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The Lorentz Force

We can now put the second part of the procedure [calculating the forces on a test charge due to known FIELDS] into a very compact form combining both the electric and the magnetic forces into one equation. If a particle with charge q and mass m moves with velocity $\Vec{v}$ in the combination of a uniform electric field $\Vec{E}$ and a uniform magnetic field $\Vec{B}$, the net force acting on the particle is the LORENTZ FORCE, which can be written (in one set of units)

$$\Vec{F} \; = \; q \, \left( \Vec{E} \; + \; {\Vec{v} \over c} \times \Vec{B} \right) ,$$ (17.9)

where (for now) we can think of c as just some constant with units of velocity.

If $\Vec{E} = 0$ and $\Vec{v}$ is perpendicular to $\Vec{B}$, the Lorentz force is perpendicular to both $\Vec{B}$ and the momentum $\Vec{p} = m \Vec{v}$. The force will deflect the momentum sideways, changing its direction but not its magnitude.^17.6 As $\Vec{p}$ changes direction, $\Vec{F}$ changes with it to remain ever perpendicular to the velocity - this is an automatic property of the cross product - and eventually the orbit of the particle closes back on itself to form a circle. In this way the magnetic field produces UNIFORM CIRCULAR MOTION with the plane of the circle perpendicular to both $\Vec{v}$ and $\Vec{B}$.

  

Figure:  Path of a charged particle with momentum $\Vec{p}$ in a uniform, static magnetic field $\Vec{B}$ perpendicular to $\Vec{p}$.

Using Newton's SECOND LAW and a general knowledge of circular motion, one can derive a formula for the radius of the circle (r) in terms of the momentum of the particle (p = mv), its charge (q) and the magnitude of the magnetic field (B). In "Gaussian units" (grams, centimeters, Gauss) the formula reads^17.7

$$r = {pc \over qB} .$$ (17.10)

It is also interesting to picture qualitatively what will happen to the particle if an electric field $\Vec{E}$ is then applied parallel to $\Vec{B}$: since $\Vec{E}$ accelerates the charge in the direction of $\Vec{E}$, which is also the direction of $\Vec{B}$, and since $\Vec{B}$ only produces a force when the particle moves perpendicular to $\Vec{B}$, in effect the "perpendicular part of the motion" is unchanged (circular motion) while the "parallel part" is unrestricted acceleration. The path in space followed by the particle will be a spiral with steadily increasing "pitch":

  

Figure:  Path of a charged particle in parallel $\Vec{E}$ and $\Vec{B}$ fields.