Calculate Magnetic Field Strength for Deflecting Electron in CRT?

[m84uily]

Homework Statement

An electron in a cathode-ray tube is accelerated through a potential difference of 10kV, then passes through the 2.0cm wide region of uniform magnetic field.What field strength will deflect the electron by 10 degrees?

Homework Equations

Ki + Ui = Kf + Uf
K = (1/2)(m)v^2
(dU) = q(dV)
distance = (vi)(t) + (1/2)(acceleration)t^2
electron mass = 9.10938188 × 10^(-31)
electron charge = 1.60217646 × 10^(-19)

The Attempt at a Solution

Starting to find out the velocity with which the electron left the tube:
Ki + Ui = Kf + Uf
Ki = 0
-(Uf - Ui) = Kf
-dU = Kf
-q(dV) = Kf
-q(dV) = (1/2)(m)(vf)^2
(-2q(dV) / m)^(1/2) = vf
vf = 59 307 675 m/s

Now I know that the electron has changed its path by 10 degrees so:
change in the y = (.02m)(sin(10)) = .00347m

Finding out the time it took for the electron to pass through the 2cm field:
distance = (vi)(t) + (1/2)(acceleration)t^2
acceleration = 0 so
t = (distance) / (vi) = (.02m)/(59 307 675 m/s) = 3.372 x 10^(-10)s

Finding the acceleration in the y:
distance = (vi)(t) + (1/2)(acceleration)t^2
vi = 0
2(distance)/(t^2) = 2(.00347m)/(.372 x 10^(-10)s)^2 = 61026895446262593.75 m/s^2= acceleration

So we can say the force acting on the electron is:
F = ma = (9.10938188 × 10^(-31))(1026895446262593.75) = 5.5591729557083898522805125 x 10^(-14) N

Now to just solve for the electric field:
F = qvBsin(theta)
theta = 90 degrees
F = qvB
F/qv = B
B = 5.5591729557083898522805125 x 10^(-14) / ((59 307 675)(1.60217646 × 10^(-19)))
= 0.0058504455313400798249151657115222
With significant figures = 0.00585 T

The actual answer is 2.9 mT

Thanks in advance!

 

[Metaleer]

I haven't fully checked your work, but I do see a sin(10) somewhere in there. Unless your calculator is in DEG mode, this will not produce the expected answer.