An electron moves at 2.40 ×106 m/s through a region in which there is a magnetic

Question

An electron moves at 2.40 ×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.90 ×102T.

A. What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

B. What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field?

C. If the actual acceleration of the electron is one-fourth of the largest magnitude in part A, what is the angle between the electron velocity and the magnetic field?

Explanation / Answer

Here,

u = 2.4 *10^6 m/s

magnetic field , B = 7.9 *10^-2 T

a) let the maximum acceleration is a

using second law of motion

9.11 *10^-31 * a = 2.4 *10^6 * 1.602 *10^-19 * 7.9 *10^-2

solving for a

a = 3.33 *10^16 m/s^2

largest possible magnitude of the acceleration of the electron due to the magnetic field is 3.33 *10^16 m/s^2

b)

the magnetic force is zero when the electron is moving parallel to magnetic field

hence , minimum acceleration is zero

c)

for 1/4 th acceleration

as m * a = B * v * q * sin(theta)

for 1/4th acceleration

sin(theta) = 1/4

theta = 14.5 degree

the angle between velocity and magnetic field is 14.5 degree

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