Answered: For problem 28.9, solve for the…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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For problem 28.9, solve for the magnetic field (For problem 28.9, solve for the magnetic field (in mT) when the accelerating voltage remains 1000 V and the voltage difference between the deflection plates (the "horizontal" par with one above the other in the picture) is 107.5 V. (5 sig figs)

•9 ILW In Fig. 28-32, an electron accelerated from rest through po-
tential difference V₁ = 1.00 kV enters the gap between two paral-
lel plates having separation d = 20.0 mm and potential difference
830
Figure 28-32 Problem 9.
Talve
d V₂
CHAPTER 28 MAGNETIC FIELDS
B = (30.0 mT) (Fig. 28-35). What are the resulting (a) electric
field within the solid, in unit-vector notation, and (b) potential dif-
ference across the solid?

Expert Solution

Step 1

The motion of the electron is in straight line therefore the forces in vertical direction due electric and magnetic field are balanced. So,

$\begin{array}{rcl} e v B & = & e E \\ v B & = & E \\ B & = & \frac{E}{v} \end{array}$

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