Magnetic fields

Bring in compass, circuit to demo field of a current.

  • Forces between currents
  • Compasses show the local direction of magnetic field lines
  • Cyclotron motion in a uniform magnetic field
  • The Lorentz force - due to electric and magnetic field
  • Current
  • Charge conservation

Playing around with currents

When currents flow in parallel wires, forces arise which are not electric. (That is, a test charge placed near the wire feels no force).

[video of forces between wires]

If two wires carry

  • parallel currents $\Rightarrow$ the wires are attracted to each other,
  • ...opposite currents $\Rightarrow$ the wires are repulsed by each other.

Playing around with compasses

  • An electric dipole in a uniform electric field experiences a torque.
  • The equilibrium position of an electric dipole is parallel to the electric field lines.

Compasses are small magnetic dipoles that, if allowed to pivot about their center, will line up parallel to magnetic field lines-- field lines for the magnetic field $\myv B$. Earth itself is a gigantic magnetic dipole (we're not completely sure why) with its axis aligned almost with Earth's axis of rotation. The component of Earth's field parallel to the surface points approximately North.

A compass placed close to a current-carrying wire points perpendicular to the wire.

Implying that there is a magnetic field perpendicular to the wire.

By symmetry, you'd expect that field to curl around the wire.

There's a right-hand rule to remember the direction of the field in relation to the current.

The roundabout way of explaining calculating this effect is to say:

Current $\Rightarrow$ Magnetic field $\Rightarrow$ Force on other moving charges

(OK, it's not *that* roundabout. This is very similar to what we did in electrostatics:

Charge $\Rightarrow$ Electric field $\Rightarrow$ Force on another charge ).

Experiments on the last link in the magnetic chain, "magnetic field $\Rightarrow$ force" can be summarized with this expression for the magnetic force on a moving charge: $$\myv F_\text{mag} = Q\,\myv v \times \myv B$$

Like other cross products, use a right-hand-rule to calculate the direction of the force:

  1. Fingers of the right hand straight out in the direction of $\myv v$,
  2. Curl those fingers towards $\myv B$ (make sure it's an angle < $180^o$), and
  3. Your thumb is pointing in the direction of the force.

Cyclotron motion

Particle with charge $Q$, mass $m$, $\myv v_0=v_0\uv y$.
Field everywhere the same: $\myv B=-B_0\uv z \perp \myv v_0$.

The particle will feel a magnetic force at right angles to its velocity, but won't be slowed down.

$\Rightarrow$ These are just the conditions for circular motion in the $xy$-plane, where the inward force $F_\text{mag}$ is just equal to $ma_\text{centripetal}$: $$|Q\myv v \times \myv B| = QvB = m \frac{v^2}{R}.$$

This corresponds to an angular speed (in radians/second) of $$\omega_c = v/R = QB/m.$$ This is called the cyclotron frequency


What if the particle started out with an arbitrary velocity $\myv v$ that has some $z$-component to it? The force on the the particle would be: $$\myv F_{\text{mag }} =Q\myv v \times \myv B = -QB \myv v \times \uv{z}.$$

  • $\Rightarrow \myv F_\text{mag} \perp \uv z$
  • $\Rightarrow (\myv F_\text{mag})_z=0$
  • There is no acceleration of the particle in the $z$ direction.

With no change of the particle's momentum in the $\uv{z}$ direction, the particle will tend to "corkscrew" around the direction of the magnetic field.

The 'solar wind' consists of ions & electrons emitted from the Sun. As these high-energy ions reach Earth and interact with Earth's magnetic field, if they spiral around magnetic field lines, where would they tend to "land" on Earth's surface?

You can see the answer because when these ions hit the atmosphere, they knock into (mostly) nitrogen and oxygen, exciting them. Then, as these molecules return to their ground state they give off light.

Mass spectrometer

If several particles have the same velocity, the radius of their cyclotron motion is proportional to their mass. The mass spectrometer, makes use of this to measure things like isotope ratios Carbon-13 / Carbon-12 ratios for dating. Or Oxygen-16 / Oxygen-18 ratios for measuring past temperatures.

Lorentz force

Putting the magnetic force together with the electric force, we get the Lorentz force law: $$\myv F = Q(\myv E + \myv v \times \myv B)$$

Work and magnetic forces

Work: $$dW= \myv F \cdot d \myv l$ where the displacement is $$d\myv l=\myv v\,dt$$ is the displacement.

Since the magnetic force is alway perpendicular to the displacement of the moving charge, magnetic forces do no work.

What about those junkyard magnets--don't the metal objects rise through the air when you turn on the magnet? Surely magnetic forces are doing work there??

Read and summarize (on Moodle) example 5.3 to see...


  • Current is charge per unit time passing a point in space.
  • A flow of positive charges to the right of 1 Coulomb / sec is the same current as 1 Coulomb / sec of negative charge flowing to the left.
  • 1 Coulomb / 1 sec $\equiv$ 1 Ampere = 1 A.

The current in a wire can be cartoonified as a constant line charge $\lambda$ (units: charge / length) moving with a velocity $v$. This combination $$I = \lambda v,$$ has the right dimensions for current....

But current also has a (vector) direction, so we should write: $$\myv I = \lambda \myv v.$$

The force on little chunk of charge, $dq$ that is moving in a magnetic field: $$d\myv F_m = dq\,\myv v\times\myv B.$$ Integrating to get the total force on a wire: $$\myv F_m = \int (\myv v \times \myv B) dq = \int (\myv v \times \myv B) \lambda d l = \int (\myv I \times \myv B) dl$$

For every segment of a wire, there is just as much current flow in as current flow out. Then $I$ is constant along the path of the wire. With $d\myv l$ pointing in the direction of the current... $$\myv F_m = I \int (d \myv l \times \myv B).$$

Electrostatics included: point charge $q$, line charge $\lambda$, surface charge $\sigma$, and volume charge densities $\rho$. If each of these "kinds" of charge is mobile, and moving with a velocity $\myv v$, the corresponding quantities are...

Current-related quantity $$\myv F_m$$
line current $\myv I = \lambda \myv v$ $$\int (\myv v \times \myv B) \lambda d l = \int (\myv I \times \myv B) dl$$
Surface current density $\myv K = \sigma \myv v$: Current per unit width perp to flow. $$\int (\myv v \times \myv B) \sigma da=\int (\myv K \times \myv B) da$$
Volume current density $\myv J = \rho \myv v$: Current per unit area perpendicular to the flow. $$\int(\myv v \times \myv B) \rho d \tau = \int (\myv J \times \myv B) d \tau$$

So, the current $dI$ flowing in the square-cross-section "extruded playdoh thing" that's shown here which has a cross-sectional area $da_\perp$, in a region where the current density is $\myv J$, is $$dI = da_{\perp} J.$$

Problem 5.5.b

A current $I$ flows down a wire of radius $a$. If it's distributed in such a way that the volume current density is inversely proportional to the distance from the axis, what is $J$?

This is two conditions on $J$:

The first is that the functional form for $J$ is $J(s)=k/s$ for $0<s<a$. (And $J=0$ for $a<s$).

The second is that the total current, integrated over a cross-section through the wire is (using cylindrical coordinates): $$I = \int dI = \int J da_\perp = \int_0^{2\pi} d \phi \int_0^a s\,ds\ k/s=2 \pi k \int_0^a ds = 2\pi k a.$$

Solving this for the constant $k$, we get: $$J(s) = \frac{I}{2\pi a} \frac{1}{s}.$$

Local charge conservation

According to the problem we just did, the current crossing some surface ${\cal S}$ is $$I = \int_{\cal S} J da_\perp = \int_{\cal S} \myv J \cdot d \myv a,$$

If we integrate over a closed surface, we can use the divergence theorem to write the integral as a volume integral over the enclosed volume. $$\oint_{\cal S} \myv J \cdot d \myv a = \int_{\cal V} (\myv \grad \cdot \myv J) d \tau.$$

This integral must be the total charge leaving $\cal V$ in a unit of time. Since charge is conserved, that's just $-\frac{d}{dt}Q_\text{enc}$ which we can write as: $$\int_{\cal V} (\myv \grad \cdot \myv J) d \tau = -\frac{d}{dt}\int_{\cal V} \rho d \tau = - \int_{\cal V} \frac{\del \rho}{\del t} d \tau.$$

But the volume $\cal V$ is completely arbitrary, so it must be the case that at every point in space: $$\myv \grad \cdot \myv J = -\frac{\del \rho}{\del t}.$$ This is called the continuity equation.

Image credits

MIT Tech-TV, Thomas Alerstam, Dystopos Nebraska Becky