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 47.4-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.7 cm from the center of the coils and is 54.2 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?

Explanation / Answer

Radius of curvature of electron path = R
theta = tan-1(27.1/10) = tan-1(2.71)
R = 1cm/sin(tan-1(2.71))
R = 1.066
By energy conservation,
KEi + Ui = KEf + Uf
Since KEi = Uf, thus we have
Ui = KEf
1/2 mv2 = qdV
v2 = 2qdV/m.............(1)
Force due to magnetic field = centripetal force
FB = Fcentripetal
qvB = mv2/R
B = mV/qR
squaring both sides
B2 = m2v2/q2R2
putting the value of v2 from (1)
B2 = m2/q2R2 * 2qdV/m
B2 = 2mdV/qR2
B = (2mdV/qR2)1/2
m = 9.11 x 10-31 kg, dV = 47.4 x 103 V, q = 1.6 x 10-19 C, R = 1.066cm = 0.01066 m
putting the values above
B = [2*9.11 x 10-31*47.4 x 103/1.6 x 10-19 *(0.01066)2]1/2
B = 68.9 x 10-3 T
B = 68.9 mT

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