Magnetic Force: Difference between revisions

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===Simple===
===Simple===


'''Question:'''  
'''Question:'''
 
An electron moves with velocity 
<math>{\vec v = \langle 3.0\times10^6,\;0,\;0\rangle\ \text{m/s}}</math> 
in a magnetic field 
<math>{\vec B = \langle 0,\;0.15,\;0\rangle\ \text{T}}</math>. 


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?  
What is the magnetic force on the electron?
Use <math>{q = -1.6\times10^{-19}\ \text{C}}</math>.


'''Solution:'''  
'''Solution:'''


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


<math>{\vec{v} = <4 \times 10^5,0,0> m/s}</math>
<math>{\vec F = q\vec v \times \vec B}.</math>


<math>{\vec{B} = <0,0,0.2> T}</math>
First, we write the velocity and magnetic field vectors in component form:


<math>q = {1.6 \times 10^{-19} C}</math>
<math>{\vec v = \langle 3.0\times10^6,\;0,\;0\rangle}</math> 
<math>{\vec B = \langle 0,\;0.15,\;0\rangle}.</math>


Therefore:  
We now compute the cross product <math>{\vec v \times \vec B}</math>. The velocity points along +x, and the field points along +y. The cross product of +x with +y points along +z. The determinant form looks like this:


<math>{\vec{F} = q\vec{v}\times\vec{B}}</math>
<math>
{
\vec v \times \vec B
=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
3.0\times10^6 & 0 & 0\\
0 & 0.15 & 0
\end{vmatrix}
=
\langle 0,\;0,\;4.5\times10^5\rangle
}.
</math>


<math>{\vec{F} = (1.6 \times 10^{-19}) <4 \times 10^5,0,0> \times <0,0,0.2>}</math>
This is the result for a hypothetical positive charge. To find the physical force, we multiply by the electron charge:


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 (middle finger) in the positive z-direction, then the resulting force vector (direction of your thumb) must be in the negative y-direction.  
<math>
{
\vec F = q(\vec v \times \vec B)
= (-1.6\times10^{-19})\langle 0,\;0,\;4.5\times10^5\rangle
= \langle 0,\;0,\;-7.2\times10^{-14}\rangle\ \text{N}
}.
</math>


Thus...
So the electron feels a force of magnitude <math>{7.2\times10^{-14}\ \text{N}}</math> in the negative z direction. The negative sign in the charge flipped the direction from +z to -z.


<math>{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = <0, -1.28 \times 10^{-14}, 0> N}</math>


===Middling===
===Middling===
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'''Question:'''
'''Question:'''


Suppose we have a situation where a positively charged particle (<math>{+ q}</math>) of mass ''m'' is in a region where a magnetic field (<math>{\vec{B}}</math>) is applied. It travels at a velocity (<math>{\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 in which this particle travels in terms of values given?
A proton moves perpendicular to a uniform magnetic field of magnitude 
<math>{B = 0.20\ \text{T}}</math>. 
The speed of the proton is
<math>{v = 3.0\times10^6\ \text{m/s}}</math>. 
 
What is the radius of the circular path that the proton follows? 
 
Use <math>{m_p = 1.67\times10^{-27}\ \text{kg}}</math> and <math>{q_p = 1.6\times10^{-19}\ \text{C}}</math>.


'''Solution:'''
'''Solution:'''


This problem may appear complicated, but it's not as hard as it seems.  
When a charged particle moves in a direction that is perpendicular to a magnetic field, the magnetic force is always perpendicular to its velocity. This constant perpendicular force causes circular motion.
 
The magnetic force is
 
<math>{|\vec F| = qvB}.</math>


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:
The centripetal force required for circular motion is


First... the magnetic force on the particle is given by the following:
<math>{|\vec F| = m\frac{v^2}{r}}.</math>


<math>{\vec{F} = q\vec{v}\times\vec{B}}</math>
Since these are the same physical force, we set them equal:


Because <math>{\vec{v}}</math> and <math>{\vec{B}}</math> are effectively perpendicular, the two vectors can be effectively combined in the following way:
<math>{qvB = m\frac{v^2}{r}}.</math>


<math>{|\vec{F}| = q|\vec{v}| |\vec{B}|}</math>
Now we solve for the radius <math>{r}</math>:


The force <math>{\vec{F}}</math> is constantly inward to generate a circular motion based path of the particle.
<math>{r = \frac{mv}{qB}}.</math>


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


<math>{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}</math>
<math>
{
r = \frac{(1.67\times10^{-27})(3.0\times10^6)}
{(1.6\times10^{-19})(0.20)}
}.
</math>


Thus, we can set the forces equal to each other:
Carrying out the arithmetic gives


<math>{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}</math>
<math>{r \approx 0.16\ \text{m}}.</math>


Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem...  
So the proton moves in a circle of radius about 0.16 meters.


<math>{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}</math>


===Difficult===
===Difficult===
Line 154: Line 193:
'''Problem:'''
'''Problem:'''


Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius <math>{R_1}</math> exists to the left of the origin a distance <math>{d_1}</math> which maintains a current <math>{I_1}</math>. Another current loop of radius <math>{R_2}</math> exists to the right of the origin a distance <math>{d_2}</math>, and it maintains a current <math>{I_2}</math>. The particle itself moves upward on the positive z-axis with a velocity <math>{\vec{v}}</math>.
A beam of positively charged particles enters a region with both an electric and a magnetic field. The electric field is
 
<math>{\vec E = \langle 0,\;E,\;0\rangle}</math>


Assume the following:
and the magnetic field is


<math>{I_1 = I_2}</math>
<math>{\vec B = \langle 0,\;0,\;B\rangle}.</math>
<math>{R_1 = 0.5R_2}</math>


<math>{d_1 = 3d_2}</math>
The particles enter with velocity


<math>{d_1, d_2 >> R_1, R_2}</math>
<math>{\vec v = \langle v,\;0,\;0\rangle}</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).  
along the +x direction.


What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the variables staed above.  
1. Determine the speed <math>{v}</math> for which the particles pass through the region without deflection. 
2. Determine the net force on the particles at that speed.


'''Solution:'''
'''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.  
Each particle feels two forces.
 
The electric force is given by
 
<math>{\vec F_E = q\vec E = \langle 0,\;qE,\;0\rangle}.</math>
 
This points in the +y direction.
 
The magnetic force is


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.  
<math>{\vec F_B = q(\vec v \times \vec B)}.</math>


''For loop 1:''
We compute the cross product. The velocity is along +x, the magnetic field along +z. The cross product +x with +z points in the negative y direction:


<math>{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>{d_1}</math> is considerably larger than <math>{R_1}</math>)
<math>
{
\vec v \times \vec B
= \langle v,\;0,\;0\rangle \times \langle 0,\;0,\;B\rangle
= \langle 0,\;-vB,\;0\rangle
}.
</math>


''For loop 2:''
Therefore


<math>{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>{d_2}</math> is considerably larger than <math>{d_2}</math>)
<math>{\vec F_B = \langle 0,\;-qvB,\;0\rangle}.</math>


Now we combine the appropriate values for radius and distance in terms of <math>{R_2}</math> and <math>{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.
This magnetic force points in the negative y direction.


<math>{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}</math>
For the particles to pass through undeflected, the net force in the y direction must be zero:


<math>{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}</math>
<math>{\vec F_E + \vec F_B = \vec 0}.</math>


<math>{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>
In component form, this gives


<math>{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}</math>
<math>{qE - qvB = 0}.</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:
We can cancel <math>{q}</math> (assuming nonzero charge) and solve for <math>{v}</math>:


<math>{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}</math>
<math>{v = \frac{E}{B}}.</math>


We can now involve our determined magnetic field that was generated by the two current carrying rings.
At this speed, the electric and magnetic forces cancel exactly, so the net force is


<math>{|\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>
<math>{\vec F_{\text{net}} = \vec 0}.</math>


The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y).  
The particles then move in a straight line without deflection.


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===
===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:
Because a current carrying wire is a collection of moving charges, the wire as a whole feels a magnetic force. For a straight segment of wire of length <math>{L}</math> with current <math>{I}</math> in a magnetic field of magnitude <math>{B}</math>, the magnitude of the force is


<math display = "block">|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus</math>
<math>{|\vec F_{mag}| = ILB\sin\theta}.</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.  
Here, <math>{\theta}</math> is the angle between the direction of the current and the magnetic field. The right hand rule still applies. Point your index finger in the direction of the current, your middle finger in the direction of the magnetic field, and your thumb will show you the direction of the force.


====Simple====
====Simple====
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B, the magnetic field on the wire, is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the magnetic force on the wire?


<math>|\vec F_{mag}| = ILBsin(45)</math>
A wire lies in the xy plane. Conventional current flows to the right. The magnetic field has magnitude <math>{B = 0.005\ \text{T}}</math> and is directed downward at a <math>{45^\circ}</math> angle relative to the wire. The current in the wire is <math>{I = 0.6\ \text{A}}</math>. What is the magnetic force on a 1 meter segment of this wire?
 
We use
 
<math>{|\vec F_{mag}| = ILB\sin 45^\circ}.</math>
 
We substitute the values:


<math>|\vec F_{mag}| = (0.6)(0.005)(sin(45))</math>
<math>{|\vec F_{mag}| = (0.6)(1.0)(0.005)\sin 45^\circ}.</math>


<math>|\vec F_{mag}| = 0.002</math> N into the page
Since <math>{\sin 45^\circ \approx \frac{\sqrt{2}}{2} \approx 0.707}</math>, we get approximately
 
<math>{|\vec F_{mag}| \approx (0.6)(0.005)(0.707) \approx 0.002\ \text{N}}.</math>
 
The detailed geometry of the right hand rule shows that the force points into the page.


====Middling====
====Middling====
A horizontal bar is falling at a constant velocity v. B, the magnetic field, points into the page. What is the the magnitude and direction of current in the bar?


<math>|\vec F_{grav}| = mg</math>
A horizontal metal bar falls downward with constant velocity in a region where the magnetic field points into the page. We know gravity pulls the bar downward. We are told the bar does not speed up or slow down, so the net force on it is zero. What is the magnitude and direction of the current in the bar?
 
First, the gravitational force on the bar is
 
<math>{|\vec F_{grav}| = mg}.</math>
 
Because the bar is not accelerating, the magnetic force must balance gravity:
 
<math>{|\vec F_{grav}| = |\vec F_{mag}|}.</math>
 
For a straight conductor of length <math>{L}</math> in a field <math>{B}</math>, the magnetic force is
 
<math>{|\vec F_{mag}| = ILB}.</math>
 
Equating the forces gives
 
<math>{mg = ILB}.</math>
 
We solve for the current:
 
<math>{I = \frac{mg}{LB}}.</math>
 
The direction of the current must be such that the magnetic force acts upward. Gravity is in the negative y direction, so the magnetic force must be in the positive y direction to balance it.
 
Using the right hand rule, we point the thumb in the positive y direction for the magnetic force. The magnetic field points into the page, which is the negative z direction, so we point the middle finger in the negative z direction. The index finger then points along the negative x direction. This is the direction of conventional current. Therefore, the current flows to the left.
 
===Circular Motion in Earth's Magnetic Field===
 
We now look at situations that connect to the circular motion of charged particles in Earth's magnetic field. In the upper atmosphere, electrons and ions spiral and circle in the geomagnetic field, which contributes to phenomena such as the auroras.
 
====Simple====
 
'''Question:'''
 
An electron in the upper atmosphere moves in a region where Earth's magnetic field can be approximated as uniform with magnitude 
<math>{B = 5.0\times10^{-5}\ \text{T}}</math>. 
The electron's velocity is initially perpendicular to the magnetic field and has magnitude 
<math>{v = 2.0\times10^6\ \text{m/s}}</math>. 
 
Find the radius of the circular orbit that the electron follows. 
 
Use <math>{m_e = 9.11\times10^{-31}\ \text{kg}}</math> and <math>{q_e = -1.6\times10^{-19}\ \text{C}}</math>. 
 
'''Solution:'''
 
When the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is
 
<math>{|\vec F| = qvB}.</math>
 
This force acts as the centripetal force for circular motion:
 
<math>{|\vec F| = m\frac{v^2}{r}}.</math>
 
Equating them:
 
<math>{qvB = m\frac{v^2}{r}}.</math>
 
We are only interested in magnitudes here, so we use <math>{|q_e| = 1.6\times10^{-19}}</math> and write
 
<math>{|q|vB = m\frac{v^2}{r}}.</math>
 
Solving for <math>{r}</math>:
 
<math>{r = \frac{mv}{|q|B}}.</math>
 
Substitute the values:
 
<math>
{
r = \frac{(9.11\times10^{-31})(2.0\times10^6)}
{(1.6\times10^{-19})(5.0\times10^{-5})}
}.
</math>
 
This expression shows that even with relatively small fields, light particles like electrons can follow quite large circular paths on the scale of meters or more, depending on the numbers.
 
The sign of the charge affects the direction of the motion, not the radius. The radius depends only on the magnitudes of the quantities.
 
====Middling====
 
'''Question:'''
 
A proton enters Earth's magnetic field with a speed 
<math>{v = 1.0\times10^7\ \text{m/s}}</math>, 
perpendicular to a local field of magnitude 
<math>{B = 5.0\times10^{-5}\ \text{T}}</math>. 
 
1. Find the radius of its circular motion. 
2. Find the period, that is, the time it takes to complete one full circle. 
 
Use <math>{m_p = 1.67\times10^{-27}\ \text{kg}}</math> and <math>{q_p = 1.6\times10^{-19}\ \text{C}}</math>.
 
'''Solution:'''
 
1. **Radius**
 
As before, the radius is given by
 
<math>{r = \frac{mv}{qB}}.</math>
 
We plug in magnitudes:
 
<math>
{
r = \frac{(1.67\times10^{-27})(1.0\times10^7)}
{(1.6\times10^{-19})(5.0\times10^{-5})}
}.
</math>
 
This gives the size of the circular orbit in meters. The specific numerical value can be found with a calculator.
 
2. **Period**
 
The period <math>{T}</math> is the time for one complete revolution. For circular motion:
 
<math>{T = \frac{\text{circumference}}{\text{speed}} = \frac{2\pi r}{v}}.</math>
 
Substituting the expression for <math>{r}</math>:
 
<math>
{
T = \frac{2\pi}{v}\left(\frac{mv}{qB}\right)
= \frac{2\pi m}{qB}
}.
</math>
 
Notice that the speed cancels. This is a nice and important result: the period of circular motion in a uniform magnetic field depends only on <math>{m}</math>, <math>{q}</math>, and <math>{B}</math>, not on the speed. So we have
 
<math>{T = \frac{2\pi m_p}{q_p B}}.</math>
 
This can be evaluated numerically by substitution. The key physics concept is that all protons in the same region with the same magnetic field complete circles in the same amount of time, regardless of their speeds, as long as their motion remains perpendicular to the field.
 
 
===Helical Motion in Magnetic Fields===
 
When the magnetic field is not perpendicular to the velocity, we can decompose the velocity into two components: one parallel to the field and one perpendicular to it. The perpendicular component produces circular motion around the field lines. The parallel component produces motion along the field lines. The combination of these two motions creates a helical path.
 
====Simple====
 
'''Question:'''
 
An electron enters a region of uniform magnetic field 
<math>{\vec B = \langle 0,\;0,\;B\rangle}</math>. 
Its velocity is
 
<math>{\vec v = \langle v_x,\;0,\;v_z\rangle}</math>,
 
so the electron is moving in the xz plane. Suppose that <math>{v_x}</math> is perpendicular to the magnetic field and <math>{v_z}</math> is parallel to the field.
 
1. Which component of the velocity causes circular motion, and which component causes motion along the field? 
2. Describe the shape of the trajectory.
 
Assume <math>{q_e = -1.6\times10^{-19}\ \text{C}}</math> and <math>{m_e = 9.11\times10^{-31}\ \text{kg}}</math>.
 
'''Solution:'''
 
We split the velocity into a component that is perpendicular to <math>{\vec B}</math> and a component that is parallel to <math>{\vec B}</math>.
 
Here, <math>{\vec B}</math> points along the z axis, so:
 
- The component <math>{v_z}</math> is parallel to <math>{\vec B}</math>.
- The component <math>{v_x}</math> is perpendicular to <math>{\vec B}</math>.
 
1. The magnetic force is proportional to <math>{\vec v \times \vec B}</math>. The force depends only on the part of <math>{\vec v}</math> that is perpendicular to <math>{\vec B}</math>. The parallel component does not feel any magnetic force.
 
So:
 
- <math>{v_x}</math> causes circular motion in the plane that is perpendicular to the field, which is the x direction around the z axis.
- <math>{v_z}</math> causes motion straight along the field lines in the z direction.
 
2. The combination of circular motion due to <math>{v_x}</math> and linear motion due to <math>{v_z}</math> produces a helix. The electron spirals around the z axis while steadily moving along it. The radius of the spiral is controlled by <math>{v_x}</math>, and the pitch of the helix (how far it moves along z per turn) is controlled by <math>{v_z}</math>.
 
====Middling====
 
'''Question:'''
 
A positively charged particle of mass <math>{m}</math> and charge <math>{q}</math> enters a uniform magnetic field 
<math>{\vec B = \langle 0,\;B,\;0\rangle}</math> 
with velocity
 
<math>{\vec v = \langle v_x,\;0,\;v_z\rangle}.</math>
 
This means the velocity has a perpendicular component <math>{v_x}</math> relative to <math>{\vec B}</math> and a parallel component <math>{v_z}</math>.
 
1. Find an expression for the radius of the circular part of the motion in terms of <math>{m}</math>, <math>{q}</math>, <math>{B}</math>, and <math>{v_x}</math>. 
2. Find the pitch of the helix, which is the distance traveled along the direction of the magnetic field in one full circular orbit, in terms of <math>{v_z}</math> and the period <math>{T}</math>. 
 
'''Solution:'''
 
1. **Radius of the circular motion**
 
Only the perpendicular component <math>{v_x}</math> contributes to the circular motion around the field lines. The radius is given by
 
<math>{r = \frac{mv_\perp}{qB}}.</math>
 
Here, <math>{v_\perp = v_x}</math>, since <math>{v_x}</math> is perpendicular to <math>{\vec B}</math>. Therefore,
 
<math>{r = \frac{mv_x}{qB}}.</math>
 
This expression gives the radius of the circular projection of the path.
 
2. **Pitch of the helical path**
 
The particle also moves along the direction of the magnetic field with speed <math>{v_z}</math>. The time for one full circle is the period <math>{T}</math>, which is the same as in pure circular motion. The distance traveled along the field in that time is
 
<math>{\text{pitch} = v_z T}.</math>
 
We can express <math>{T}</math> in terms of known quantities. From circular motion in a magnetic field, the period is
 
<math>{T = \frac{2\pi m}{qB}}.</math>
 
This does not depend on the speed. Therefore, the pitch is
 
<math>{\text{pitch} = v_z \left(\frac{2\pi m}{qB}\right)}.</math>


<math>|\vec F_{grav}| = \vec F_{mag}</math>     because there is no gravitational acceleration, the net force must equal zero.  
This shows that the radius depends on <math>{v_x}</math>, while the pitch depends on <math>{v_z}</math>. By changing the ratio of these two components, we can make the helix tighter or more stretched out.


<math>mg = I(L\times\vec B)</math>


<math>I = \frac{mg}{LB}</math>
These example categories illustrate how the basic law 


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.  
<math>{\vec F = q\vec v \times \vec B}</math>  


I, the conventional current, flows to the left.
leads to a wide range of physical behavior: straight line motion if there is no field or no velocity, circular orbits when the motion is perpendicular to the field, helical motion when there is a mixture of parallel and perpendicular components, and forces on entire current carrying wires when many charges move together.


==Application (i.e. What Does This Have To Do With Anything?)==
==Application (i.e. What Does This Have To Do With Anything?)==

Revision as of 22:08, 22 November 2025

Claimed by Ritwic Verma, Fall 2025

The Main Idea

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.

By applying forces to charged particles depending on both their charges and the properties of the electric field itself, an electric field acts as a mediator in interactions between charged particles. A charged particle encounters a force because of the existence of an electric field while it is in one. This force is closely correlated with the particle's charge and the strength of the electric field. The equation \( \mathbf{F} = q \cdot \mathbf{E} \) mathematically expresses this relationship, with \( \mathbf{F} \) representing the force, \( q \) representing the particle's charge, and \( \mathbf{E} \) representing the electric field vector.

The sign of the particle's charge determines the force's direction. The force acts in the direction of the electric field if the charge is positive. In contrast, the force acts in the opposite direction of the electric field when the charge is negative. This differentiation emphasizes that the force and the electric field are both vectors.

Crucially, the existence of the electric field is independent of the mobility or stationary state of the source charge that generates it. This notion follows from the fact that an electric field is created when there are changes in the distribution of charge, whether such changes are brought about by the motion of charges or by other factors.

Magnetic forces are on moving particles, not stationary particles which means that the calculation of magnetic force MUST relate to the particle's velocity (we see this quantitatively with the Biot-Savart Law).

If the source charge is moving, it also generate a magnetic field; so not only is velocity involved in calculation of the magnetic force on a moving particle, or collection of moving particles (as we see in a rod or a wire), but this phenomenal relationship includes magnetic field as well.

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.

The excited molecules and atoms, having absorbed the energy imparted by the electrons, undergo a subsequent relaxation process. As they return to lower energy levels, a radiant emission of light ensues — the hallmark Northern Lights spectacle. The distinctive hues and vibrant colors characterizing the display are contingent upon the specific molecules and atoms involved in these collisions. The variance in colors is a testament to the diversity of atmospheric constituents participating in this cosmic ballet. This celestial dance, orchestrated by the interaction between charged particles and the Earth's magnetic field, not only enchants observers but also serves as a captivating reminder of the intricate interplay between astrophysical phenomena and the Earth's atmospheric composition.

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].

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, and equation (2) gives the magnitude of the magnetic force. 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. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. 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.


A Computational Model

The following Glowscript model displays a moving particle's path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force.

Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle's velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic force always point in the positive z direction?

No, the direction of the magnetic force consistently changes since the direction of the particle's velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle's velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle.


[Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]


However, consider the case where the initial direction of the electron's velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron's path, and thus the overall path of the electron resembles that of a helix.


[Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]


Also, take note of the iterative calculations made in the code. Within the code, we must initalize values for the initial velocity and momentum, position, mass, and charge of the particle, and magnetic field present in the location of the electron. In the iterative calculations, we must update the value of the magnetic force, as it is constantly changing directions since the electron's velocity is also changing in direction. Similarly, a net force causes a change in momentum, so we must update the momentum and velocity of the particle by utilizing the momentum principle where the derivative of momentum with respect to time is equivalent to the net force acting upon the particle. Furthermore, we update the particle's position and extend and append the trail with the particle's current location to display the path.

Examples

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

Simple

Question:

An electron moves with velocity [math]\displaystyle{ {\vec v = \langle 3.0\times10^6,\;0,\;0\rangle\ \text{m/s}} }[/math] in a magnetic field [math]\displaystyle{ {\vec B = \langle 0,\;0.15,\;0\rangle\ \text{T}} }[/math].

What is the magnetic force on the electron? Use [math]\displaystyle{ {q = -1.6\times10^{-19}\ \text{C}} }[/math].

Solution:

The magnetic force on a moving charge is given by

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

First, we write the velocity and magnetic field vectors in component form:

[math]\displaystyle{ {\vec v = \langle 3.0\times10^6,\;0,\;0\rangle} }[/math] [math]\displaystyle{ {\vec B = \langle 0,\;0.15,\;0\rangle}. }[/math]

We now compute the cross product [math]\displaystyle{ {\vec v \times \vec B} }[/math]. The velocity points along +x, and the field points along +y. The cross product of +x with +y points along +z. The determinant form looks like this:

[math]\displaystyle{ { \vec v \times \vec B = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3.0\times10^6 & 0 & 0\\ 0 & 0.15 & 0 \end{vmatrix} = \langle 0,\;0,\;4.5\times10^5\rangle }. }[/math]

This is the result for a hypothetical positive charge. To find the physical force, we multiply by the electron charge:

[math]\displaystyle{ { \vec F = q(\vec v \times \vec B) = (-1.6\times10^{-19})\langle 0,\;0,\;4.5\times10^5\rangle = \langle 0,\;0,\;-7.2\times10^{-14}\rangle\ \text{N} }. }[/math]

So the electron feels a force of magnitude [math]\displaystyle{ {7.2\times10^{-14}\ \text{N}} }[/math] in the negative z direction. The negative sign in the charge flipped the direction from +z to -z.


Middling

Question:

A proton moves perpendicular to a uniform magnetic field of magnitude [math]\displaystyle{ {B = 0.20\ \text{T}} }[/math]. The speed of the proton is [math]\displaystyle{ {v = 3.0\times10^6\ \text{m/s}} }[/math].

What is the radius of the circular path that the proton follows?

Use [math]\displaystyle{ {m_p = 1.67\times10^{-27}\ \text{kg}} }[/math] and [math]\displaystyle{ {q_p = 1.6\times10^{-19}\ \text{C}} }[/math].

Solution:

When a charged particle moves in a direction that is perpendicular to a magnetic field, the magnetic force is always perpendicular to its velocity. This constant perpendicular force causes circular motion.

The magnetic force is

[math]\displaystyle{ {|\vec F| = qvB}. }[/math]

The centripetal force required for circular motion is

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

Since these are the same physical force, we set them equal:

[math]\displaystyle{ {qvB = m\frac{v^2}{r}}. }[/math]

Now we solve for the radius [math]\displaystyle{ {r} }[/math]:

[math]\displaystyle{ {r = \frac{mv}{qB}}. }[/math]

We substitute the given values:

[math]\displaystyle{ { r = \frac{(1.67\times10^{-27})(3.0\times10^6)} {(1.6\times10^{-19})(0.20)} }. }[/math]

Carrying out the arithmetic gives

[math]\displaystyle{ {r \approx 0.16\ \text{m}}. }[/math]

So the proton moves in a circle of radius about 0.16 meters.


Difficult

Problem:

A beam of positively charged particles enters a region with both an electric and a magnetic field. The electric field is

[math]\displaystyle{ {\vec E = \langle 0,\;E,\;0\rangle} }[/math]

and the magnetic field is

[math]\displaystyle{ {\vec B = \langle 0,\;0,\;B\rangle}. }[/math]

The particles enter with velocity

[math]\displaystyle{ {\vec v = \langle v,\;0,\;0\rangle} }[/math]

along the +x direction.

1. Determine the speed [math]\displaystyle{ {v} }[/math] for which the particles pass through the region without deflection. 2. Determine the net force on the particles at that speed.

Solution:

Each particle feels two forces.

The electric force is given by

[math]\displaystyle{ {\vec F_E = q\vec E = \langle 0,\;qE,\;0\rangle}. }[/math]

This points in the +y direction.

The magnetic force is

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

We compute the cross product. The velocity is along +x, the magnetic field along +z. The cross product +x with +z points in the negative y direction:

[math]\displaystyle{ { \vec v \times \vec B = \langle v,\;0,\;0\rangle \times \langle 0,\;0,\;B\rangle = \langle 0,\;-vB,\;0\rangle }. }[/math]

Therefore

[math]\displaystyle{ {\vec F_B = \langle 0,\;-qvB,\;0\rangle}. }[/math]

This magnetic force points in the negative y direction.

For the particles to pass through undeflected, the net force in the y direction must be zero:

[math]\displaystyle{ {\vec F_E + \vec F_B = \vec 0}. }[/math]

In component form, this gives

[math]\displaystyle{ {qE - qvB = 0}. }[/math]

We can cancel [math]\displaystyle{ {q} }[/math] (assuming nonzero charge) and solve for [math]\displaystyle{ {v} }[/math]:

[math]\displaystyle{ {v = \frac{E}{B}}. }[/math]

At this speed, the electric and magnetic forces cancel exactly, so the net force is

[math]\displaystyle{ {\vec F_{\text{net}} = \vec 0}. }[/math]

The particles then move in a straight line without deflection.


Magnetic Forces in Wires

Because a current carrying wire is a collection of moving charges, the wire as a whole feels a magnetic force. For a straight segment of wire of length [math]\displaystyle{ {L} }[/math] with current [math]\displaystyle{ {I} }[/math] in a magnetic field of magnitude [math]\displaystyle{ {B} }[/math], the magnitude of the force is

[math]\displaystyle{ {|\vec F_{mag}| = ILB\sin\theta}. }[/math]

Here, [math]\displaystyle{ {\theta} }[/math] is the angle between the direction of the current and the magnetic field. The right hand rule still applies. Point your index finger in the direction of the current, your middle finger in the direction of the magnetic field, and your thumb will show you the direction of the force.

Simple

A wire lies in the xy plane. Conventional current flows to the right. The magnetic field has magnitude [math]\displaystyle{ {B = 0.005\ \text{T}} }[/math] and is directed downward at a [math]\displaystyle{ {45^\circ} }[/math] angle relative to the wire. The current in the wire is [math]\displaystyle{ {I = 0.6\ \text{A}} }[/math]. What is the magnetic force on a 1 meter segment of this wire?

We use

[math]\displaystyle{ {|\vec F_{mag}| = ILB\sin 45^\circ}. }[/math]

We substitute the values:

[math]\displaystyle{ {|\vec F_{mag}| = (0.6)(1.0)(0.005)\sin 45^\circ}. }[/math]

Since [math]\displaystyle{ {\sin 45^\circ \approx \frac{\sqrt{2}}{2} \approx 0.707} }[/math], we get approximately

[math]\displaystyle{ {|\vec F_{mag}| \approx (0.6)(0.005)(0.707) \approx 0.002\ \text{N}}. }[/math]

The detailed geometry of the right hand rule shows that the force points into the page.

Middling

A horizontal metal bar falls downward with constant velocity in a region where the magnetic field points into the page. We know gravity pulls the bar downward. We are told the bar does not speed up or slow down, so the net force on it is zero. What is the magnitude and direction of the current in the bar?

First, the gravitational force on the bar is

[math]\displaystyle{ {|\vec F_{grav}| = mg}. }[/math]

Because the bar is not accelerating, the magnetic force must balance gravity:

[math]\displaystyle{ {|\vec F_{grav}| = |\vec F_{mag}|}. }[/math]

For a straight conductor of length [math]\displaystyle{ {L} }[/math] in a field [math]\displaystyle{ {B} }[/math], the magnetic force is

[math]\displaystyle{ {|\vec F_{mag}| = ILB}. }[/math]

Equating the forces gives

[math]\displaystyle{ {mg = ILB}. }[/math]

We solve for the current:

[math]\displaystyle{ {I = \frac{mg}{LB}}. }[/math]

The direction of the current must be such that the magnetic force acts upward. Gravity is in the negative y direction, so the magnetic force must be in the positive y direction to balance it.

Using the right hand rule, we point the thumb in the positive y direction for the magnetic force. The magnetic field points into the page, which is the negative z direction, so we point the middle finger in the negative z direction. The index finger then points along the negative x direction. This is the direction of conventional current. Therefore, the current flows to the left.

Circular Motion in Earth's Magnetic Field

We now look at situations that connect to the circular motion of charged particles in Earth's magnetic field. In the upper atmosphere, electrons and ions spiral and circle in the geomagnetic field, which contributes to phenomena such as the auroras.

Simple

Question:

An electron in the upper atmosphere moves in a region where Earth's magnetic field can be approximated as uniform with magnitude [math]\displaystyle{ {B = 5.0\times10^{-5}\ \text{T}} }[/math]. The electron's velocity is initially perpendicular to the magnetic field and has magnitude [math]\displaystyle{ {v = 2.0\times10^6\ \text{m/s}} }[/math].

Find the radius of the circular orbit that the electron follows.

Use [math]\displaystyle{ {m_e = 9.11\times10^{-31}\ \text{kg}} }[/math] and [math]\displaystyle{ {q_e = -1.6\times10^{-19}\ \text{C}} }[/math].

Solution:

When the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is

[math]\displaystyle{ {|\vec F| = qvB}. }[/math]

This force acts as the centripetal force for circular motion:

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

Equating them:

[math]\displaystyle{ {qvB = m\frac{v^2}{r}}. }[/math]

We are only interested in magnitudes here, so we use [math]\displaystyle{ {|q_e| = 1.6\times10^{-19}} }[/math] and write

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

Solving for [math]\displaystyle{ {r} }[/math]:

[math]\displaystyle{ {r = \frac{mv}{|q|B}}. }[/math]

Substitute the values:

[math]\displaystyle{ { r = \frac{(9.11\times10^{-31})(2.0\times10^6)} {(1.6\times10^{-19})(5.0\times10^{-5})} }. }[/math]

This expression shows that even with relatively small fields, light particles like electrons can follow quite large circular paths on the scale of meters or more, depending on the numbers.

The sign of the charge affects the direction of the motion, not the radius. The radius depends only on the magnitudes of the quantities.

Middling

Question:

A proton enters Earth's magnetic field with a speed [math]\displaystyle{ {v = 1.0\times10^7\ \text{m/s}} }[/math], perpendicular to a local field of magnitude [math]\displaystyle{ {B = 5.0\times10^{-5}\ \text{T}} }[/math].

1. Find the radius of its circular motion. 2. Find the period, that is, the time it takes to complete one full circle.

Use [math]\displaystyle{ {m_p = 1.67\times10^{-27}\ \text{kg}} }[/math] and [math]\displaystyle{ {q_p = 1.6\times10^{-19}\ \text{C}} }[/math].

Solution:

1. **Radius**

As before, the radius is given by

[math]\displaystyle{ {r = \frac{mv}{qB}}. }[/math]

We plug in magnitudes:

[math]\displaystyle{ { r = \frac{(1.67\times10^{-27})(1.0\times10^7)} {(1.6\times10^{-19})(5.0\times10^{-5})} }. }[/math]

This gives the size of the circular orbit in meters. The specific numerical value can be found with a calculator.

2. **Period**

The period [math]\displaystyle{ {T} }[/math] is the time for one complete revolution. For circular motion:

[math]\displaystyle{ {T = \frac{\text{circumference}}{\text{speed}} = \frac{2\pi r}{v}}. }[/math]

Substituting the expression for [math]\displaystyle{ {r} }[/math]:

[math]\displaystyle{ { T = \frac{2\pi}{v}\left(\frac{mv}{qB}\right) = \frac{2\pi m}{qB} }. }[/math]

Notice that the speed cancels. This is a nice and important result: the period of circular motion in a uniform magnetic field depends only on [math]\displaystyle{ {m} }[/math], [math]\displaystyle{ {q} }[/math], and [math]\displaystyle{ {B} }[/math], not on the speed. So we have

[math]\displaystyle{ {T = \frac{2\pi m_p}{q_p B}}. }[/math]

This can be evaluated numerically by substitution. The key physics concept is that all protons in the same region with the same magnetic field complete circles in the same amount of time, regardless of their speeds, as long as their motion remains perpendicular to the field.


Helical Motion in Magnetic Fields

When the magnetic field is not perpendicular to the velocity, we can decompose the velocity into two components: one parallel to the field and one perpendicular to it. The perpendicular component produces circular motion around the field lines. The parallel component produces motion along the field lines. The combination of these two motions creates a helical path.

Simple

Question:

An electron enters a region of uniform magnetic field [math]\displaystyle{ {\vec B = \langle 0,\;0,\;B\rangle} }[/math]. Its velocity is

[math]\displaystyle{ {\vec v = \langle v_x,\;0,\;v_z\rangle} }[/math],

so the electron is moving in the xz plane. Suppose that [math]\displaystyle{ {v_x} }[/math] is perpendicular to the magnetic field and [math]\displaystyle{ {v_z} }[/math] is parallel to the field.

1. Which component of the velocity causes circular motion, and which component causes motion along the field? 2. Describe the shape of the trajectory.

Assume [math]\displaystyle{ {q_e = -1.6\times10^{-19}\ \text{C}} }[/math] and [math]\displaystyle{ {m_e = 9.11\times10^{-31}\ \text{kg}} }[/math].

Solution:

We split the velocity into a component that is perpendicular to [math]\displaystyle{ {\vec B} }[/math] and a component that is parallel to [math]\displaystyle{ {\vec B} }[/math].

Here, [math]\displaystyle{ {\vec B} }[/math] points along the z axis, so:

- The component [math]\displaystyle{ {v_z} }[/math] is parallel to [math]\displaystyle{ {\vec B} }[/math]. - The component [math]\displaystyle{ {v_x} }[/math] is perpendicular to [math]\displaystyle{ {\vec B} }[/math].

1. The magnetic force is proportional to [math]\displaystyle{ {\vec v \times \vec B} }[/math]. The force depends only on the part of [math]\displaystyle{ {\vec v} }[/math] that is perpendicular to [math]\displaystyle{ {\vec B} }[/math]. The parallel component does not feel any magnetic force.

So:

- [math]\displaystyle{ {v_x} }[/math] causes circular motion in the plane that is perpendicular to the field, which is the x direction around the z axis. - [math]\displaystyle{ {v_z} }[/math] causes motion straight along the field lines in the z direction.

2. The combination of circular motion due to [math]\displaystyle{ {v_x} }[/math] and linear motion due to [math]\displaystyle{ {v_z} }[/math] produces a helix. The electron spirals around the z axis while steadily moving along it. The radius of the spiral is controlled by [math]\displaystyle{ {v_x} }[/math], and the pitch of the helix (how far it moves along z per turn) is controlled by [math]\displaystyle{ {v_z} }[/math].

Middling

Question:

A positively charged particle of mass [math]\displaystyle{ {m} }[/math] and charge [math]\displaystyle{ {q} }[/math] enters a uniform magnetic field [math]\displaystyle{ {\vec B = \langle 0,\;B,\;0\rangle} }[/math] with velocity

[math]\displaystyle{ {\vec v = \langle v_x,\;0,\;v_z\rangle}. }[/math]

This means the velocity has a perpendicular component [math]\displaystyle{ {v_x} }[/math] relative to [math]\displaystyle{ {\vec B} }[/math] and a parallel component [math]\displaystyle{ {v_z} }[/math].

1. Find an expression for the radius of the circular part of the motion in terms of [math]\displaystyle{ {m} }[/math], [math]\displaystyle{ {q} }[/math], [math]\displaystyle{ {B} }[/math], and [math]\displaystyle{ {v_x} }[/math]. 2. Find the pitch of the helix, which is the distance traveled along the direction of the magnetic field in one full circular orbit, in terms of [math]\displaystyle{ {v_z} }[/math] and the period [math]\displaystyle{ {T} }[/math].

Solution:

1. **Radius of the circular motion**

Only the perpendicular component [math]\displaystyle{ {v_x} }[/math] contributes to the circular motion around the field lines. The radius is given by

[math]\displaystyle{ {r = \frac{mv_\perp}{qB}}. }[/math]

Here, [math]\displaystyle{ {v_\perp = v_x} }[/math], since [math]\displaystyle{ {v_x} }[/math] is perpendicular to [math]\displaystyle{ {\vec B} }[/math]. Therefore,

[math]\displaystyle{ {r = \frac{mv_x}{qB}}. }[/math]

This expression gives the radius of the circular projection of the path.

2. **Pitch of the helical path**

The particle also moves along the direction of the magnetic field with speed [math]\displaystyle{ {v_z} }[/math]. The time for one full circle is the period [math]\displaystyle{ {T} }[/math], which is the same as in pure circular motion. The distance traveled along the field in that time is

[math]\displaystyle{ {\text{pitch} = v_z T}. }[/math]

We can express [math]\displaystyle{ {T} }[/math] in terms of known quantities. From circular motion in a magnetic field, the period is

[math]\displaystyle{ {T = \frac{2\pi m}{qB}}. }[/math]

This does not depend on the speed. Therefore, the pitch is

[math]\displaystyle{ {\text{pitch} = v_z \left(\frac{2\pi m}{qB}\right)}. }[/math]

This shows that the radius depends on [math]\displaystyle{ {v_x} }[/math], while the pitch depends on [math]\displaystyle{ {v_z} }[/math]. By changing the ratio of these two components, we can make the helix tighter or more stretched out.


These example categories illustrate how the basic law

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

leads to a wide range of physical behavior: straight line motion if there is no field or no velocity, circular orbits when the motion is perpendicular to the field, helical motion when there is a mixture of parallel and perpendicular components, and forces on entire current carrying wires when many charges move together.

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

James Clerk Maxwell

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.

Additionally, the topic of magnetic force can't be explored without magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn't discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other.

Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only act on moving charges.

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.

In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions.

In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with help from a US Navy navigational satellite.

See also

Further reading

CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider

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

A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation

Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful

An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies

Footage from space of Aurora Borealis

Magnetic force fields generated in copper (with more advanced and complex applications)

References

1. Chabay, R.W; Sherwood, B.A.; Matter and Interactions. 2015. 4. 805-812.

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