Section 2-5 - Magnetism: Effects & Causes

The Magnetic Field

We can make artificial lodestones from iron (and other materials), which we call magnets.  Investigating the interactions between magnets, we find that:
Like poles repel,
Unlike poles attract.

In order for a compass to work, what must be at (or near) the earth's Geographic North Pole?

This is very similar to the rules that govern the interactions of electric charges, so perhaps some analogies can be drawn.  For example, can we model the interactions of two poles with a magnetic field, somewhat similar to the electric field?  Since there are no magnetic monopoles to use as magnetic test charges, we will instead use magnetic dipoles as our test objects.  We saw in a previous section that electric dipoles will tend to align with the electric field.  Let's place some small test compasses in the vicinity of a larger magnet:

It seems that we could indeed think of the magnetic dipoles as aligning with a magnetic field, which resembles fairly closely the electric field for two equal but opposite charges:

So let's define the magnetic induction field (symbol B, unit the Tesla; see Note, below) and arbitrarily choose its direction as follows: consider the test dipoles to be little arrows with the point at the north end that point in the direction of the magnetic field:

We see that the field runs from the north pole to the south pole, except inside the magnet, where it runs from south pole to north pole.  This is somewhat different from the situation with charges, in which the electric field ran from positive charges to negative charges.  Here, we see that the magnetic field lines instead form closed loops.

Gauss's Law for Magnetism

Interactions between Currents and Magnetic Fields

Now consider the effects of a magnetic field on individual charged particles.  A cathode ray tube (CRT) such as those in televisions shoot a stream of electrons toward a screen, which is covered with a fluorescent material which glows as the electrons strike.  We orient the magnetic field in different ways in the vicinity of the electrom beam, and find a similar deflection.  Suppose that the beam is coming out toward you and that the magnetic field (from a horseshoe magnet) is pointing toward your left.  Based on the previous discussion, which way would you expect the force on the electrons to be directed, as evidenced by the their deflection?

Let's guess a relationship for the force:
We expect that the force is proportional to the charge on the particles: F ~ q. This could be verified by using alpha particles instead of electrons or protons.
We could vary the speed of the particles, ands we would find that F ~ v.
We can add magnets to vary B, and we find that F ~ B.
Also, the angular dependence F~ sinq_v,B, so that
F_M = qvB sinq_v,B  (RHR), or F = qv x B

Note: Some define the magnetic field line as that path along which a charged particle can move and experience no magnetic force.

Is this consistent with our previous result?  In a naive way, we can write that, for a wire,
F_M = ILBsinq_I,B (RHR) = [q/t]LB sinq_I,B (RHR) = q[L/t]B sinq_I,B (RHR) = qvB sinq_I,B (RHR) = qvB sinq_v,B (RHR) or F = qv x B.
where we understand that the direction of the product of q and v is in the direction of the current (that is, if q is positive, v and I and parallel, but if q is negative, then qv and I are parallel).

Let's consider a special case of a charged particle (mass  m and charge +q) entering a region of uniform magnetic field (B) so that initially, v and B are perpendicular.

There will be a magnetic force acting on the particle, F_M = qvB (remember that we stipulated that q_v,B is 90^o).  This force will be direcred as shown, and will deflect the path of the particle from its original direction to the one shown.  Once it has arrived at its 'new position' with its new velocity, there will still be a magnetic force, but that force will be directed at a slightly different angle, as shown.  This causes another deflection and the particle moves to the third position shown, et c.
What can we say about the speed of the particle?

Let's look at this same situation in a slightly different way to obtain an interesting result:
F_M = F_C
qvB = mrw²
qrwB = mrw²
qB = mw
w = qB/m.
That is, the angular velocity (or, more interestingly and equivalently, the frequency) of a charged particle in a magnetic field is independent of the energy (speed) of the particle!

This result is the foundation for several interesting devices, notably the mass spectrometer.  Suppose that we wish to identify a material from its molecular mass.  We place the material in an oven which vaporizes it and spews it out at high speed.  Unfortunately, there is a distribution of speeds (the Maxwell distribution). After passing through various collimators to produce a well defined beam of particles, they molecules are directed through an electron stripper, which removes one (or more) electrons.  The now charged particles travel through a velocity selector, which is a region of crossed electric and magnetic fields.

There is an electric force qE on the particle toward the bottom of the page and a magnetic force qvB directed toward the top of the page (check it!).  The electric force is the same regardless of the speed of the particle, but the magnetic force is proportional to the speed of the particles.  For fast particles, F_M is greater than F_E, and the particles will be deflected upward.  For slow particles, F_E is greater than F_M, and the particles are deflected downward.  However, for particles with just the right speed, the forces cancel and the particles travel in a straight line.  This will occur when
F_E = F_M
qE = qvB  (assume q is not zero)
E = vB
v = E/B.
Now, the particles enter the main spectrometer, a region of uniform B-field, and their paths are bent into circular arcs as shown.  Since all particles now have the same speed, and (we hope) the same charge, and they are experiencing the same B, we find that the radius of each orbit is proportional to the mass of the particle:
r = [v/qB] m.
By running a detector along one edge of the chamber and recording the number of particles arriving as a function of distance d (=2r), a spectrum of the masses of the particles can be taken and the materials identified.
m = [qB_SPECTB_SELECT/2E_SELECT]d.
Often there are complications: for example, some particles will be doubly or triply charged, giving rise to extraneous lines at m/2 and m/3, and the molecules are often broken into fragments which also register as peaks in the spectrum.

This process was initially used to separate uranium early in the wartime effort to construct the atomic bomb.  There are several isotopes of uranium, the most common being U-235 and U-238.  U-235 is useful for bombs, while U-238 is useful in power reactors, but will squelch the necessary fast chain reaction in a bomb.  They can not be separated chemically, since they have the same electronic structure.  A method was developed to use the mass spectrometer to separate the isotopes, but the production rate was eventually deemed to be too slow.

Consider these questions:
What shape path will a charged particle take when injected into a uniform electric field with its initial velocity not perpendicular to the field?

Biot-Savart Law

Example:
Find the B-field (magnitude and direction) at the centre of a circular loop of radius R carrying current I as shown.

Consider this current arrangement:

The straight portions of the wire extend out to infinity and the curved jag is a semi-circle of radius R.  Find the magnetic field at P.

Ampère's Law

Suppose that we have a current which is producing a magnetic field B.  Let's draw an imaginary closed loop in space in the vicinity of the current.  Break the loop up into many very short lengths, dl.  As we follow the path along the loop, we look at the component of the magnetic field which is parallel (or anti-parallel) to the loop and multiply it by the length of the little bit of the loop:
B_|| dl.
This is analogous to choosing an imaginary surface and looking at the perpendicular component of the electric field in Gauss's law:
E_perp dA
Now, let's add up all these terms all the way around the loop; if the B-field component is in the same direction in which we are traversing the loop, we'll call the term positive and if the component is opposite to the direction we are going, we'll call it negative.  This is again analogous to calling the flux positive if the E-field points from the inside of the gaussian surface to the outside, and calling it negative if the E-field points from the outside in.  When the sum is complete, the result will be proportional to the net current passing through the loop:
B_|| dl = m_o I_enclosed 
or more explicitly mathematically:
B ^.dl = m_o I_enclosed 

Just as for Gauss's Law, this is always true, but it may be useful for finding B only in certain symmetric situations.  In Section 2-1, we tried to find surfaces for which either the E-field was along the surface (so that E_perp = 0) or if not so, then exactly perpendicular and constant in magnitude.  Similar hopes and dreams run wild here: we'd like to choose loop paths such that either B is perpendicular to the path (and so there is no contribution to the sum) or if not, then preferably along the path and constant in magnitude.

Consider the example of an infinite, straight, current-carrying wire.  First, we need to at least consider the Biot-Savart Law (and the RHR) to get an idea of what the B-field looks like:

It seems clear that the B-field will form circular loops around the wire, as shown.  Furthermore, symmetry arguments can be made that the magnitude of the field at points at any given distance r from the wire will be the same.  It seems that a good choice for our imaginary loop might be a circle of radius r which lies along one of the field loops.  In that case, we write that
B_|| dl = m_o I_enclosed 
Since B is parallel to the loop at all points, we can replace B_|| with plain ole B:
B dl = m_o I_enclosed 
Also, since we argued that the magnitude of B is the same everywhere on the loop, we can pull it out of the integral:

Bdl = m_o I_enclosed 
The integral has been reduced to the length of the loop, i.e., the circumference of the circle:
B 2pr = m_o I_enclosed.
B = [m_o I_enclosed]/2pr,
and in this case, I_enclosed is just the given current I, so
B = m_o I/2pr.
It's acceptable to have r in the answer since r corresponds to the real distance from the wire at which we want to find the field.

Consider the infinitely long solenoid, which is a wire coiled up along a cylindrical form, somewhat like a 'slinky.'  Let's draw this schematically by slicing the solenoid in half along its length:

The lower vectors represent the current coming under the bottom of the solenoid, while the upper vectors represent the current rolling back over the top of the solenoid.  We pick a point P inside the solenoid on its central axis at which to find the field.  Consider one little bit of the current, here labeled '1.'  The B-field from that bit of current is in the direction shown and is labeled B₁.  Likewise for a bit of current exactly opposite it and the same distance from point P, labeled '2.'  We see that the net field is directed to the left, since the vertical components will cancel.  This will be true for all such pairs we pick, so the net field is to the left.  Also, we can say that the magnitude of B will be the same all along that central axis, since the solenoid is infinite in length and we can shift it to the right or left any amount and the system should appear to be the same.  This is true if the points we look at are on the central axis of the solenoid; what if we look at a point off axis?  We shall contend that the B-field is to the left everywhere inside the solenoid, based again on symmetry arguments.
What about the field outside of the solenoid?  We contend that the B-field there is to the right and very weak, weak enough to neglect.  Remember that B-field lines from closed loops, so the lines running inside the solenoid towards the left end have to loop back on the outside to the right end.  Since the solenoid is infinitely long, we might expect that the lines are very spread out from each other on the 'return trip,' and since the strength of the field is related to how closely the lines are spaced, we can say that the field is very weak.
Let's choose a rectangular path of length L along which to calculate our sum:

Label the segments of the path as shown.  The sum can be written as:
B dl = m_o I_enclosed  =  ₍₁₎ B_|| dl +₍₂₎ B_|| dl + ₍₃₎ B_|| dl + ₍₄₎ B_|| dl  = m_o I_enclosed.
Along (1), B is parallel to the path and has constant magnitude, so B_|| = B, and so it can be factored out of the integral to give B ₍₁₎ dl = BL.
Along (2), B is either zero or it is perpendicular to the path, so the contribution is zero.
Along (3), B is zero, so the contribution is zero.
Along (4), B is either zero or it is perpendicular to the path, so the contribution is zero.
So, we're left with
BL = m_o I_enclosed.
Now, work on the right hand side of the relationship.  I_enclosed depends on how many turns of the solenoid are enclosed; let's just say N turns.  Then,
BL = m_oNI.
B = m_oNI/L.
Now, we have a problem.  When we discussed Gauss's Law, we pointed out that there should be no quantities in our final answer that are dependent on the dimensions of our imaginary surface.  In this result, we have two quantities that depend on the size of our imaginary loop, namely the length L and the number of turns enclosed, N.  What is the resolution to this paradox?

Lastly, let's look at the example of a flat, infinite sheet of current.  The current at any spot is directed parallel to some given axis (in the figure, it's all out of the page). Since the total amount of current is infinite (so long as it's not zero), we should define a current density, the amount of current per unit length: J = I/L.  This is not the usual current density J, the current per unit area.  Look at some point a distance r from the sheet and try to determine the appearance of the B-field.

Consider two currents, I₁ and I₂, each producing a contribution to the total field; we see that the vertical components add and the horizontal components cancel.  In addition, since we can slide the sheet up or down (or for that matter, into or out of the page) some distance and it will still look the same, we can say that the magnitude of the B-field is the same value at any point a given distance r from the sheet.  What's more, the field on the other side of the sheet at distance r will have that same magnitude, although we can see quickly with the Right-Hand-Rule that the direction is reversed.

Let's choose a rectangularly shaped loop about which to sum.  Since we as yet have no idea how the B-field strength varies with distance from the sheet, we'll draw the loop symmetrically as shown; then at least the field magnitudes along parts (1) and (3) will be the same.
So,
B_|| dl = m_o I_enclosed =  ₍₁₎ B_|| dl +₍₂₎ B_|| dl + ₍₃₎ B_|| dl + ₍₄₎ B_|| dl  = m_o I_enclosed.
Along (1), B is parallel to the path and has constant magnitude, so B_||  = B and it can be factored out of the integral to give B ₍₁₎ dl  = BL.
Along (2), B is perpendicular to the path, so the contribution to the sum is zero.
Along (3), B is parallel to the path and has constant magnitude, so B_||  = B and it can be factored out of the integral to give B ₍₃₎ dl  = BL.
Along (4), B is perpendicular to the path, so the contribution to the sum is zero.
So, we're left with
2BL = m_o I_enclosed.
Now, work on the right hand side of the relationship.
I_enclosed = J L, so
2BL = m_oJ L.
B = m_oJ/2.
Note that the answer is independent of r, the distance from the sheet.

Magnetic Moment

Why does a bar magnet act like a current loop?  See below.

Let's consider a rectangular loop (length l and width w) carrying current I, in a uniform magnetid field, B, the plane of which is inclined to the direction of the field by some angle q, which we will measure between the magnetic field and a line perpendicular to the plane of the loop:

Consider a more schematic view of this loop:

In this figure, the current running up the back leg is not shown.  From the previous discussion, we might well expect that the current will cause to form a north pole somewhere up and to the right of the loop (and a south pole down and to the left) and that, as a result, the loop will begin to swing clockwise to align itself with the magnetic field.  Let's examine this in closer detail:
The loop comprises four legs or straight sections, each carrying current I, in a magnetic field B.  There will therefor be a magnetic force on each leg,
F_M = ILB sinq_I,B (RHR):

F_top = IwB sin 90^o = IwB (up),
F_front = IlB sinf (out of the page),  f is the complement of our original q.
F_bottom = IwB sin 90^o = IwB (down),
F_rear = IlB sin (180^o - f)  = IwB sinf (into page).

Now, the net force on the loop is clearly zero, but what about the torque?  On the front and rear sides, the net torque will be zero, since, for every bit of current on the front side experiencing a magnetic force, there is an oppositely directed current bit experiencing a force in the other direction which as acting along the same line as the first force.  However, on the top and bottom legs, the forces do not act along the same line, and so there will be some net torque.  Remembering that
t = rF sinq_r,F (RHR),
divining that r is l/2 (since the loop will most likely rotate about its centre of mass, although in fact this doesn't really matter) and that q_r,F is the same as our original q,
t_top = [l/2][IwB] sinq (into the page)
t_bottom = [l/2][IwB] sinq (also into the page),
and the total torque is
t_total = lIwB sinq (into the page).
If there had been N turns in the loop, the torque would have been N times larger, so let's work that into the relationship.  Also, lw is the area of the loop, and it can be shown that the result valid for this specific case is valid for any flat loop of area A:
t_total = NIAB sinq (into the page).
Now, we will do what we did in a similar situation back in Section 1.  We will define a new vector, m, as the magnetic dipole moment; its magnitude will be NIA and its direction will be from the centre of the loop toward the 'north pole' formed by the current.  We see, then, that the theta in our result is the angle between B and m, and so a Right-Hand-Rule seems appropriate. Try B x m; that points out of the page, so let's make it t = m x B, or,
t = mB sinq_m,B (RHR).

Loops which are not flat can of course also have magnetic moments, but they must be calculated a bit more carefully.  Here is an example:

Consider a circular loop of radius R which has been bent at a right angle along a diameter.  What is the magnetic moment of this current distribution?

One last bit of advice: don't confuse the magnetic dipole moment m with out magnetic constant, m_o.

Now, let's look at this a different way:  we can consider a torque acting to rotate our current loop, thus causing an angular acceleration, or we can develop a potential energy function, and consider said PE to be converted into kinetic energy as the loop swings around.  We will make use of the result we obtained for the electric dipole and guess that
PE_dipole = - mB cosq_m,B.
Careful experimentation confirms that this guess is acceptable.

Causes of Magnetic Effects in Materials

Consider these questions:

What happens if a magnet gets broken?  Do we get a (separate) north pole and a south pole?

No, we wouild get two magnets:

Why are not all pieces of iron magnets?

The magnetic forces causing this ordering are short-range forces, that is, the ordering is local only.  In most iron, there are regions called magnetic domains in which a majority of the moments are aligned, but the overall average magnetization is zero (or close to it).

 How can we make a magnet?

We can place the iron in a strong external field.  Then, just like little compass needles, the magnetic moments will align with that field.  When the field is removed, the moments are already aligned in such as manner that their interactions reinforce each other, which will preserve the alignment in that direction.

How can we destroy a magnet?

Any process which acts to disrupt the interaction between atoms should do the trick. First, heating the iron introduces thermal energywhen the thermal energy is larger than the associated magnetic energy, the spins will align randomly.  The temperature above which ferromagnetism disappears is called the Curie temperature.  Above the Curie temperature, ferromagnetic materials become paramagnetic.

Paramagnetism occurs when there is relatively little or no interaction between adjacent magnetic moments, and so they are arranged randomly.  When exposed to an external magnetic field, the moments will align along the field, just as would compass needles, but they will randomize again when the field is removed.

Anti-ferromagnetism is a bit more common than ferromagnetism.  In this case, the adjacent moments do align in opposite directions.  An example is NiO.

If an external B-field is applied to an anti-ferromagnet, what will happen?

In ferrimagnetism, there are two (or more) types of magnetic atoms which have different magnetic moments which are anti-parallel, and which therefor don't quite cancel:

This means that these materials can be very magnetic, and indeed, most commercial ceramic magnets (like the ones on your refrigerator) are ferrimagnetic.
In fact, out prototype natural magnet, lodestone (magnetite Fe₃O₄) is a very complicated ferrimagnet, with three species of magnetic atoms, all iron.  First, there are doubly ionized iron atoms which occupy what are called 6-f sites in the crystal lattice, and there are trebly ionized irons which occupy two different types of locations in the crystal lattice, 6-f and 4-f sites.  A schematic of this system might look like this:

 

[Note: the notes on dimagnetism are not complete; in particular, you should ignore the formulas, which still contain errors!]

Unlike the effects list above, diamagnetism is an orbital effect, rather than a spin effect.  Here is a simple classical model:  Consider two electrons in the same orbital of an atom.  We consider them to be orbiting in opposite directions (that is, one has positive and the other negative orbital angular momentum).  The magnetic moment of each is calculated from
m = IA = [q/t][pr²] = [qv/d]pr² = [qv/2pr]pr² = qvr/2
and the directions of the moments are given by the RHR, and so add to zero.
FIGURE
The centripetal force for each is provided by the coulomb attraction from the positive nucleus:
k_eqQ/r² = mv²/r.
Now, let's add a magnetic field B into the page.  There will then be a magnetic force acting on each electron (qvB), inward on the clockwise moving electron (1) and outward on the counterclockwise moving electron (2); these additional forces change the centripetal forces, moving (1) into a tighter orbit and (2) electron into a bigger orbit:
k_eqQ/r₁² + qvB = mv²/r₁
k_eqQ/r₂² - qvB = mv²/r₂
We'll assume that v doesn't change, although that may well not be true.
Continuing along this line requires solving a messy quadratic, but all we need is to see that the two orbit radiuses will be different, r₁<r₂.  The magnetic moment due to orbital motion will be given by
m = IA = [q/t]pr² = q[v/2pr]pr² = qvr/2
So, m1 is
Since the radiuses of the orbits are now different, so are the magnetic moments; the net magnetic moment actually points opposite to the applied magnetic field.  As a result, the total magnetic field inside a diamagnetic material is reduced, in a manner similar to what happens to the electric field in a dielectric.  In fact, in the same way that conductors act to eliminate any electric fields in their interiors, superconductors are perfectly diamagnetic, so that the total magnetic field in a superconductor is zero.