# Doubt regarding Force exerted by a magnetic field on a moving charge

1. Apr 14, 2009

### ApuroopS

a charged particle ( say 'q' ) moving with velocity 'V' in a region containing a magnetic field 'B' experiences a force 'F' given by:

F = q(VxB)

when
V = 0,
then force = 0

my doubt is,
is this velocity relative to the magnetic field ?
( i.e. when the velocity relative to the magnetic field becomes = 0, then force due to the magnetic field on the particle = 0 )

2. Apr 14, 2009

### Born2bwire

In a sense, yes it is relative to the B field, but this is hidden in the cross product. The magnitude of the velocity affects the amount of force experienced in the B field but the direction of travel with respect to the polarization of the magnetic flux density also matters. If the charge is moving with the magnetic flux's field lines, then there is no force while the force is maximized if the charge moves normal to the field lines. If you want to have the equation in strictly scalar form then,

$$\left| \mathbf{F} \right| = q \left| \mathbf{v} \right| \left| \mathbf{B} \right| \sin \theta$$

Where $$\theta$$ is the angle between the velocity and magnetic flux density vectors.

3. Apr 14, 2009

### Bob S

If the observer were moving at a relativistic velocity with respect to a constant E or B field, or a constant E or B field were moving with respect to the observer, then there is an electromagnetic transformation of the longitudinal component that will produce a B or E component. To see the transformations, visit the LBL Particle Data Group website at

http://pdg.lbl.gov/2008/reviews/contents_sports.html

then click on the constants, units, etc. category, then click on the electromagnetic relations section, then look at the last four lines of the table.