5. Magnetostatics

5. Magnetostatics 5A. Current and Conservation of Charge Current and Current Density • Charge can not only be at rest, it can also move from one region to another • If a charge density  moving with velocity v, we define the current density • If we have several components, this generalizes to • It is possible to have current with no net charge density • The integral of current density over area is called current: • Units are amps: A = C/s • For example, a wire with current I(t) flowingin the +z direction through it would have I

Conservation of Charge • Charge is conserved – it can neither be created nor destroyed • Change in charge in a volume is due to current flowing out, so • Use divergence theorem • Write charge as integral of charge density • Rearrange: • If true for any volume V, then we must have • Local conservation of charge • We are going to start with magnetostatics: no dependence on time, so • If we’re working with wires, this means all currents go in loops • Or come in from/out to infinity V J

Magnetic Flux and Forces, Force on a Wire • In addition to electric fields, there is also magnetic flux density* • Unit is Tesla, T = N/m/A • Magnetic flux density has no effect on static charges • But they do cause forces on moving charges • Add this to the electric forceto get the Lorentz force • On a moving charge density, the magnetic force is • If you have several charge densities, this adds up to • There is also a torque on current: • Add it up for the current distribution *Confusingly, also called magnetic field

Sample Problem 5.1 A charge q has mass m and arbitrary initial velocity in a constant magnetic flux density B in the z-direction. Find the velocity of the particle at all times. • Force is: • From graduatemechanics: • Equating these,we have • The final equation is easy to solve • To solve the others, take another derivative: • General solution of this equation is: • Substitute to get vy: • Matchvelocitiesat t = 0:

Force and Torque on a Wire • Consider a thin wire carrying a current I • Break the volume integralinto an integral along the wireand across the wire • If the wire is thin, B willbe constant across the wire • And I will be constant alongthe wire • Note: for closed loop in constant magnetic field: I

Sample Problem 5.2 A region in the xy-plane has a magnetic flux density . Find the total force on a circular loop of wire of radius a in the xy-plane carrying a counter-clockwise current I centered at height h above the wire, where h > a. y • For convenience, place center of loop on y-axis at y = h • Measure positions on the loop by angle  • Then the coordinate ofa point on the circle is: • We now need the force: • In this case, dl is just dx, so • Not hard to see that the cos integral vanishes • Let Maple dothe other one: a h x

5B. Creating Magnetic Fields The Biot-Savart Law* • Not only do currents feel magnetic flux density, they also generate it • A wire carrying a current at x'generates a magnetic flux density B(x) : • Perpendicular to the direction dl' the current is travelling • Perpendicular to the separation between the wire and the current x – x' • The formula is • Integrate it up to getthe total magnetic fieldfrom the entire wire: • The Biot-Savart law • The constant 0 is called themagnetic permeability of free space • Its value is exactly: • The Coulomb is defined to make this true B x x – x' dl' x' I *Not the BORlaw

Sample Problem 5.3 Find the magnetic field from a finite wire segment with current I at an arbitrary point. x • Put the wire on the z-axis, and work in cylindrical coordinates • Put the measurement point at z = 0, so • Let wire run from z = a to z = b • Then magnetic field is: • Those last two terms can be interpreted in terms of angles • Note that this formula should not be used alone, since current isn’t in loops! • But you can use it for an infinite wire:  a b z = a z = b I

Magnetic Flux Density from a Wire • A current carrying wire produces a magnetic flux density that makes loops around it • Can find direction from right-hand rule • Point thumb in direction of current • Fingers curl in direction of magnetic flux density • Note that the magnetic field “curls” around the wire • As we’ll see, this implies the curl of B is the current density J I

Current Density B-field, and Gauss’s Law for B • Current is really integral of current density • Let’s find the divergence of this: • Distribute the derivative using product rule • No derivative on J because it’s with respect to x, not x', so • Remaining term vanishes (homework problem 0.1), so • This is called Gauss’s Law for magnetic flux density

Ampere’s Law, Derivative Version (1) • Let’s findcurl of B • Use product rule, keeping in mind that J is not a function of x: • First term was done first day of class • Second term, use the fact that • So far we have

Ampere’s Law, Derivative Version (2) • On second term,consider theproduct rule • Because we are doing magnetostatics: • Integrate over volume using the divergence theorem • At infinity, left side is zero, so right side is zero • This is the same as the last term above • The remaining term in the integral is trivial

Integral Version of These Rules 1 A 2 A 4 A 5 A 7 A • Integrating over volume, it is easy to get an integralversion of Gauss’s law for magnetic fields • Integrate Ampere’s law over an arbitrary surface • Use Stoke’s theorem on the left • The right side is the current passing through the surface • The sense of this rule is governed by the right-hand rule: • Curl fingers in the direction of the loop integration • Thumb points in direction of I • For example, consider the problem at right: • In this case, we’d have

Sample Problem 5.4 A wire of radius a is centered along the z-axis, and carries current I in the +z direction distributed uniformly over its cross-section. Find the magnetic field everywhere. • We expect magnetic field to makecounter-clockwise loops around the z-axis • It should depend only ondistance from the axis • To find the field outside, draw Ampere loop at radius  > a: • This loop contains all the current, so • To find the field inside, draw Ampere loop at radius  < a: • This time theenclosed current is • Therefore, I

5C. The Vector Potential Vector Potential and Gauge Transformations • The magnetic field satisfies: • Anything without a divergence can written as the curl of something • This something is called the vector potential A • Is A unique? • Suppose two A’s havethe same curl, then • Anything with vanishingcurl is a divergence, so • This is our first example of a gauge transformation •  is called a gauge function • We need to pick a gauge condition to get rid of this ambiguity

Coulomb Guage • A common choice of gauge condition would be Coulomb gauge: • Can we always do this? • Suppose we have Anot in Coulomb gauge • We want to find a gauge transformation such that A'willbe in Coulomb gauge • So we want: • This equation has a unique solution • Hence we can always change to Coulomb gauge • In Coulomb gauge: • This is just three versions of the Poisson equation • We know the solutions:

5D. Magnetic Dipoles Field Far from a Current Distribution (1) • Suppose we are far from a current distribution: • We want to find the vector potential in this limit • We approximate:

Sample Problem 5.5 Show that for a localized static current distribution, • Consider the expression • Use the divergence theorem • Currents vanish at infinity • Use product rule • Static currents have no divergence • Write out dot product as a sum • Since every component vanishes:

Field Far from a Current Distribution (2) • For any localized distribution • Proof by sample problem • We can also show • Proof by homework problem • Which letsus rewrite: • Or • Now average them:

Field Far from a Current Distribution (3) • Define the magnetic dipole moment: • Then we have: • The magneticfield is:

Contact Term (1) • Shrink the dipole source to a point, then these formulas should work for r > 0 • Consider integral of magnetic field over a spherical volume of radius R • Can show: • Proof by quantum homework problem • Multiply by mj, sum on j, and rearrange • We therefore have

Contact Term (2) • However, if we naïvely integrate the magnetic flux density: • How did we mess up? • Our formulas only work for r > 0 • Need to add a “contact term” – delta function right at x = 0

Dipole Moment of a Loop A I • Consider a loop of wire in the xy-plane. • Let current I flow counter-clockwise • Dipole moment is • Clearly, onlythe z-component exists: • Use Stokes’s Theorem • Generalize: • Direction of normal governed by right-hand rule

Force on a Localized Current (1) • Consider magnetic force on a localized current • If B is slowly varying, Taylor expand B around x = 0: • Recall: • Write out the components of F • Use anti-symmetry of j l • This is a cross product: • We have:

Force on a Localized Current (2) • Compare to: • We therefore have: • Fancy identity with Levi-Civita symbol: • We therefore have: • But recall that • Putting it all together, we have

Torque on a Localized Current • Consider magnetic torque on a localized current • Treat magnetic field as constant • Expand double cross-product • Use the fact that • This means you can swap x and J if you throw in a minus sign, or even average the two resulting from swapping • First two terms are a double cross product • Recall: • So the torque is

Energy of a Current Loop in a B-field? • It seems pretty easy to get the energy of a loop something like: • We can use this to get the torque or the force Is this the energy required to bring in a current loop from infinity? • If we are dealing with a current maintained by (say) a battery, then as we move the loop, the magnetic flux through the loop changes • According to Faraday’s Law (not yet discussed) there will be an electromotive force resisting this change in flux • Therefore we have to restore the current using a battery • So this doesn’t work • For fundamental magnetic dipoles (like an electron, for example) this does work

5D. Macroscopic Magnetostatics Smoothing and Background Magnetic Dipoles • As with electric fields, there can be large microscopic magnetic fields • To get around these, do a space smoothing. • Then we would still have: • We can therefore still write • And still use Coulomb gauge if we want • Divide the current into two pieces: • The second term is due to microscopic dipoles we aren’t interested in • The vector potential will be due to a combination of these two:

Smoothing the Background Dipoles • Because of the smoothing, thedipoles are effectively smeared out • If we have several types of magnetic dipoleswith dipole moment mi with number density ni(x), define • Then our equation becomes: • Use the product rule on the last term: • First term integrates to a surface term that vanishes if M vanishes at infinity • So we have:

The Magnetic Field • When only the first term is present, we know thatit leads to a magnetic flux density satisfying • It is clear that background magnetic dipoles can be included by J  J +   M • Therefore, we now have • Define the magnetic field H by • Units A/m • We therefore have

Boundary Conditions and Constitutive Relation Comment: Two of these equations will have to be modified soon • For electric field and electric displacement, we found boundary conditions, appropriate wherever free charges are not present • By analogy, boundary conditions for B and H: • We also need some sort of constitutiverelationship between B and H • We will often assume this relationship is linear • The proportionality constant  is called the magnetic permeability • Units N/A2 or H/m or Wb/A/m • We can also write this in terms of the magnetic susceptibility m • Dimensionless

Types of Materials There are generally three different types of materials: • Diamagnetic materials are molecules with no unpaired electrons • They develop currents that oppose the background magnetic field, so m < 0 • Effects are small, |m| < 10-4 • Paramagnetic materials are molecules with unpaired electrons • Electrons flip their spins to line upwithmagnetic field, so som> 0 • Effects are small, |m| < 10-2 • Ferromagnetic materials have electrons in“domains” with their spins aligned • In the presence of magnetic fields,the domains all line up m > 0 • Effects large, m > 104 not uncommon

Ferromagnets, Saturation, and Hysteresis • Because the domains are large, thermal effects do not typically cause the spins to flip randomly • Even a modest magnetic field can get most of the spins to align • The magnetic field becomes saturated – it reaches its maximum value • Non-linear behavior • If you turn the field back off, the domains will not easily go back to being random • Magnetization can then exist even in the absence of a background field • In particular, the magnetization is a function not only of the magnetic field, but also the history • This phenomenon is called hysteresis

Sample Problem 5.6 Suppose you have an infinite magnetic cylinder of arbitrary cross-section shape, with uniform magnetization M along it. What is the magnetic field and magnetic flux density everywhere? • Magnetic field, from symmetry, will always point in the z-direction and be independent of z: • Since we have no current, we have • Hence H must be independent of x and y: • But if it must vanish when x or y are infinite • We now find B everywhere: M

Bound Currents • Recall the vector field fromcurrents plus magnetism: • It is clear that there are bound currentsassociated with the magnetization • If you are in the interior of a linear medium, then we have • Most places there are no free currents most places • However, on the surface of a magnetic material there will be a surface current L – L

Magnetic Surface Currents z y – L x L • Consider small loop sketched at right, half in and half out of the magnetic material • Consider the integral • Use Stokes’s theorem • Substitute Jb • Current is integral of charge density • For integral of magnetization, the sides are short, and M = 0 in vacuum • In a similar fashion, we can put the loop in the y-direction • This suggests a surface currentK flowing along the surface: • Units A/m • Generalizing, the surface current will be:

5E. Solving Magnetostatic Problems Strategies for a Uniform Linear Medium What sorts of problems might we have with magnetic materials? • We could have linear materials where  is constant • We can have hard magnetic materials where M is constant • Many problems have no explicit current J Strategy with a uniform linear medium • Because B = 0, we can always write • For example, for Coulomb gauge in a linear medium, we would find • We already know how to solve this:

Vector Potential with Hard Magnets • Suppose we have hard magnets with fixed M • We know general solution is • Typically M will be constant on themagnet, but will result in surface currents • We therefore have: • Then we get the magnetic flux density from

Magnetic Potential • Suppose we have no currents • Then the magnetic field can be written as a gradient: • For example, suppose we have a hard magnet, M fixed • For example, if M is constant, then M exists only on the surface, where it becomes like a “magnetic surface charge” • We find:

Sample Problem 5.7 A long but finite bar magnet has uniform magnetization M along it and cross sectional area A. Find the magnetic field not near either end. • We use the formula • Only contribution is on the ends • Because we aren’t near the end, x – x' is roughly constant over the end • We therefore have • Magnetic field is M

Solving New Problems from Old • Suppose we have no currents • And we have a linear medium • Compare with problems from electrostatics: • Any equation we have already solved for electrostatics can be solved for magnetostatics • Just make the substitutions

Sample Problems 5.8 A sphere of radius a and magnetic permeability  is placed in a uniform magnetic field of magnitude H0. Find the magnetic potential m everywhere. • We did this problem already for electric fields. • Take over the results, with trivial modifications • Let’s also work out the magnetic field and magnetic flux density inside: • The magnetization inside is:

Comment on Sphere in a Background Field: • We know this solution satisfies • To this add a constant magnetic field: • The new fields still satisfy the same equations • The new potentials are • Themagnetizationis the same • We therefore have the magneticpotential from a magnetic sphere • Fields inside, for example, are • And fields outside are

Faraday’s Law – + Electromotive Force (EMF) and Magnetic Flux • Suppose we have some source of force on charges that transport them • Suppose it is capable of doing work W on each charge • It will keep transporting them until thework required is as big as the work it can do q • The voltage difference at this point is the electromotive force (EMF) • Denoted E • Magnetic flux through a surface is the integralof the magnetic flux density over the surface • Unit is Weber, Wb = Tm2

Motional EMF • Suppose you have the following circuit in the presence of a B-field • Charges inside the cylinder • Now let cylinder move • Moving charges inside conductor feel force • Force transport charges – it is capable of doing work • This force is like a battery - it produces EMF v  B W v L B • v is the rate of change of the width W • We can relate this to the change in magnetic flux Right hand rule for Faraday’s Law: EMF you get is right-handed compared to direction you calculated the flux

Electric Fields from Faraday Magnet • We can generate electromotive force – EMF – by moving the loop in and out of magnetic field • Can we generate it by moving the magnet? Faraday’s Law works whether the wire is moving or the B-field is changing* • How can there be an EMF in the wire in this case? • Charges aren’t moving, so it can’t be magnetic fields • Electric fields must be produced by the changing B-field! • The EMF is caused by an electric field that points around the loop *Ultimately, this leads to Einstein’s Theory of Relativity

Differential Version of Faraday’s Law • Note we are no longer doing statics! • Use Stokes’s theorem on the first term • Write the flux as a space integral • Bring the time derivative inside • Must convert it to a partial derivative • The only way this can be true for anysurface S is if the argument in []’s vanishes • This replaces   E = 0 • We now have three correct Maxwell’s Equations • And one incorrect one

Energy in Magnetic Fields Power Delivered to a Charge • We would like to find how much work we needto do to build up a set of steady-state currents • We will assume in the final state, there are only currents and magnetic fields • But at intermediate steps, since the magnetic fields are changing, there must be electric fields • But we will assume the changing magnetic fields are the only source of the electric fields • We are implicitly assuming the changes are slow • We first consider the Lorentz force on a charge • The power (change in work)done on a single charge is • Now consider many charges:

Power in Terms of Current • Now imagine we have a number densityni(x) of charges of magnitude qi • Compare to the expressionfor current density: • We therefore have: • This formula is usefulin its own right

Rewriting the Work to Create a Current • Now, use • So we have: • Use the product rule: • So we now have: • Use divergence theorem and Faraday’s Law: • Assuming fields at infinity vanish, we have

5. Magnetostatics