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## Homework Statement

An electron is moving through a magnetic field whose magnitude is 9.10x10^-4 T. The electron experiences only a magnetic force and has an acceleration of magnitude 3.60x10^14 m/s^2. At a certain instant, it has a speed of 6.30x10^6 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.

## Homework Equations

Magnetic Force

F = qvBsin(theta)

F=ma

## The Attempt at a Solution

F=ma --> mass of electron x acceleration = 3.27x10^-16N

F = qvBsin(theta)

3.27x10^-16N / [(charge of electron) x (6.30x10^6 m/s) x (9.10x10^-4 T)] = sin (theta)

solving like this would make sin theta = 3.56 x 10^11, which is not reasonable at all....What am i doing wrong?