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
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Answer:
The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.
Explanation:
Use the frequency f, speed v, and wavelentgh L relationship:
We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:
The fundamental frequency (first harmonic) is 250 Hz
The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz
the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz
1 ys is the smallest unit of time.