# Magnetic Fields. Finding the angle of a moving charge.

1. Mar 14, 2009

### zeldajae

1. The problem statement, all variables and given/known data

An electron is moving through a magnetic field whose magnitude is 8.70 X 10-4 T. The electron experiances only a magnetic force and has an acceleration of magnitude 3.50 X 1014 m/s2. At a certain instant, it has a speed of 6.80 X 106 m/s. Determine the angle $$\theta$$ (less than 90o) between the electron's velocity and the magnetic field.

2. Relevant equations

F=ma

B = F / qo v (sin$$\theta$$)

3. The attempt at a solution

We know the mass of the electron= 1.673 X 10-27
So we plug that in the equation F=ma=(1.673 X 10-27)(3.5 X 1014 m/s2=5.86 X 10-13

Then we use the magnetic field equation and solve for sin$$\theta$$

sin$$\theta$$ = F / q v B = (5.86 x 10-13)/((8.7 X 10-4)(6.8 X 106)(1.6 X 10-19)) = 618.608

That is wrong because the books answer is 19.7o. I know I am supposed to take inverse of the answer to find the angle. But I did my calculation correct and I am still getting the same answer...is there anything that I am doing wrong?

Thanks

Last edited: Mar 14, 2009
2. Mar 14, 2009

### gabbagabbahey

Really?!

3. Mar 14, 2009

### zeldajae

Sorry, The mass of an electron is 9.1 X 10-31

After all the calculation, I get the right answer! Thanks for letting me know! I did a silly mistake!