# Electrodynamics/Lorentz Force

Paulo Lacombe Spring 2018

In order to find the force acting on a charged particle inside a magnetic and/or electric field we need to use the Lorentz force rule. When adding a charged particle to a field the field must exert a force on the particle based on the magnitude and direction of the field and the particle.

Because the Lorentz force involves both an electric and magnetic field it is considered part of electromagnetism. Lorentz force also completes the net forces acting on a particle together with other forces such as gravitational force.

## Mathematical Model

### Complete Formula

$\mathbf {F} =q(\mathbf {E} +\mathbf {v} \times \mathbf {B} )$

F is the force acting on the particle

q is the charge of the particle

E is the electric field

v is the velocity

B is the magnetic field

x is the cross product. In this case the result is the vector cross product between the particle velocity and the magnetic field.

#### Electric Force

$\mathbf {F} =q*\mathbf {E}$

#### Magnetic Force

$\mathbf {F} =q(\mathbf {v} \times \mathbf {B} )$

### Analysis

Based on the equation we can conclude that the electric force on the particle is in the same direction as the electric field as long as the particle is positive, the force is in the opposite direction if the particle is negative. The magnetic force will be perpendicular to the velocity and magnetic field, in other words the normal to the magnetic force and the velocity. This force generally causes the particle to move in a spiral as long as the magnetic force is not parallel to the velocity. In that case the force would be equal to zero.

In order to determine the direction of the magnetic force in a physical manner we can use the Right Hand Rule. By placing your four main fingers in the direction of the magnetic field and your thumb in the direction of the velocity, the direction of the force is pointed out of your palm as shown in the diagram.

## Example

### Problem 1

Find the magnitude and direction of the magnetic force on the antiproton.

${\vec {v}}_{p}=v<0,-1,0>$  ${\hat {r}}={{\vec {r}} \over |r|}={<-d,-d,0> \over {\sqrt {(-d)^{2}+(-d)^{2}}}}={1 \over {\sqrt {2}}}<-1,-1,0>$

${\vec {B_{p}}}={\mu _{0} \over 4\pi }*{q{\vec {v_{p}}}\times {\widehat {r}} \over r^{2}}={\mu _{0} \over 4\pi }*{e<0,-v,0>\times {1 \over {\sqrt {2}}}<-1,-1,0> \over 2d^{2}}$

${\vec {B_{p}}}={\mu _{0}ev \over 8\pi {\sqrt {2}}d^{2}}*<0,0,-1>$

${\vec {F_{B}}}={\mu _{0}ev \over 8\pi {\sqrt {2}}d^{2}}*<1,0,0>\times <0,0,-1>={\mu _{0}ev \over 8\pi {\sqrt {2}}d^{2}}*<0,1,0>$