Magnetic Force
Edited by Brandon Kang Fall 2017
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history) KALEY PARCHINSKI 11/6/16 JULIA LEONARD Spring 2017
The Main Idea
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can 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 (i.e. not the source charge but the actual charged particle of our system of study) must be moving. "If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest." 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. We will combine the two in later sections.
The Main Idea - Aurora Borealis Edition
The Aurora Borealis or more commonly called, 'The Northern Lights' is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we've all done this admittedly, kinda fun experiment in chemistry).
However, the part that we're interested in is when the electrons collide with the magnetic field of the earth.
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.
For our purposes we're going to focus on two things: 1- The circular orbit of the electrons in the Earth's magnetic field 2- The helical orbit of the electrons in the Earth's magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights.
Let's imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation (1). However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth's magnetic field) then the charged particle will continue to move in a circular path in the x-z plane.
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.
(7) [math]\displaystyle{ {\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)} }[/math]
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.
[[1]]
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} = \frac {m v} {q 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 at this exact position? 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! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, 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]
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y).
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.
Magnetic Forces in Wires
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation:
[math]\displaystyle{ |\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus }[/math]
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F.
Simple
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?
[math]\displaystyle{ |\vec F_{mag}| = ILBsin(45) }[/math]
[math]\displaystyle{ |\vec F_{mag}| = (0.6)(0.005)(sin(45)) }[/math]
[math]\displaystyle{ |\vec F_{mag}| = 0.002 }[/math] N into the page
Middling
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?
[math]\displaystyle{ |\vec F_{grav}| = mg }[/math]
[math]\displaystyle{ |\vec F_{grav}| = \vec F_{mag} }[/math] because there is no gravitational acceleration, the net force must equal zero.
[math]\displaystyle{ mg = I(L\times\vec B) }[/math]
[math]\displaystyle{ I = \frac{mg}{LB} }[/math]
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.
I, the conventional current, flows to the left.
Application (i.e. What Does This Have To Do With Anything?)
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle.
As a chemist (The Astrochemist, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force.
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics of the early universe. None of this would be possible without the dynamics of magnetic force!
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits.
History
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861.
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820.
Hendrik Lorentz provided the actual "Lorentz Force Law" of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands.
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry).
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite.
See also
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
Further reading
Argonne National Laboratory information regarding the Advanced Photon Source
National Oceanic and Atmospheric Administration's explanation of the Northern Lights
Secrets of the Polar Aurora - NASA
National Geographic - Heavenly Lights
External links
Footage from space of Aurora Borealis
References
1. Chabay, R.W; Sherwood, B.A.; Matter and Interactions. 2015. 4. 805-812.