The picture tube in an old black-and-white television uses magnetic deflection c

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 48.4-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 11.1 cm from the center of the coils and is 46.8 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? mT

I got 72.238 mT and it was incorrect

Explanation / Answer

The accelerating potential provides the KE of motion

1/2 m v^2 = q V

v = sqrt (2 q V/m)

Later the magnetic force balances the net centripital force

q v B = m v^2/R

B = mv/qR = m sqrt (2 q V/m)/qR

B = 1/R sqrt (2 m V/q)

Using trigonometry

theta = tan^-1(w/a) = tan^-1(46.8/2/11.1) = 64.62 deg

d/R = sin62.62 =. R = 1/sin62.62 = 1.13 cm

B = 1/0.0113 x sqrt (2 x 9.1 x 10^-31 x 48.4 x 10^3/1.6 x 10^-19) = 0.06566 = 65.6629 mT

Hence, B = 65.6629 mT

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