Chapter 29: Q. 67 (page 796)

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

Short Answer

Expert verified

The particle's speed at a later time $t$ is $V = \sqrt{{v_{0}}^{2} + {(\frac{q E_{0} t}{m})}^{2}}$

Step by step solution

Introduction

The given is $\overset{⇀}{E} = E_{0} \hat{k}, \overset{⇀}{B} = B_{0} \hat{k}, \overset{⇀}{V} = V_{0} \hat{i}, t = 0$ . The objective is to find the particle's speed at a later time $t$ .

Step 1

$\overset{⇀}{F_{E}} = q \overset{⇀}{E} = q E_{0} \hat{k} \overset{⇀}{F_{M}} = q \overset{⇀}{V} \times \overset{⇀}{B} = q V_{0} B_{0} (- j)$

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

Step 2

$\overset{⇀}{F_{M}} = q V_{0} B_{0} \hat{r} \Leftrightarrow F_{E} = q E_{0} \hat{k} \sum_{\overset{⇀}{F}} = m \overset{⇀}{a} F_{1 z} = q V_{0} B_{0} = m a_{r} = \frac{m {v_{0}}^{2}}{2} F_{1 t} = 0$

Finally, we come across $\overset{⇀}{v} (t)$ :

$\overset{⇀}{V} (t) = V_{0} \hat{r} + (0 + a_{z} t) \hat{k} \overset{⇀}{V (t)} = V_{0} \hat{i} + a_{z} t \hat{k} V = \sqrt{{V_{0}}^{2} + (a_{z} t)} V = \sqrt{{V_{0}}^{2} + {(\frac{q E_{0} t}{m})}^{2}}$

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A particle of charge qand mass mmoves in the uniform fields role="math" localid="1648977372617" E⇀=E0k^andB⇀=B0k^. At t=0, the particle has velocity role="math" localid="1648977203095" V0⇀=V0i^. What is the particle's speed at a later time t?

Short Answer

Step by step solution

Introduction

Step 1

Step 2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Modern Physics

Magnetism and Electromagnetic Induction

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Study anywhere. Anytime. Across all devices.