What Angle Does the Electron's Velocity Make with the Magnetic Field?

[Dart82]

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?

 

[ranger]

I didnt punch in the numbers, but you need to solve for theta not sine theta (assuming the last step you did was the finial step). Example in a right triangle, arcsine(opp/hyp) = theta.

 

[Dart82]

i understand what you are saying, however when i use the numbers i have calculated in the arcsine (opp/hyp) = theta equation, i am getting numbers that don't correspond to reasonable angles. This leads me to believe i am screwing up something in the F=ma or F=qvBsin(theta) department. in other words arcsine(F/qvB) should equal theta...right?

 

[Dick]

If it's any encouragement, I plugged your numbers in and got a perfectly reasonable value of sin(theta). Your theory is fine, your plugging is a problem.

 

[Dart82]

ok, i'll check it out again

 

[Dart82]

amazing...i finally got it! what in the world was i doing to get such wacky #'s. Thanks guys.
i just realized i was putting in the wrong exponent for the charge of the electron. DOH!