Physics

# Magnetic Fields

[latexpage]

## Notations:

• Out of the screen/paper:
• Into the screen/paper:

## Magnetic field created by a current-carrying wire

Consider the following current-carrying wire:

• General equation describing this system:

$\left| \overrightarrow { B } \right|= \frac{\mu I}{2 \pi r}$

•  Variables:
• $\overrightarrow{B} =$ magnetic field created by moving charge
• $\left| \overrightarrow { B } \right| =$ magnitude of the magnetic field at a point in space outside the wire (SI unit: $\mathrm{T = N} / \mathrm{Am}$)
• $\mu =$ permeability the material of the wire is made of (SI unit: $\mathrm{H} / \mathrm{m}$)
• $I =$ electric current (SI unit: $\mathrm{A = C} / \mathrm{s}$)
• $r =$ distance from the current-carrying wire (SI unit: $\mathrm{m}$)

## Magnetic Force: two currents in the same direction

Consider two identical current-carrying wires with charge flowing through them in the same direction:

• Variables
• $R$ = distance between the two wires
• $\overrightarrow{B}_i$ and $\left\| \overrightarrow{L}_i \right\|$ denote the magnetic field and length of wire $i$ (respectively)
• Importantly, for the purposes of this article, the direction $\overrightarrow{L}_i$ is defined as the direction current $I_i$ is traveling
• Note: we treat $I_i$ as a scalar value
• Let $\overrightarrow{F}_{12}$ denote the magnetic force of $I_1$ on $I_2$
• Then, $\overrightarrow{F}_{12} = I_2 \overrightarrow{L}_2 \times \overrightarrow{B}_1$
• Using right-hand rule convention,  this implies that the direction of $\overrightarrow{F}_{12}$ is pointed purely in the horizontal direction towards the right
• Similarly, if we let $\overrightarrow{F}_{21}$ deonote the magnetic force of $I_2$ on $I_1$ then…
• $\overrightarrow{F}_{21} = I_1 \overrightarrow{L}_1 \times \overrightarrow{B}_2$
• Using the right-hand rule convention, this implies that the direction of $\overrightarrow{F}_{21}$ is pointed purely in the horizontal direction towards the left 3
• Summary of this example:
• $\overrightarrow{B}_1$ is exerting a downward force on $I_2$
• $\overrightarrow{B}_2$ is exerting an upward force on $I_1$
• $\overrightarrow{F}_{12}$ and $\overrightarrow{F}_{21}$ directions indicate that both wires will become attracted towards eachother
• This will cause $R$ (i.e., the distance between the wires) to decrease
• As $R$ decreases , the wires wil accelerate towards eachother

## Induced current in a wire

Consider the following magnetic field $\overrightarrow{B}$ (purple) popping out of the page and a wire laid over the field so that it is overlapping $\overrightarrow{B}$ for a distance $L$

• Let $Q$ (orange) be a charge positioned at the lower end of the wire (SI units: Coulombs)

Nothing happens when $Q$ is stationary

• This is because the force $\overrightarrow{F}$ due to the magnetic field is equal to the cross product of $Q$ times the velocity $v$ of the charge and $\overrightarrow{B}$:

$\overrightarrow{F} = Q \overrightarrow{v} \times \overrightarrow{B}$

• Recall: the magnitude of the cross product can be written as…

$\left| \overrightarrow{F} \right| = Q \left| \overrightarrow{v} \right| \cdot \overrightarrow{B} \sin {\theta}$

In our diagram, the wire and the magnetic field are perpendicular to eachother

• Recall: $\sin {90^\circ} = 1$
• Thus, for this instance,

$\left| \overrightarrow{F} \right| = Q \left| \overrightarrow{v} \right| \cdot \overrightarrow{B}$

• For our stationary charge, $\left| \overrightarrow{v} \right| = 0$
• Thus, $\left| \overrightarrow{F} \right| = 0$

The work done by a magnetic field per charge in a coductive wire can be described by the following equation:

$\frac{W}{Q} = L(\overrightarrow{v} \times \overrightarrow{B})$

• Importantly, the output for this equation has units in Joules/Coulomb
• Recall: $1 \mathrm{Volt} = 1 J / C$
• **Thus, the work done by the magnetic field on a charge causes the charge to move AS IF there is a potential differe4nce of $1 \mathrm{V}$ over the length of the wire that overlaps with $\overrightarrow{B}$
• In this context, we aren’t saying the charge is moving due to a difference in potential energy (like it does in a simple DC circuit)
• Even though differences in potential energy give rise to voltage, in teh context of our example, it is the magnetic field giving rise to the voltage
• Specifically, when a voltage is induced from a magnetic field, it is said that the magnetic field is exerting an electromotive force or emf
• emf has units in J/C
• For circuit analysis, this has the same effect as a potential (or voltage) difference
• We can define the emf mathematically as $\mathrm{emf} = L \overrightarrow{v} \times \overrightarrow{B}$
• Recall: we can pull constants out of a cross product to rewrite this equation as $\mathrm{emf} = L(\overrightarrow{v} \times \overrightarrow{B})$

## Sources:

This site uses Akismet to reduce spam. Learn how your comment data is processed.