Magnetostatics and Magnetic Fields

Magnetostatics

Magnetostatics deals with magnetic fields that are static in nature. A key difference between electric and magnetic fields is:

• Electric charges can be isolated (exist as monopoles).

• Magnetic fields always have dipoles; magnetic monopoles do not exist. $<br>abla <br>eq 0$ (magnetic field divergence not equal to zero) implies magnetic monopoles don't exist, where E = Electric Field. $<br>eq$ doesn't exist.

Definition of a Magnetic Field

The magnetic field at a point P is defined by considering a charge $q$ at point P moving with velocity $v$.

1. The magnetic force $F_B$ at P on the charged particle is perpendicular to $v$ at P.

2. By varying the direction of $v$, we find a particular direction (and its opposite) where $F_B = 0$. The direction of the magnetic field $B$ is defined along this line.

3. Varying $v$ such that it is perpendicular to $B$ maximizes the force, $F<em>{B,max}$. The magnitude of $B$ is defined as: $B = {F</em>{B,max} \over qv}$

The magnetic force is defined by the vector cross product:

$F_B = qv \times B$

When the sign of the charge is switched, the direction of the magnetic force reverses.

Units of Magnetic Field

The SI unit for magnetic field is the tesla (T):

$1 \text{ tesla} = 1 \text{ T} = 1 \frac{\text{Newton}}{\text{Coulomb} \cdot \text{meter/second}} = 1 \frac{\text{N}}{\text{C} \cdot \text{m/s}} = 1 \frac{\text{N}}{\text{A} \cdot \text{m}}$

Another unit is the gauss (G):

$1 \text{ T} = 10^4 \text{ G}$

Properties of Magnetic Force

• $F_B$ is always perpendicular to $v$ and $B$.

• $F_B$ cannot change the particle's speed (and thus kinetic energy).

• $F_B$ cannot do work on the particle.

• $F_B$ can alter the direction of $v$.

Example: Magnetic Force on a Proton

Given:

• Velocity of proton, $v = 6 \times 10^7 \text{ m/s}$

• Magnetic field, $B = 0.5 \text{ T}$

• Angle between $v$ and $B$ is $90^\circ$

Magnitude of the force:

$F_B = qvB \sin{\theta} = (1.6 \times 10^{-19} \text{ C}) (6 \times 10^7 \text{ m/s}) (0.5 \text{ T}) (1) = 4.8 \times 10^{-12} \text{ N}$

Magnetic Force on a Current-Carrying Wire

A current-carrying wire experiences a magnetic force when placed in a magnetic field because the current consists of moving charged particles.

• A wire suspended between magnetic poles deflects left or right depending on the current direction.

To calculate the force, consider a segment of wire of length $l$ with conduction electrons drifting at velocity $\vec{v_d}$.

$\vec{v_d} = \frac{\vec{l}}{t}$

where $\vec{l}$ is the length vector directed along the current.

The magnetic force on the wire segment is:

$F<em>B = q(\vec{v</em>d} \times B) = I t (\frac{\vec{l}}{t} \times B) = I (\vec{l} \times B)$

If $\vec{l}$ and $\vec{B}$ are at 90 degrees:

$F_B = IlB$

Arbitrary Wire Shape

For a wire of arbitrary shape, sum the forces on small segments $d\vec{l}$.

$dF_B = I (d\vec{l} \times B)$

The total magnetic force is:

$F<em>B = I \int</em>a^b (d\vec{l} \times B)$

where a and b are the endpoints of the wire.

Curved Wire in Uniform Magnetic Field

$F_B = I \int (d\vec{l} \times B) = I (\int d\vec{l}) \times B = I \vec{L} \times B$

$\vec{L}$ is the length vector from point a to b.

Production of Magnetic Fields

Moving charges (currents) are the source of magnetic fields. The magnetic field at a point P due to a current in a wire can be calculated by summing the contributions from small wire segments $d\vec{l}$.

• $d\vec{l}$ has magnitude equal to the segment length and points in the current direction.

• The infinitesimal current source is $Id\vec{l}$.

Biot-Savart Law

The Biot-Savart law gives the magnetic field contribution $d\vec{B}$ from a current source $Id\vec{l}$:

$d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times \hat{r}}{r^2}$

where:

• $\mu<em>0$ is the permeability of free space: $\mu</em>0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$

• $r$ is the distance from the current source to the field point P.

• $\hat{r}$is the unit vector from the source to the field point.

The total magnetic field is:

$B = \int d\vec{B} = \int \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times \hat{r}}{r^2}$

In terms of vectors from source to field point:

$\hat{r} = \frac{\vec{r} - \vec{r'}}{|\vec{r} - \vec{r'}|}$

$B = \int d\vec{B} = \int \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^3}$

Similar to Coulomb's Law:

$E(r) = \frac{1}{4 \pi \epsilon_0} \frac{dq}{|r-r'|^2} \frac{r-r'}{|r-r'|}$

Magnetic Field Due to a Finite Straight Wire

1. Consider a source element $d\vec{l'} = dx' \hat{x}$ carrying current $I$ in the x direction.

2. The location of the source element is $\vec{r'} = x' \hat{x}$.

3. The field point is located at $(x, y) = (0, z)$, so the field vector is $\vec{r} = z \hat{z}$.

4. The separation vector is $\vec{r} - \vec{r'} = z \hat{z} - x' \hat{x}$.

5. Magnitude: $|\vec{r} - \vec{r'}| = \sqrt{z^2 + x'^2}$

6. Unit vector: $\hat{r} = \frac{\vec{r} - \vec{r'}}{|\vec{r} - \vec{r'}|} = \frac{z \hat{z} - x' \hat{x}}{\sqrt{z^2 + x'^2}}$

7. The cross product is:
   $d\vec{l'} \times \hat{r} = (dx' \hat{x}) \times (z \hat{z} - x' \hat{x}) = z dx' (\hat{x} \times \hat{z}) - x' dx' (\hat{x} \times \hat{x})$

8. Since $\hat{x} \times \hat{z} = -\hat{y}$ and $\hat{x} \times \hat{x} = 0$:
   $d\vec{l'} \times \hat{r} = -z dx' \hat{y}$

9. The magnetic field contribution is:
   $d\vec{B} = \frac{\mu<em>0 I}{4\pi} \frac{d\vec{l'} \times \hat{r}}{|\vec{r} - \vec{r'}|^2} = \frac{\mu</em>0 I}{4\pi} \frac{-z dx' \hat{y}}{(z^2 + x'^2)^{3/2}}$

Direction is out of page.

To obtain the total contribution from a line of length 2L, integrate from -L to L:

$B = \int d\vec{B} = -\hat{y} \frac{\mu<em>0 I z}{4\pi} \int</em>{-L}^{L} \frac{dx'}{(z^2 + x'^2)^{3/2}}$

$B = \frac{\mu_0 I L}{4 \pi z \sqrt{L^2 + z^2}}$

For an infinitely long wire (L >> z):

$B = \frac{\mu_0 I}{2 \pi z}$

In this limit, the system possesses cylindrical symmetry, and the magnetic field lines are circular.

Ampere's Law

Electric currents generate magnetic fields whose field lines form closed loops.

• Compass needles near a current-carrying wire deflect.

• In the absence of current ($I = 0$), all compass needles point in the same direction.

• When $I \neq 0$, needles deflect along the tangential direction of the circular path.

Divide a circular path of radius $r$ into small length vectors $d\vec{l}$.

$\oint \vec{B} \cdot d\vec{l} = B \oint dl = B (2 \pi r) = \mu_0 I$

This result is obtained by choosing a closed path (Amperian loop) that follows a magnetic field line.

Ampere's law in magnetism is analogous to Gauss's law in electrostatics. To apply them, the system must possess certain symmetry. Ampere's law can be readily applied to an infinite wire due to cylindrical symmetry. For a finite wire, the Biot-Savart law must be used.

Applications of Ampere's Law

Magnetic Field of a Straight Wire

What is the field outside a long straight wire of radius R carrying a current I of uniform current density?

• For r > R, the Amperian loop encircles the entire current, i.e., $I_{enc} = I$.

Applying Ampere's Law:

$\oint \vec{B} \cdot d\vec{l} = B(r) (2 \pi r) = \mu_0 I$

$B = \frac{\mu_0 I}{2 \pi r}$

• For r < R, the current encircled is proportional to the area enclosed:

$I_{enc} = I \frac{\pi r^2}{\pi R^2}$

Applying Ampere's Law:

$\oint \vec{B} \cdot d\vec{l} = B(2 \pi r) = \mu_0 I \frac{r^2}{R^2}$

$B = \frac{\mu_0 I r}{2 \pi R^2}$

The magnetic field is zero at the center of the wire increases linearly with $r$ until $r = R$. Outside the wire, the field falls off as $1/r$.

Magnetic Field of a Solenoid

A solenoid is a long coil of wire tightly wound in a helical form.

• For an ideal solenoid (infinitely long, tightly packed turns), the magnetic field inside is uniform and parallel to the axis, vanishing outside.

To compute the magnetic field $B$, consider a rectangular Amperian path of length $\ell$ traversing in a counterclockwise manner.

$\oint \vec{B} \cdot d\vec{l} = \int<em>1 \vec{B} \cdot d\vec{l} + \int</em>2 \vec{B} \cdot d\vec{l} + \int<em>3 \vec{B} \cdot d\vec{l} + \int</em>4 \vec{B} \cdot d\vec{l}$

The contributions along sides 2 and 4 are zero because $\vec{B}$ is perpendicular to $d\vec{l}$. $B = 0$ along side 1 because the magnetic field is zero outside.

The current enclosed is $I_{enc} = NI$, where $N$ is the number of enclosed turns. Applying Ampere's law:

$\oint \vec{B} \cdot d\vec{l} = B \ell = \mu_0 NI$

$B = \frac{\mu<em>0 NI}{\ell} = \mu</em>0 n I$

where $n = N/\ell$ represents the number of turns per unit length.