Magnetic Force

Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)

The Main Idea

We know so far that an electric field will act on a charged particle in a specific manner. In effect, a charged particle in the vicinity of any electric field will undergo a force based upon the magnitude and sign of the charged particle. This electric field is generated regardless of whether the charge is stationary or moving.

In the case of a moving charged particle, it is also known to generate a magnetic field. Whenever any charge maintains some velocity, it will necessarily produce a magnetic field. This applies whether you're dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk.

Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest must be moving. These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest.

A Mathematical Model

Suppose we have a moving particle. It has a charge given by q. It has a velocity given by [math]\displaystyle{ {\vec{v}} }[/math]. It is also in the presence of a magnetic field given by [math]\displaystyle{ {\vec{B}} }[/math]. The force that this particle will experience is given by the following:

(1) [math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

Therefore, for a particle at rest ([math]\displaystyle{ {\vec{v} = \vec{0}} }[/math]), the particle will experience a force given by [math]\displaystyle{ {\vec{F} = \vec{0}} }[/math].

The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).

Note that the above equation (1) denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way:

(2) [math]\displaystyle{ {\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)} }[/math]

In equation (2), the angle [math]\displaystyle{ {\theta} }[/math] represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It's important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter.

What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire.

Recall... (1) [math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

Thus, for some section of charge [math]\displaystyle{ {\Delta q} }[/math]... (3) [math]\displaystyle{ {\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})} }[/math]

... hence, for n charged particles, A cross sectional area, and sectional length [math]\displaystyle{ {\Delta L} }[/math], we have... (4) [math]\displaystyle{ {\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})} }[/math]

... and now, by re-arranging the terms to collectively represent some current I, we have... (5) [math]\displaystyle{ {\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})} }[/math]

Equation (5) can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!

Recall that a moving charged particle generates a magnetic field [math]\displaystyle{ {\vec{B}} }[/math] given by the following:

(6) [math]\displaystyle{ {\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}} }[/math]

Equation (6) involves the vector [math]\displaystyle{ {\hat{r}} }[/math] which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It's important to remember that a charge won't enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.

A Computational Model

The following link contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.

Examples

We can now consider several example problems related to this topic.

Simple

Question:

A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle?

Solution:

This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have...

[math]\displaystyle{ {\vec{v} = \lt 4E5,0,0\gt m/s} }[/math]

[math]\displaystyle{ {\vec{B} = \lt 0,0,0.2\gt T} }[/math]

q = 1.6E-19 C

Therefore:

[math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

[math]\displaystyle{ {\vec{F} = (1.6E-19 C) \lt 4E5,0,0\gt m/s \times \lt 0,0,0.2\gt T} }[/math]

The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction.

Thus...

[math]\displaystyle{ {\vec{F} = (1.6E-19 C)(4E5 m/s * 0.2 T} }[/math]

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