# Right Hand rule to find direction of the force

1. May 7, 2007

### t_n_p

1. The problem statement, all variables and given/known data
http://img71.imageshack.us/img71/8453/untitleddt6.jpg [Broken]

Imagine that current flow from east to west. what is the direction of the force on the current when the magnetic field B is (i) to the East and (ii) Vertically up?

3. The attempt at a solution
I've done a bit of research mainly from this website http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html but I am still unclear about how to go about the question.

For (i) the magnetic field is to the east, so my fingers are to the right hand side, but my thumb is in the direction of the velocity. Where exactly is the velocity?

Last edited by a moderator: May 2, 2017
2. May 7, 2007

### Yura

i dont know where you got velocity from... (might be something but i cant remember learning anything on it to do with this) and i cant seem to manage to relate the question the the image..

if you hold your palm flat down on a table top and have your thumb at a rightangle with all your fingers (each of your fingers that isnt a thumb, =\ is parallel)
your thumb is in the direction of the current.
your fingers are pointing in the direction of the magnetic field and if there was an arrow shooting out of the palm of your hand, it would be pointing in the direction of the force.

so your question asks (im gonna ignore the image because im not sure what you're asking) the current flow from east to west.

the first one (im not sure if im right here so someone correct me if im wrong)
has the current flowing in the same direction as the field applied. so there will be no resulting force

second one, position your hand, in the form i stated above somewhere, so that your fingers (magnetic field) point to the sky and your thumb points to the west. im geographically impaired so you're going to have to figure out which direction, north or south, that an arrow shooting out of your palm is going to point.

i hope this helps..

3. May 7, 2007

### mbrmbrg

Is there a reason you have to use $$\vec{F_B}=q\vec{v}\times\vec{B}$$ rather than $$\vec{F_B}=i\vec{L}\times\vec{B}$$?