Moving Point Charge

(Grace Townsend Spring 2017)

Claimed by Blake Heard, Fall 2017

The Main Idea

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Magnetic field created by a moving charge. [1]

As we have learned, a point charge creates an Electric Field that is given by Coulomb's Law:

However, when a point charge moves with some velocity, it not only creates an electric field, but it also creates a magnetic field that curls around the charge.

A Mathematical Model

The magnetic field of a moving point charge has units in Teslas (T) and can be found using a derivation of the Biot-Savart Law for magnetic fields:

where: [math]\displaystyle{ \frac{\mu_0}{4*\pi} = 1*10^{-7} \frac{T * m^2}{C * \frac{m}{s}} }[/math]

In this equation, q represents the scalar charge of the particle, [math]\displaystyle{ \vec v }[/math] is the vector velocity of the moving particle, and [math]\displaystyle{ \hat r }[/math] is a unit vector that points from the charge to the observation location. [math]\displaystyle{ \vec v \times \hat r }[/math] is a cross product that can be computed by the formula:

This cross product can also be expressed in terms of magnitude: [math]\displaystyle{ \left\vert \vec v \times \hat r \right\vert = \left\vert \vec v \right\vert \left\vert \hat r \right\vert \sin \theta }[/math]

[math]\displaystyle{ \theta }[/math] in this formula represents the angle between [math]\displaystyle{ \vec v }[/math] and [math]\displaystyle{ \hat r }[/math] when they are placed tail to tail.

The easiest way to determine the direction of [math]\displaystyle{ \vec v \times \hat r }[/math] is the Right Hand Rule. However, it is important to remember to take into account the charge of the particle from the Biot-Savart Law. If the particle is positively charged, the magnetic field will point in the same direction as [math]\displaystyle{ \vec v \times \hat r }[/math], while the magnetic field will point in the opposite direction as [math]\displaystyle{ \vec v \times \hat r }[/math] if the particle is negatively charged.

A Computational Model

VPython can easily model the magnetic field created by a moving charged particle. Check out this code: Magnetic Field for a Positively Charged Particle. As the particle travels towards the observation location, it creates a magnetic field that grows in magnitude the closer the particle is to the observation location. As the proton gets farther away, the magnetic field decreases in magnitude as expected.

Now compare with this code: Magnetic Field for a Negatively Charged Particle. This negatively charged particle is the exact same as the positively charged particle in all but its sign. Looking carefully reveals that this change in sign has reversed the direction of the magnetic field from what it was in the positively charged particle demo.

Examples

Simple

At a particular instant, a proton at the origin has a velocity, [math]\displaystyle{ \vec v }[/math] of < 4e4, -3e4, 0 > m/s. Calculate the magnetic field at the observation location < 0.03, 0.06, 0 > m, due to the moving proton.

Solution:

1. First, find [math]\displaystyle{ \vec r }[/math] and [math]\displaystyle{ \hat r }[/math]. Subtracting the observation location from the source location gives us

[math]\displaystyle{ \vec r = }[/math] < 0.03, 0.06, 0 >.

The unit vector can then be found with the formula [math]\displaystyle{ \hat r = \frac{\vec r}{\left\vert \vec r \right\vert} }[/math], which results in

[math]\displaystyle{ \hat r = }[/math] < 0.447, 0.894, 0 >.

2. Once we have both [math]\displaystyle{ \vec v }[/math] and [math]\displaystyle{ \hat r }[/math], we can take their cross product and solve the Biot-Savart Law. The resulting cross product is < 0, 0, 49200 >.

3. Finally, we can finish solving the Biot-Savart Law by multiplying the charge of a proton, [math]\displaystyle{ 1.6*10^{-19} }[/math], to our cross product from step 2 and dividing that by [math]\displaystyle{ \left\vert \vec r \right\vert^2 }[/math]. To finish, multiply by the constant: [math]\displaystyle{ \frac{\mu_0}{4*\pi} = 1*10^{-7} \frac{T * m^2}{C * \frac{m}{s}} }[/math]

The final answer will be [math]\displaystyle{ \lt 0, 0, 1.75*10^{-19}\gt }[/math] T

Medium

The electron in the figure below is traveling with a speed of [math]\displaystyle{ \left\vert \vec v \right\vert = 4*10^6 }[/math] m/s. What is the magnitude of the magnetic field at location A if [math]\displaystyle{ \left\vert \vec r \right\vert = 7*10^{-10} }[/math] m and [math]\displaystyle{ \theta = 57^\circ }[/math]?

Solution:

1. First find [math]\displaystyle{ \vec v }[/math] by multiplying [math]\displaystyle{ \left\vert \vec v \right\vert }[/math] by [math]\displaystyle{ \cos \theta }[/math] and by [math]\displaystyle{ \sin \theta }[/math]. Multiplying by [math]\displaystyle{ \cos \theta }[/math] will give the x-component of [math]\displaystyle{ \vec v }[/math], while multiplying by [math]\displaystyle{ \sin \theta }[/math] will give the y-component of [math]\displaystyle{ \vec v }[/math].

2. Now find [math]\displaystyle{ \left\vert \vec v \right\vert }[/math].

3. Multiply this by the magnitude of the charge for an electron, as well as by [math]\displaystyle{ \mu_0 }[/math] and then divide this by [math]\displaystyle{ r^2 }[/math]

The final answer will be 0.11 T

Complex

An electron is moving horizontally to the right with speed [math]\displaystyle{ 6*10^6 }[/math] m/s. What is the magnetic field due to this moving electron at the indicated locations in the figure? Each location is d = 8 cm from the electron, and the angle θ = 33°. Give both magnitude and direction of the magnetic field at each location.

Solution:

1. Find the [math]\displaystyle{ \vec r }[/math] for each location, using [math]\displaystyle{ \theta }[/math] to calculate the x and y components.

2. For each [math]\displaystyle{ \vec r }[/math], take the cross product, [math]\displaystyle{ \vec v \times \hat r }[/math], where the [math]\displaystyle{ \left\vert \vec v \right\vert }[/math] is given.

3. Multiply each respective cross product by the magnitude of charge and [math]\displaystyle{ \mu_0 }[/math].

4. In order to find the direction of the magnetic fields, use the Right Hand Rule. One way to do this is to point your thumb in the direction of the velocity, your pointer finger in the direction of r hat, and look which way your palm is facing in order to find the direction of the magnetic field.

At location P2 and P5 the magnetic field will be zero. P1 will be into the page, P3 will be out of the page, P4 will be out of the page, and P6 will be into the page, all with a magnitude of [math]\displaystyle{ 8.17*10^{-18} }[/math] T.

Important Applications

A single moving point charge represents the most simple situation of charges moving in space to produce a magnetic field. In reality, this situation rarely occurs, however understanding how a single moving point charge interacts to produce a field will allow you to understand how sets of moving charges produce a field in space as well.

We can use the magnetic field of a single moving charge formula to derive a more practical formula of a short current carrying wire:

Where dl is a vector pointing in the same direction as the conventional current This formula is more applicable to many real life situations where we can have macroscopic objects that can be used to calculate the magnetic field, such as the Magnetic Field of a Long Straight Wire or a Magnetic Field of a Loop.

We can apply the magnetic field formula to numerous situations by integration.

History

Oliver Heaviside first derived this relationship from Maxwell's Equations in 1888. The Biot-Savart Law was named after Jean-Baptiste Biot and Felix Savart who, in 1820, showed a needle deflection from a current carrying wire, thus relating electricity and magnetism.

See also

Biot-Savart Law for Currents

Magnetic Force

Right-Hand Rule

References

http://maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html http://ruh.li/3DMathVectors.html

Moving Point Charge