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David23454
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Homework Statement
An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 2.4-cm-wide region of uniform magnetic field in the figure(Figure 1) .
What field strength will deflect the electron by θ = 13 degrees?
Homework Equations
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Δx=vt+0.5aΔt2
KE=0.5mv2
U=qΔV
F=qvBsin(θ)
F=ma
B=ma/qvsin(θ)
Ki+Ui=Kf+Uf
The Attempt at a Solution
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Since Ki=0, then Kf=-ΔU
Since U=qΔV, then Kf=-qΔV
Since Kf=0.5mv2, then 0.5mv2=-qΔV
Using 0.5mv2=-qΔV, and inputing values 0.5(9.11×10-31)v2)=-(-1.60×10-19)(10,000)
So, v after acceleration by ΔV is 5.931×107 m/s.
Δx=viΔt+0.5axΔt2
Since ax=0, because the force should be acting to change only the ay
t=4.047×10-10 s
so, Δy=viΔt+0.5ayΔt2
and since Δy=0.024tan(13)=5.54×10-3
then, ay=6.765×1016
Since: F=qvBsin(θ) and F=ma, then
B=ma/qvsin(θ), and θ=90, (I think?)
so
B=(9.11×10-31)(6.756×1016)/(1.60×10-19)(5.931×107)=6.49×10-3 T
This answer is incorrect. Could someone show me where I went wrong? Thanks!