An electron in a TV camera tube is moving at 8.00 times 10^6 m/s in a magnetic f

Question

An electron in a TV camera tube is moving at 8.00 times 10^6 m/s in a magnetic field of strength 71 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 0/10 Minimum force? Submit Answer Tries 0/10 At one point the acceleration of the electron is 8.071 times io16 m/s2. What is the angle between the electron velocity and the magnetic field? (deg) Submit Answer Tries 0/10

Explanation / Answer

magnetic force Fm = q ( v X B) = qvB sin@

maximum force will be when angle between them @ =90 deg

Fmax = 1.6 x 10^-19 x 8 x 10^6 x 71 x 10^-3 = 9.09x 10^-14 N

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minimum force will when @= 0

Fmin = 0

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a = F/m

so a = qvBsin@ /m

8.071 x 10^16 = 1.6 x 10^-19 x 8 x 10^6 x 71 x 10^-3 x sin@ / (9.109 x 10^-31)

sin@ = 0.809

@ = 54 deg

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