**Lorentz Force:** When a charged particle is moving in space where both electric and magnetic fields exist simultaneously, it experiences a force known as Lorentz Force.

Whenever the moving charge enters a uniform magnetic field it experiences a force known as Lorentz Force. Expression of this magnetic force is given below:

\begin{align*} F_B = qvB\sin{\theta} \qquad \text{Force on a Charged Particle}\end{align*}

where \begin{align*}q\end{align*}

Note that for problems where the direction of the particle and the direction of the magnetic field are perpendicular then

\begin{align*} F_B = qvB \end{align*}

also, recall that

\begin{align*} F &= qE \\
E &= V \Delta x \qquad \text{ for a constant electric field}\end{align*}

#### Guidance

As moving charges create magnetic fields, so they experience forces from magnetic fields generated by other materials. The magnitude of the force experienced by a particle traveling in a magnetic field depends on the charge of the particle \begin{align*} (q) \end{align*}

There is a **second right hand rule** that will show the direction of the force on a positive charge in a magnetic field: point your index finger along the direction of the particle’s velocity. If your middle finger points along the magnetic field, your thumb will point in the direction of the force.

**NOTE:** For negative charge reverse the direction of the force (or use your left hand)

For instance, if a positively charged particle is moving to the right, and it enters a magnetic field pointing towards the top of your page, it feels a force going out of the page, while if a positively charged particle is moving to the left, and it enters a magnetic field pointing towards the top of your page, it feels a force going into the page.

##### Example 1: Find the Magnetic Field

**Question:** An electron is moving to the east at a speed of \begin{align*}1.8\times10^6\;\mathrm{m/s}\end{align*}

**Answer:** There are two parts to this question, the magnitude of the electric field and the direction. We will first focus on the magnitude.

To find the magnitude we will use the equation \begin{align*}F_B=qvB\sin\theta
\end{align*}

We were given the force of the magnetic field \begin{align*}(2.2\times10^{-12}\;\mathrm{N})\end{align*}

\begin{align*}F_B=qvB\sin\theta\end{align*}

Solving for \begin{align*}B\end{align*}:

\begin{align*}B = \frac{F_B}{qv\sin\theta}\end{align*}

Now, plugging the known values we have:

\begin{align*}B=\frac{F_B}{qv\sin\theta}=\frac{2.2 \times 10^{-12}\:\mathrm{N}}{1.6\times10^{-19}\:\mathrm{C} \times 1.8\times10^6\:\mathrm{m/s}\times1}=7.6\:\mathrm{T} \end{align*}

Now we will find the direction of the field. We know the direction of the velocity (east) and the direction of the force due to the magnetic field (up, out of the page). Therefore we can use the second right hand rule (we will use the left hand, since an electron's charge is negative). Point the pointer finger to the right to represent the velocity and the thumb up to represent the force. This forces the middle finger, which represents the direction of the magnetic field, to point south. Alternatively, we could recognize that this situation is illustrated for a *positive* particle in the right half of the drawing above; for a negative particle to experience the same force, the field has to point in the opposite direction: south.

##### Example 2: Circular Motion in Magnetic Fields

Consider the following problem: a positively charged particle with an initial velocity of \begin{align*} \vec{v}_1 \end{align*}, charge \begin{align*} q \end{align*} and mass \begin{align*} m \end{align*} traveling in the plane of this page enters a region with a constant magnetic field \begin{align*} \vec{B} \end{align*} pointing into the page. We are interested in finding the trajectory of this particle.

Since the force on a charged particle in a magnetic field is always perpendicular to both its velocity vector and the field vector (check this using the second right hand rule above), a *constant* magnetic field will provide a centripetal force --- that is, a constant force that is always directed perpendicular to the direction of motion. Two such force/velocity combinations are illustrated above. According to our study of rotational motion, this implies that as long as the particle does not leave the region of the magnetic field, it will travel in a circle. To find the radius of the circle, we set the magnitude of the centripetal force equal to the magnitude of the magnetic force and solve for \begin{align*} r \end{align*}:

\begin{align*}F_c = \frac{mv^2}{r} = F_B = qvB\sin \theta = qvB\end{align*}

Therefore,

\begin{align*}r = \frac{mv^2}{qvB} \end{align*}

In the examples above, \begin{align*} \theta \end{align*} was conveniently 90 degrees, which made \begin{align*} \sin \theta = 1\end{align*}. But that does not really matter; in a constant magnetic fields a different \begin{align*} \theta \end{align*} will simply decrease the force by a constant factor and will not change the qualitative behavior of the particle.

#### Watch this Explanation

Watch the video at: http://www.youtube.com/watch?v=JjO_VvBxzig

Watch the video at: http://www.youtube.com/watch?v=AKKpvVvvxkY

### Magnetic Force on a Current Carrying Conductor

We can extend the analysis for force on a moving charge to force on a current carrying element as \begin{align*}Idl\end{align*} is equivalent to \begin{align*}qv\end{align*}. By replacing \begin{align*}Idl\end{align*} by \begin{align*}qv\end{align*} in the expression of force, we obtain an expression for the force on the current element. When the current is traveling through a magnetic field while inside a wire, the magnetic force is still exerted but now it is calculated as the force on the wire rather than on the individual charges in the current.

The equation for the force on the wire is given as \begin{align*}F = BIL\end{align*}, where \begin{align*}B\end{align*} is the strength of the magnetic field, \begin{align*}I\end{align*} is the current in amps and \begin{align*}L\end{align*} is the length of the wire in the field and perpendicular to the field.

**Example Problem:** A wire 0.10 m long carries a current of 5.0 A. The wire is at right angles to a uniform magnetic field. The force the field exerts on the wire is 0.20 N. What is the magnitude of the magnetic field?

**Solution:** \begin{align*}B=\frac{F}{IL}=\frac{0.20 \ N}{\left(5.0 \ A \right) \left(0.10 \ m\right)}=0.40 \ \frac{N}{A \cdot m}\end{align*} (\begin{align*}N/A \cdot m\end{align*} are also known as **Tesla**)