(8c28p68) An electron in a TV camera tube is moving at 7.40x10° m/s in a magneti
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Question
(8c28p68) An electron in a TV camera tube is moving at 7.40x10° m/s in a magnetic field of strength 67 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force? Submit Answer Tries 0/99 Minimum force? Submit Answer Tries o/99 At one point the acceleration of the electron is 8.576x1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg) Submit Answer Tries 0/99Explanation / Answer
Given :-
B = 67 mT = 67 x 10^-3 T
q = 1.6 x 10^-19 C
v = 7.40 x 10^6 m/s
Maximum force,
F = qvB
F = (1.6 x 10^-19 x 7.40 x 10^6 x 67 x 10^-3 T)
F = 7.933 x 10^-14 N
Minimum force, F = 0
(b)
a = 8.576 x 10^16 m/s^2
m = 9.11 x 10^-31 kg
F = ma
F = 8.576 x 10^16 x 9.11 x 10^-31
F = 7.813 x 10^-14 N
F = Bvq sin?
sin? = F / qvB
sin? = 0.98484
? = 80 degrees
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