What is the force on an electron moving in a magnetic field?

[airkapp]

electrons and magnetism. - need help.

An electron moves with velocity v = (4.0i – 6.0ij) x 104 m/s in a magnetic field B= (-0.80i + 0.60j) T. Determine the magnitude and direction of the force on the electron.

F=ma: qvB=m*v^2/r : r= mv/qB

r = (9.1E^-31kg * (4.0i – 6.0ij) x 104 m/s) / (1.6E^-19 C * (-0.80i + 0.60j) T)

= (3.8E-28i + -5.7E-28j) / (-1.28E-19i + 9.6E-20j)T

= (-3E-9i - 5.9E-9j)T


qvB=m*v^2/r

= 9.1E^-31kg (3.8E-28i * -5.7E-28j)2 / (-3E-9i - 5.9E-9j)T


=(-4.44E-77i + 5.011E-77)T N southward.


I'm totally stuck on this problem. This is my work but I'm pretty sure it's wrong. I'm not sure how to handle the i, j, k component problems. or is this the way... take the i and j components of the velocity and the magnetic field, and find two separate force components, Fi and Fj. Then assume i and j are perpendicular vectors and then with Fi and Fj find the
magnitude of the force. Finally, find the angle in the i-j plane in this
way, knowing that the force would come into the k plane but not at a
right angle to both i and j?? is that correct?

thanks, any help would be appreciated.

Air K

 

[MathStudent]

I haven't even looked at your work, but the equation for the force of a point charge due to a magnetic field is:

$$\vec{F_B} = q\vec{v} \ X \ \vec{B}$$

where "X" is the cross product.
You can calculate F in terms of components i, j, k. (do you know how to do cross product? )

Then finding the direction and magnitude of F can be done the same way you find the magnitude and direction of any vector.

The equations you gave are for the radius of an electron in uniform circular motion due to a magnetic field and are completely irrelevant to the question.

 

[airkapp]

I haven't even looked at your work, but the equation for force is

$$\vec{F_B} = q\vec{v} \ X \ \vec{B}$$

where "X" is the cross product.
You can calculate F in terms of components i, j, k. (do you know how to do cross product? )

Then finding the direction and magnitude of F can be done the same way you find the magnitude and direction of any vector.

hmm, I've done cross product before I'm just not sure if my method is right. I think I posted the same formula as you just did. Will my answer be in terms of I, J, K?

Don't I need to find "r" to solve the problem?

 

[MathStudent]

cross product: http://mathworld.wolfram.com/CrossProduct.html

note: look to my edited post about your equations.

 

[vincentchan]

use this one
$$\vec{F}=q\vec{v}\times\vec{B}$$
you have v and B already, don't over complicate the problem
do you know cross product?

 

[MathStudent]

hmm...why didnt I think of that

 

[airkapp]

cross product: http://mathworld.wolfram.com/CrossProduct.html

note: look to my edited post about your equations.

thanks for the refresher link. I think I got it now.