel{\rightharpoonup }{F}}=m\stackrel{\rightharpoonup }{a} \\ {F}_{1z}=q{V}_0{B}_0=m{a}_r=\frac{m{v_0}^2}{2} \\ {F}_{1t}=0 \end{gathered}$$

Finally, we come across $\stackrel{\rightharpoonup }{v}\left(t\right)$:

$$\begin{gathered} \stackrel{\rightharpoonup }{V}\left(t\right)=V_0\hat{r}+(0+a_{z}t)\hat{k} \\ \stackrel{\rightharpoonup }{V\left(t\right)}=V_0\hat{i}+a_{z}t\hat{k} \\ V=\sqrt{{V_0}^2+\left(a_{z}t\right)} \\ V=\sqrt{{V_0}^2+{\left(\frac{q{E}_{0}t}{m}\right)}^2} \end{gathered}$$