- #1

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

[/B]

Δx=vt+0.5aΔt

^{2}

KE=0.5mv

^{2}

U=qΔV

F=qvBsin(θ)

F=ma

B=ma/qvsin(θ)

K

_{i}+U

_{i}=K

_{f}+U

_{f}

## The Attempt at a Solution

[/B]

Since K

_{i}=0, then K

_{f}=-ΔU

Since U=qΔV, then K

_{f}=-qΔV

Since K

_{f}=0.5mv

^{2}, then 0.5mv

^{2}=-qΔV

Using 0.5mv

^{2}=-qΔV, and inputing values 0.5(9.11×10

^{-31})v

^{2})=-(-1.60×10

^{-19})(10,000)

So, v after acceleration by ΔV is 5.931×10

^{7}m/s.

Δx=v

_{i}Δt+0.5a

_{x}Δt

^{2}

Since a

_{x}=0, because the force should be acting to change only the a

_{y}

t=4.047×10

^{-10}s

so, Δy=v

_{i}Δt+0.5a

_{y}Δt

^{2}

and since Δy=0.024tan(13)=5.54×10

^{-3}

then, a

_{y}=6.765×10

^{16}

Since: F=qvBsin(θ) and F=ma, then

B=ma/qvsin(θ), and θ=90, (I think?)

so

B=(9.11×10

^{-31})(6.756×10

^{16})/(1.60×10

^{-19})(5.931×10

^{7})=6.49×10

^{-3}T

This answer is incorrect. Could someone show me where I went wrong? Thanks!