Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a p
hosphor screen. The magnetic fields were created by coils of wire on either side of the tube carrying large currents. In one such TV set, the phosphor screen is 51.2 cm wide, and is 11.1 cm away from the center of the magnetic deflection coils (that is, the center of the region of magnetic field). The electron beam is first accelerated through a 22,000 V potential difference before it enters the magnetic field region, which is 1.00 cm wide. The field is approximately uniform and perpendicular to the velocity of the electrons. If the field were turned off, the electrons would hit the center of the screen. What magnitude of magnetic field (in mT) is needed to deflect the electrons so that they hit the far edge of the screen
1 answer:
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Refer to the diagram shown below.
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)