Q. 67 A particle of charge q聽and mass... [FREE SOLUTION]

Chapter 29: Q. 67 (page 796)

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

Step 1

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

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

Step 2

$\overset{鈬赌}{F_{M}} = q V_{0} B_{0} \hat{r} 鈬� F_{E} = q E_{0} \hat{k} \underset{\overset{鈬赌}{F}}{鈭�} = m \overset{鈬赌}{a} F_{1 z} = q V_{0} B_{0} = m a_{r} = \frac{m {v_{0}}^{2}}{2} F_{1 t} = 0$

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

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

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A particle of charge qand mass mmoves in the uniform fields role="math" localid="1648977372617" E鈬赌=E0k^andB鈬赌=B0k^. At t=0, the particle has velocity role="math" localid="1648977203095" V0鈬赌=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

Atoms and Radioactivity

Geometrical and Physical Optics

Energy Physics

Electricity

Physics of Motion

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.