Magnetic Force

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 charged particle. It has charge q, velocity [math]\displaystyle{ \vec{v} }[/math], and it is in the presence of a magnetic field [math]\displaystyle{ \vec{B} }[/math]. The force that this particle experiences is given by:

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

For a particle at rest ([math]\displaystyle{ \vec{v}=\vec{0} }[/math]), the magnetic force is:

(2) [math]\displaystyle{ \vec{F}=\vec{0} }[/math]

Force is measured in newtons (N), magnetic field in tesla (T), charge in coulombs (C), and velocity in meters per second (m/s).

Equation (1) involves the vector cross product. Therefore, the magnetic force on a moving charged particle is always perpendicular to the plane spanned by [math]\displaystyle{ \vec{v} }[/math] and [math]\displaystyle{ \vec{B} }[/math]. The magnitude of the force is:

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

Here, [math]\displaystyle{ \theta }[/math] is the angle between the velocity and magnetic field vectors. - If [math]\displaystyle{ \theta = 90^\circ }[/math], then [math]\displaystyle{ \sin\theta = 1 }[/math], and the force magnitude is maximized. - If [math]\displaystyle{ \theta = 0^\circ }[/math] or [math]\displaystyle{ 180^\circ }[/math], then the vectors are parallel or anti-parallel, [math]\displaystyle{ \sin\theta = 0 }[/math], and the magnetic force is zero.

Because the magnetic force is always perpendicular to the instantaneous velocity, it does no work on a charged particle; it only changes the direction of motion.

Magnetic Force on a Current-Carrying Wire

Consider a small segment of charge [math]\displaystyle{ \Delta q }[/math] moving with drift velocity [math]\displaystyle{ \vec{v} }[/math]. From (1), the force on this segment is:

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

If the segment contains n charged particles in a wire of cross-sectional area A and length [math]\displaystyle{ \Delta L }[/math], then:

(5) [math]\displaystyle{ \Delta\vec{F} = nA\,\Delta L\,(\vec{v}\times\vec{B}) }[/math]

Using the definition of current [math]\displaystyle{ I = nqAv }[/math], the force on the wire segment becomes:

(6) [math]\displaystyle{ \Delta\vec{F} = I\,(\vec{\Delta L}\times\vec{B}) }[/math]

For a wire of arbitrary shape, the total force is obtained by integration:

(7) [math]\displaystyle{ \vec{F} = \int I\,d\vec{L}\times\vec{B} }[/math]

This relation applies to any current distribution in a magnetic field.

Magnetic Field of a Moving Charge (Biot–Savart Law)

A moving charge produces a magnetic field according to:

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

Here, [math]\displaystyle{ \hat{r} }[/math] is the unit vector from the charge to the observation point, and [math]\displaystyle{ r }[/math] is the distance to that point. A charge does not exert a magnetic force on itself, but different moving charges interact through their magnetic fields.

Circular and Helical Motion of Charged Particles

If a charged particle enters a region where [math]\displaystyle{ \vec{v}\perp\vec{B} }[/math], then the magnetic force acts as a centripetal force. The radius of the resulting circular motion is:

(9) [math]\displaystyle{ r = \frac{mv}{|q|B} }[/math]

The angular frequency (cyclotron frequency) is:

(10) [math]\displaystyle{ \omega = \frac{|q|B}{m} }[/math]

If the magnetic field is strong enough (e.g., Earth's magnetic field acting on charged solar particles), the particle may remain trapped in circular orbits.

If the velocity is not perpendicular to the magnetic field, decompose:

(11) [math]\displaystyle{ \vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp} }[/math]

- [math]\displaystyle{ \vec{v}_{\perp} }[/math] causes circular motion. - [math]\displaystyle{ \vec{v}_{\parallel} }[/math] produces uniform motion along the field lines.

The resulting motion is helical:

(12) [math]\displaystyle{ \text{pitch} = v_{\parallel} T = v_{\parallel}\left(\frac{2\pi}{\omega}\right) }[/math]

This helical trajectory of charged particles spiraling along Earth's magnetic field lines is a key component in the formation of the auroras (Northern and Southern Lights).

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.

In addition to examining how individual charged particles behave in a magnetic field, it is equally important to understand how an entire current-carrying wire experiences magnetic forces. A current is simply a collection of moving charges, and each charge experiences a magnetic force. When these individual forces are combined, the total magnetic force acting on a segment of wire inside a magnetic field is given by:

[math]\displaystyle{ \vec F = I\, \vec L \times \vec B }[/math]

Here:

• [math]\displaystyle{ I }[/math] is the conventional current through the wire
• [math]\displaystyle{ \vec{L} }[/math] is a vector whose direction represents the direction of current and whose magnitude is the length of the wire segment inside the magnetic field
• [math]\displaystyle{ \vec{B} }[/math] is the magnetic field vector

Because the force is described by a cross product, the resulting magnetic force is always perpendicular to both the direction of the current and the direction of the magnetic field. This means that changing either the orientation of the wire or the direction of the magnetic field will change the direction — and sometimes even the sign — of the resulting magnetic force.

The following GlowScript model demonstrates exactly how the direction of the magnetic field and the orientation of the wire determine the force on the wire. Sliders allow you to independently change the x, y, and z components of the magnetic field vector, as well as rotate the wire in the plane. As these quantities change, the magnetic force vector updates instantly to reflect the geometry of the cross product in real time.

This simulation provides an intuitive way to understand:

• how the magnitude of the force depends on the angle between [math]\displaystyle{ \vec{L} }[/math] and [math]\displaystyle{ \vec{B} }[/math]
• how the direction of the force rotates smoothly as either vector is rotated
• why the right-hand rule naturally follows from the properties of the cross product

You can interact with this model here:

[Magnetic Force on a Current-Carrying Wire ]

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.