# Maxwell equations and their applications

###### ppt on maxwell equations in thermodynamics and maxwell equations applications
Dr.GriffinWood,United Kingdom,Teacher
Published Date:23-07-2017
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Maxwell’s Equations Physics 2415 Lecture 29 Michael Fowler, UVa Today’s Topics • Maxwell’s equations • The speed of light Equations for Electricity and Magnetism • Gauss’ law for electric fields E dA q / 0  the electric flux out of a volume = (charge inside)/ . 0 • Gauss’ law for magnetic fields B dA 0  • There is no such thing as magnetic charge: magnetic field lines just circulate, so for any volume they flow out of, they flow back into it somewhere else. Equations for Electricity and Magnetism • Electrostatics: (no changing fields) Ed 0  around any closed curve: this means the work done against the electric field from A to B is independent of path, the field is conservative: a potential energy can be defined. • Faraday’s law of induction: in the presence of a changing magnetic field, the above equation becomes: Edd// dt BdAd dt  B  the integral is over an area “roofing” the path. A changing magnetic flux through the loop induces an emf. Equations for Electricity and Magnetism • Mangnetostatics: B d I 0  around any closed curve: I is the total current flow across any surface roofing the closed curve of integration. • But is this the whole story? • Fields changing in time changed the electrostatic equation, what about this magnetostatic equation? • Let’s look at a particular case… Spherical Current • At t = 0, a perfectly spherical ball of charge is placed at the center of a very large spherical conductor. The charge flows away equally in all directions. What is the magnetic field generated? • 1) It points outwards equally in all directions • 2) Same but pointing inwards • 3) It circles around the initial sphere • 4) No magnetic field is produced by these currents Those Spherical Currents… • Cannot produce a magnetic field • The configuration has perfect spherical symmetry—it would not be changed by turning it through an angle about any axis. • The only fields satisfying this would point in or out along radii everywhere—but that could only happen with a net magnetic charge (N or S) at the center. So, no field at all… Ampère’s Law and Spherical Currents • Imagine in 3D currents flowing spherically outward symmetrically from a ball of charge injected into a large conducting medium. • Imagine a circular curve, like a crown, placed above the source. Clearly some of the current flows through a surface roofing this loop, so  I is nonzero. 0 • But around the loop, because the Bd 0  field B is zero everywhere • So Ampère’s law is not the whole story… Another Ampère’s Law Paradox • Suppose now a capacitor is being charged by a steady I current in a wire. • Consider Ampère’s law for a circular contour around the wire—it’s supposed to be the same for any surface S roofing the circle, but we could choose S , going 2 between the plates, so no current crosses it Maxwell’s Solution • Maxwell knew Faraday had generalized the electrostatic law to include a time-varying magnetic field by adding the changing flux through the curve: Edd// dt BdAd dt  B  • He noticed that when Ampère’s law failed, there was a time-varying electric field through the surface roofing the curve, and suggested including it like this: d  Bd I EdA I d / dt  0 0 0 0 0 E  dt  • (Writing , the electric flux.) EdA E Why the Two Surfaces Give the Same Result • The current I flowing I through surface S is the rate 1 of change of charge on the top capacitor plate, I = dQ/dt. • If the plates are close, all the electric field from the top plate will point down, none E dA Q / will cross S , so 0 1  SS 12 and d d dQ E  EdA I 00  Bottom line: the rate of change of electric dt dt dt S 2 flux through S = current through S . 2 1Ampère’s Law and Charge Conservation • Ampère’s law cannot work by I itself for all surfaces spanning a circle like this: • The surface S is drawn to 2 avoid the current. This is only possible because charge is piling up. The rate of change 1/ of electric flux just equals 0 times how fast the charge is piling up, from Gauss’ law. • This must equal the ingoing current— so the integral over S = that over S . 1 2Maxwell’s Equations • The four equations that together give a complete description of electric and magnetic fields are known as Maxwell’s equations: E dA q / B dA 0 0   Maxwell himself called this term the Edd / dt “displacement B  current”: it produces magnetic d field like a current. E Bd I 0 0 0  dt online notes Magnetic Field In Charging Capacitor • Taking the field between plates • . I uniform, use d Bd I Ed A 0 0 0  dt • For a disc surface between the plates, there is no current I through the surface, there is a changing electric field uniform over the area, generating a circular magnetic field. d d dQ E • For the total plate area,  EdA I 00  dt dt dtMagnetic Field In Charging Capacitor d dQ • For the total area  EdA I 0  • . dt dt I • For a disc of radius r, the total R changing electric field is given by 2 dr r  E dA I 0 2  dt R d • Now use Bd EdA 00  dt  I r 0 B to find between plates. 2 2 RCharging Capacitor and Betatron • Recall that in the betatron a uniform magnetic field increasing in strength in time generated a circling electric field that could be used to accelerate charged particles. • In the charging capacitor we’ve been looking at, a uniform electric field increasing in strength in time generates a circling magnetic field. • In regions of space where there are no charges, changing electric and magnetic fields are related to each other in a very symmetric way. Clicker Question (Review) • Suppose you have an infinite uniform plane of electric charge. What is its electric field? A. Parallel to the plane, of uniform strength. B. Parallel to the plane, decreasing strength with distance from the plane. C. Perpendicular to the plane, of uniform strength throughout space. D. Perpendicular to the plane, decreasing strength with distance from the plane. Clicker Answer • C: Perpendicular to the plane, of uniform strength throughout space. • An infinite plane is of course an idealization: but for a uniform plane charge distribution of finite size, the electric field has very close to uniform strength for distances from the plane less than the linear size of the charge distribution. Clicker Question • Suppose now the uniformly charged plane is set in motion with constant velocity. This means we have a plane of electric current. • The magnetic field generated by this current: A. Is perpendicular to the plane. B. Is parallel to the plane, and in the same direction as its velocity. C. Is parallel to the plane, and perpendicular to the velocity direction. Clicker Answer • Is parallel to the plane, and perpendicular to the velocity direction. ˆ  Idr 0 dB • Remember the Biot Savart law: 2 4 r • The magnetic field from a small piece of current is perpendicular to the current direction—but in this moving plane, all current flow is in the same direction, so all fields are perpendicular to that direction.

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