Magnetic Fields. Finding the angle of a moving charge.

[zeldajae]

Homework Statement

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 90^o) between the electron's velocity and the magnetic field.

Homework Equations

F=ma

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

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 19.7^o. 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

 

[gabbagabbahey]

We know the mass of the electron= 1.673 X 10^-27

Really?!

 

[zeldajae]

Really?!

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!