5A. Current and Conservation of Charge Current and Current Density

5A. Current and Conservation of Charge Current and Current Density
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) flowing in 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 force to 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 graduate mechanics: 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: Match velocities at t = 0:

Force and Torque on a Wire
Consider a thin wire carrying a current I Break the volume integral into an integral along the wire and across the wire If the wire is thin, B will be constant across the wire And I will be constant along the 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 of a 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 do the 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 get the total magnetic field from the entire wire: The Biot-Savart law The constant 0 is called the magnetic permeability of free space Its value is exactly: The Coulomb is defined to make this true x B x – x' dl' x' I *Not the BOR law

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 find curl 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 the product 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
Integrating over volume, it is easy to get an integral version 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 5 A 2 A 1 A 4 A 7 A

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 make counter-clockwise loops around the z-axis It should depend only on distance 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 the enclosed current is Therefore, I

Vector Potential and Gauge Transformations
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 have the same curl, then Anything with vanishing curl 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 A not in Coulomb gauge We want to find a gauge transformation such that A' will be 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:

Field Far from a Current Distribution (1)
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:

For any localized distribution Proof by sample problem We can also show Proof by homework problem Which lets us rewrite: Or Now average them:

Define the magnetic dipole moment: Then we have: The magnetic field 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, only the 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: 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, the dipoles are effectively smeared out If we have several types of magnetic dipoles with 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 that it 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 constitutive relationship 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 up with magnetic 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 from currents plus magnetism:
It is clear that there are bound currents associated 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 – L y 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 current K 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 the magnet, 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 The magnetization is the same We therefore have the magnetic potential from a magnetic sphere Fields inside, for example, are And fields outside are

Electromotive Force (EMF) and Magnetic Flux
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 the work 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 integral of the magnetic flux density over the surface Unit is Weber, Wb = Tm2

Right hand rule for Faraday’s Law:
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

*Ultimately, this leads to Einstein’s Theory of Relativity
Electric Fields from Faraday We can generate electromotive force – EMF – by moving the loop in and out of magnetic field Can we generate it by moving the magnet? 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 any surface 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 need to 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 density ni(x) of charges of magnitude qi Compare to the expression for current density: We therefore have: This formula is useful in 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

And the Energy Density Is …
As with electric fields, we are now going to assume linear behavior: In which case we can write So we have Integrate this over time Assume that W = 0 when there are no fields We can also think of this as an energy density

Another Expression for Energy
With electric fields, there are two equivalent ways to write the energy Can we do something comparable with magnetic energy? Since B has no divergence, we can write it as We therefore have Use one of our product rules: Substitute in: Use   H = J And the divergence theorem Assuming fields vanish at infinity, we have

Sample Problems 5.9 Suppose we have a static localized distribution of currents producing a magnetic field. Show that the expression for total energy is unchanged due to a gauge transformation. A gauge transformation is given by If we perform a gauge transformation, then we have Use the product rule: We therefore have: For static currents,   J = 0 Use the divergence theorem Since currents are localized, they vanish at infinity

5A. Current and Conservation of Charge Current and Current Density

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5A. Current and Conservation of Charge Current and Current Density

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