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 4 \times 10^5,0,0\gt m/s} }[/math]

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

[math]\displaystyle{ q = {1.6 \times 10^{-19} C} }[/math]

Therefore:

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

[math]\displaystyle{ {\vec{F} = (1.6 \times 10^{-19}) \lt 4 \times 10^5,0,0\gt \times \lt 0,0,0.2\gt } }[/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} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = \lt 0, 1.28 \times 10^{-14}, 0\gt N} }[/math]

Middling

Question:

Suppose we have a situation where a positively charged particle ([math]\displaystyle{ {+ q} }[/math]) of mass m is in a region where a magnetic field ([math]\displaystyle{ {\vec{B}} }[/math]) is applied. It travels at a velocity ([math]\displaystyle{ {\vec{v}} }[/math]). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?

Solution:

This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it's not as hard as it seems.

Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:

First... the magnetic force on the particle is given by the following:

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

Because [math]\displaystyle{ {\vec{v}} }[/math] and [math]\displaystyle{ {\vec{B}} }[/math] are effectively perpendicular, the two vectors can be effectively combined in the following way:

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

The force [math]\displaystyle{ {\vec{F}} }[/math] is constantly inward to generate a circular motion based path of the particle.

Recall that for circular motion with a constant inward force, the force is given by:

[math]\displaystyle{ {|\vec{F}| = m \frac{|\vec{v}|^2} {r}} }[/math]

Thus, we can set the forces equal to each other:

[math]\displaystyle{ {|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}} }[/math]

Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem...

[math]\displaystyle{ {r = \frac {m v^2} {q v B}} }[/math]

Difficult

Problem:

Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius [math]\displaystyle{ {R_1} }[/math] exists to the left of the origin a distance [math]\displaystyle{ {d_1} }[/math] which maintains a current [math]\displaystyle{ {I_1} }[/math]. Another current loop of radius [math]\displaystyle{ {R_2} }[/math] exists to the right of the origin a distance [math]\displaystyle{ {d_2} }[/math], and it maintains a current [math]\displaystyle{ {I_2} }[/math]. The particle itself moves upward on the positive z-axis with a velocity [math]\displaystyle{ {\vec{v}} }[/math]

Assume the following:

[math]\displaystyle{ {I_1 = I_2} }[/math]

[math]\displaystyle{ {R_1 = 0.5R_2} }[/math]

[math]\displaystyle{ {d_1 = 3d_2} }[/math]

[math]\displaystyle{ {d_1, d_2 \gt \gt R_1, R_2} }[/math]

The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right).

What is the net force exerted on the particle? Determine an expression in terms of any of the above state variables.

Solution:

We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging.

Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.

For loop 1:

[math]\displaystyle{ {B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}} }[/math] (This approximation can be used because of the fourth assumption made above, where [math]\displaystyle{ {d_1} }[/math] is considerably larger than [math]\displaystyle{ {R_1} }[/math])

For loop 2:

[math]\displaystyle{ {B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}} }[/math] (This approximation can be used because of the fourth assumption made above, where [math]\displaystyle{ {d_2} }[/math] is considerably larger than [math]\displaystyle{ {d_2} }[/math])

Now we combine the appropriate values for radius and distance in terms of [math]\displaystyle{ {R_2} }[/math] and [math]\displaystyle{ {d_2} }[/math], so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.

[math]\displaystyle{ {B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}} }[/math]

[math]\displaystyle{ {B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}} }[/math]

[math]\displaystyle{ {B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}} }[/math]

[math]\displaystyle{ {B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}} }[/math]

Now let's pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... but wait! The particle involved here is an electron! That means that the electron would experience a force in the positive y-direction. Therefore, we can say:

[math]\displaystyle{ {|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|} }[/math]

We can now involve our determined magnetic field that was generated by the two current carrying rings.

[math]\displaystyle{ {|\vec{F_{net}}| = e |\vec{v}| |\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}|} }[/math]

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