Q51E When a particle of charge q >... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q51E (page 915)

When a particle of charge q > 0 moves with a velocity ofat 45.0掳 from the +x-axis in the xy-plane, a uniform magnetic field exerts a force along the -z-axis (Fig. P27.51). When the same particle moves with a velocitywith the same magnitude asbut along the +z-axis, a forceof magnitude F₂ is exerted on it along the +x-axis. (a) What are the magnitude (in terms of q, v₁, and F₂) and direction of the magnetic field? (b) What is the magnitude ofin terms of F₂?

Short Answer

Expert verified

The magnitude and direction of magnetic field is ${\overset{鈫�}{B}}_{y} = 鈭� \frac{F_{2}}{v_{1} q} \hat{y}$
The magnitude of is ${\overset{鈫�}{F}}_{1} is F_{1} = \frac{F_{2}}{\sqrt{2}}$

Step by step solution

Magnetic force and electric force

The magnetic force is given by

$\overset{鈫�}{F_{B}} = q (\overset{鈫�}{v} 脳 \overset{鈫�}{B})$

The electric force is given by

$\overset{鈫�}{F_{E}} = q \overset{鈫�}{E}$

Determine the magnetic field

(a)

The magnetic force is given by

$\overset{鈫�}{F_{B}} = q (\overset{鈫�}{v} 脳 \overset{鈫�}{B})$

Now the magnetic force is

$\begin{aligned} {\overset{鈫�}{F}}_{2} = F_{2} \hat{x} \\ {\overset{鈫�}{v}}_{2} = v_{1} \hat{k} \\ F_{2} \hat{x} = {qv}_{1} \hat{k} 脳 \overset{鈫�}{B} \\ = 鈭� {qB}_{y} v_{1} \hat{x} + {qB}_{x} v_{1} \hat{y} + 0 \hat{z} \\ B_{x} = 0 \\ F_{2} = 鈭� {qv}_{1} B_{y} \end{aligned}$

Therefore, the magnetic field is ${\overset{鈫�}{B}}_{y} = 鈭� \frac{F_{2}}{V_{1} q} \hat{y}$

Determine the magnitude of F1→

(b)

The magnitude of $\overset{鈫�}{F_{1}}$ is

role="math" localid="1668264415915" $\begin{aligned} {\overset{鈫�}{F}}_{1} = 鈭� F_{1} \hat{z} \\ {\overset{鈫�}{v}}_{1} = v_{1} \cos 鈦� (45) \hat{x} + v_{1} \sin 鈦� (45) \hat{y} \\ 鈭� F_{1} \hat{z} = q {\overset{鈫�}{v}}_{1} 脳 \overset{鈫�}{B} \\ = {qv}_{1} \sin 鈦� (45) B_{z} \hat{x} 鈭� {qv}_{1} \cos 鈦� (45) B_{z} \hat{y} + ({qv}_{1} \cos 鈦� (45) B_{y} 鈭� {qv}_{1} \sin 鈦� (45) B_{x}) \hat{z} \\ Now, \\ B_{z} = 0 \\ B_{x} = 0 \\ 鈭� F_{1} = B_{y} {qv}_{1} \cos 鈦� (45) \\ F_{1} = 鈭� B_{y} {qv}_{1} \cos 鈦� (45) \\ = 鈭� (鈭� \frac{F_{2}}{v_{1} q}) {qv}_{1} \cos 鈦� (45) \\ F_{1} = F_{2} \cos 鈦� (45) = \frac{F_{2}}{\sqrt{2}} \end{aligned}$