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 85.7 V. 5 sig. figs.

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For problem 9, solve for the magnetic field (in mT) when the accelerating voltage remains 1000 V and the voltage difference between the voltage deflection plates (the "horizontal" par with one above the other in the picture) is 85.7 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?
Transcribed Image Text:•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?
QUESTION 2
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 85.7 V. 5 sig.
figs.
Transcribed Image Text:QUESTION 2 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 85.7 V. 5 sig. figs.
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