Question

The picture tube in an old black-and-white television uses magnetic deflection coils rather than electric deflection plates. Suppose an electron beam is accelerated through a 54.0-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.2 cm from the center of the coils and is 48.6 cm wide. When the field is turned off, the electron beam hits the center of the screen. Ignoring relativistic corrections, what field magnitude is necessary to deflect the beam to the side of the screen?

Answer 1

The electrons leave the gun with a speed dependent on the
potential difference. From the work-energy theorem the speed of
the electrons is

v = sqrt (2W/m)
= sqrt(2eV/m)

where m is the electron mass, e the elementary charge and V the
accelerating potential difference.
When the electrons enter the magnetic field, they move
along a circular path due to the magnetic interaction. The
radius of the path can be determined from Newton’s second law
– the centripetal acceleration is proportional to the magnetic
force

r = mv / eBsin90

The radius of the circular path (and the width of the magnetic
field) affects the deflections angle

sin α = D/r

That angle also determines where the electron beam hits the
screen
y = L tan α

Hence

B = mv / er

= m sqrt(2eV/m) / e D/sin arctany/L

= sqrt(2meV) / sqrt(e).D * sin arctan y/L

For the beam to reach a side of the screen, the magnetic field
strength must be

Bmax = (sqrt (2*(9.1*10^-31)(54*10^3)) / (sqrt (1.6*10^-19) * 0.01) ) * sin arctan (24.3/10.2)
= 0.0722 = 72.2mT