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MickOtto
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Homework Statement
The Picture tube in an old black and white TV uses magnetic deflection coils rather than electric plates. Suppose an electron beam is accelerated from rest through a 50kV potential difference and then through a region with uniform magnetic field 1cm wide. The screen is located 10cm from the coils and is 50cm wide. When the field is off, the electron beam hits the centre of the screen. Ignoring relativistic corrections, what field magnitude is needed to deflect the beam to the side of the screen.
m_e = 9.11E-31 kg;
q_e = 1.6E-19 C;
delta V = 50E3 V;
d_1 = 0.01m
d_2 = 0.1m
d_3 = 0.5m
Homework Equations
F = qv x B = qvBsin(theta) = qvB (assuming the magnetic field is at right angles);
1/2mv^2 = q*deltaV;
F = ma;
v_f = v_i + at;
v=d/t
The Attempt at a Solution
I will say the velocity from the potential difference is positive x and screen to be in the yz plane. I also assume the magnetic field is pointed in the positive z direction, giving the velocity from magnetic field in the y direction.
First I got the velocity from potential difference.
1/2mv^2=qV
=> v = sqrt(2q_eV/m_e)
Then I worked out the velocity in y from magnetic field, I'll call this v'.
F = qvB = ma
=> a=qvB/m.
v'_f = v'_i + at v'_i = 0;
v'_f = at =qvBd_1/mv = qBd_1/m_e (using t = d_1/v)
Now, time taken for beam to travel 0.1m in x is
t = d/v = 0.1/sqrt(2q_eV/m_e) = 7.5E-10
Beam must travel d_3/2 = 0.25m in this time, so v' = d/t becomes
0.25/7.5E-10 = 3.3E8.
We have v' = qBd_1/m_e.
Rearranging I got
B = m_e v/q_e d_1.
Plugging in numbers I get B =0.18 T.
The answer the book gives is 70 mT.
In the solutions the book works out R as a radius of curvature from Pythagorus and uses qvB=mv^2/R. My problem with this is that the magnetic field only exists for 0.01m, not 0.1m which is what they use.
Any help/pointing out the obvious and making me feel like an idiot will be greatly appreciated.
Thanks,
Michael.