# The Motion of a Charged Particle in Electric and Magnetic Field

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Please review my solution to the problem and explain in detail what I may be doing wrong and what concepts I may not be applying correctly. I am not sure if I am apply the crossed fields concept correct in presuming that B is perpendicular.

The problem states:

An electron has an initial velocity of (12.0j + 15.0k) km/s and a constant acceleration of (2.00 x 1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If B = (400μT)i, find the electric field E.

We know:

This is a crossed fields problem with the two fields perpendicular to each other.

B is perpendicular to v, out of the page.

Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.

q = 1.6 x 10-19 C

v = (12.0j + 15.0k) km/s = square root(12.02 + 15.02) = 19.21 km/s = 1.921 x 104 m/s

B = 400μT = 400 x 10-6 T = 400 x 10-6 N/Cm/s

Φ = 90º

E = qvBsinΦ

E = (1.6 x 10-19 C)(1.921 x 104 m/s)(400 x 10-6 N/Cm/s)(sin90º)

E = 1.229 x 10-18 N

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##### Solution Summary

An electron moves with constant acceleration in presence of electric and magnetic field both. The electric field is calculated.

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