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 45.0-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.3 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?

I have asked this twice and have gotten the wrong answer both times. The answer is not 64.5, 68, 69, or 70 mT. Please, whoever gets this right I will give great feedback.

Explanation / Answer

First we need the straight forward velocity of the electrons from the conservation of energy

KE = PE

.5mv2 = qV

(.5)(9.11 X 10-31)(v2) = (1.6 X 10-19)(45000)

v = 1.26 X 108 m/s

Then how long will it have to hit the screen...

d = vt

.103 = (1.26 X 108)(t)

t = 8.19 X 10-10 sec

In that amount of time, the electron will need to travel half the screen length

d = vt

.243 = v(8.19 X 10-10)

v = 2.97 X 108 m/s

That means the magnetic field must accelerate the electron to that speed in the 1 cm distance it acts

Apply vf2 = vo2 + 2ad

(2.97 X 108)2 = 0 + 2(a)(.01)

a = 4.41 X 1018 m/s2

Then the magnetic force is F = qvB

qvB = ma

(1.6 X 10-19)(1.26 X 108)(B) = (9.11 X 10-31)(4.41 X 1018)

B = .199 T which is 199 mT

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.