3. Let B(x, y, z) be an arbitrary magnetic field. For a charged particle with charge q travelling at velocity v, the force on the particle is F = qB X v. Note that this is different from gravitational and electrostatic fields, for which the force at a point is simply proportional to the field itself: F here is not simply a standard vector field, since its magnitude and direction don't depend just on the position (that would be true for B but not F). Consequently, we cannot talk Instead, F depends on the velocity v at which the particle travels, and that is not a vector field but dependent on the motion of the particle v, which is not defined for an arbitrary position but instead is only defined along the trajectory of the particle.about a curl of F, or about F being a conservative vector field (it isn't a vector field! - different particles at the same point will be subject to different forces if they travel at different velocities). Nonetheless, we can show that work done by a magnetic field is always zero. Let C be an arbitrary path followed by the particle. C does not need to be a closed loop. Show that if F is given in terms of B and velocity v as above, then the amount of work ,F . dr done in moving the particle along that path is always zero. Hint: parameterize the curve C' in terms of time elapsed t when computing F . dr. C