An electron moves at 2.30 Times 10^6 m/s through a region in which there is a ma

Question

An electron moves at 2.30 Times 10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.90 Times 10^-2 T. What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? If the actual acceleration of the electron is 1/4 of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Explanation / Answer

a) F = m a

F = q v B sin phi

a = q v B / m = (1.6 * 10-19 * 2.3 * 106 * 7.9 * 10-2) / (9.11 * 10-31)

= 3.19 * 1016 m/s2

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When magnetic field is parallel to the velocity, acceleration is minimal

m a = q v B sin 0

a = 0 m/s2

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a = 1/4 a = 1/4 ( q v B / m ) = Q v B sin theta / m

sin theta = 1/4

theta = sin-1 ( 1/4)

= 14.47 deg

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