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Advanced EM - Master in Physics 2011-2012 1 Magnetic potential and field of a SOLENOID Infinite length N spires/cm Current I Radius R The problem -for.

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Presentation on theme: "Advanced EM - Master in Physics 2011-2012 1 Magnetic potential and field of a SOLENOID Infinite length N spires/cm Current I Radius R The problem -for."— Presentation transcript:

1 Advanced EM - Master in Physics 2011-2012 1 Magnetic potential and field of a SOLENOID Infinite length N spires/cm Current I Radius R The problem -for A x –is mathematically similar to finding the electrostatic potential Φ with a surface charge ρ 0 sinθ. This type of surface charge is what is given by two parallel cylinders of uniform and equal, but of opposite sign, charge densities, with their axes displaced by an infinitesimal distance d: the calculation is shown below. Let V be the potential of a cylinder of radius R and charge density uniform. Two such cylinders of the same radius and charge density opposite in sign have a superficial charge density : A Their potential has abehaviour J

2 Advanced EM - Master in Physics 2011-2012 2 Owing to the azimutal symmetry of the system, we can calculate the Φ for x=0, where y=r. We get a r dependence of Φ ~r inside the wire and ~(1/r) outside. Then, A being always along the direction of the θ axis, it turns around the solenoid axis like the current does. It varies with r like r for r R. Once we put all the coefficients right, we obtain: Having A written as a function of r we can calculate its curl, i.e. B. Having B z finite inside the solenoid and at its radius suddenly zero is not realistic. But… also an infinite solenoid is not quite realistic. And anyway the coil has a finite thickness. Note at this point an important physical fact: outside the solenoid the vector potential is not zero nor constant, BUT: θ

3 Advanced EM - Master in Physics 2011-2012 3 BUT: While not being trivial still has curl equal zero. On the other hand it is quickly verified that this A has zero divergence. We are then led by this paradoxical result to ask ourselves where the Physics is, in the field B or in the potential A. The answer has been given by the Bohm – Aharonov experiment, 1956. Which has already been described.

4 Advanced EM - Master in Physics 2011-2012 4 EM fields varying in time This is the term added by Maxwell to preserve charge conservation. It allows the equations to become “Wave equations”. With this new term the electrical and magnetic fields are now coupled through their dependence on time. Without it, a variation of B with time would have caused a variation of E. End of the story. Now, a variation of E changes B as well: the process is self-sustaining. Now a question comes out immediately: beforehand (in electrostatics) we knew that we could express E as the gradient of a potential and B as the curl of a “vector potential”. How does that change now? This still holds, then this is still valid We then insert it in the second Maxwell equation and obtain }{ 14 0 1 4 tcc tc         Ε JΒ Β Β Ε Ε  

5 Advanced EM - Master in Physics 2011-2012 5 This equation is satisfied, by the formula on the right: the 3 rd and 2 nd EofM are satisfied since the condition on the curl of E is satisfied, and this determines E up to the gradient of any function! We call that function Φ ; it will be determined by the 1 st EofM which, upon substitution of this value for E becomes We still have to choose the gauge. For the time-dependent case we choose the Lorentz gauge, defined by: Vector formula

6 Advanced EM - Master in Physics 2011-2012 6 Inserting now the Lorentz gauge condition in the last 2 equations in Φ and in A they become much simpler: Which is (are, actually) the classical wave equations. With wave velocity = c !!! These equations are of course NOT independent: beside the Lorentz gauge which has been used to obtain them, and constrains Φ and A, J and ρ are also related by the charge conservation. In formulas:

7 Advanced EM - Master in Physics 2011-2012 7 Faraday’s law The second EofM, also called Faraday’s induction law, usually given in its differential form: can also be written in integral form (using the Stokes theorem in the passage from line to surface integral) if we define the “electromotive force” E Now, let the closed line over which we calculate the line integral be a real electric wire. The electromotive force E is generated by the variation (in time) of the flux of B through any surface enclosed by the closed line. Such variation can be generated in three different ways: Varying the current through the magnets. Moving the circuit. Moving the magnets. Matter of fact, what is actually measurable is the force on a charge in the wire, so the equation becomes:

8 Advanced EM - Master in Physics 2011-2012 8 In the laboratory reference system, if the circuit is moving with velocity v, a free charge inside it will feel the force 1.What changes in time is the current running in the magnets. Then: Case verified 2. The circuit moves with velocity v while the magnetic field does not change. V Then a free charge (electron) will feel the Lorentz force: ( V х dl ) is a vector which is normal to the surface element dS (yellow in the drawing) covered by the circuit in the time dt. Then: E = -(1/c) · [ time variation of the flux of B due to the change of surface covered by the circuit (in blue and in yellow in the drawing)]. Let us now study these 3 different cases and see what we get from them.

9 Advanced EM - Master in Physics 2011-2012 9 3. The magnets are moved: the currents which generate B do not change, the circuit sits still, but the magnets are moved. This case is a sort of bridge between the two previous ones: from the mathematical point of view in the formulas it is B that has changed; but it is also identical to the second case, because if what has moved is the circuit or the magnets the result is the same. BUT… But, the choice of the reference system, causes the choice in the mathematics of either of the term in the calculation of E. The result is, however, the same! Another remark on the case a): currents varying with time. Let us imagine a transformer. An electromotive force is generated on the circuit C when the current is changed on A. But, where C is located there is hardly a magnetic field. There is, however a significant vector potential. C A

10 Advanced EM - Master in Physics 2011-2012 10 The conservation of energy We have already seen the mathematical expression for the charge conservation Since we know that energy is also conserved, we would like to find another similar equation for the energy conservation – of the electromagnetic field, of course. We are therefore looking for two physical quantities, let us call them U and S, which can be related to energy density (of the EM field) and to energy flow, which satisfy the following equation. The term on the right-hand side of this equation has the form: What we shall do now is to try and find a formula – in which only the fields appear - which looks like the first term in the equation above which states the conservation, i.e. an equation which has the form: “(the derivative of a scalar plus the divergence of a vector) = - E·J”. We shall do that starting from E·J, by substituting J with its expression taken from the 4 th EofM:

11 Advanced EM - Master in Physics 2011-2012 11 Here goes the calculation: we shall use the vector formula The calculation goes as follows: This is Poynting’s theorem. Note that it equals E·J to the sum of a time derivative of a scalar + the divergence of a vector, precisely what we were looking for. It is therefore immediate to interpret this formula as follows: And the equation for the conservation of energy is:

12 Advanced EM - Master in Physics 2011-2012 12 So, we have now, in regime of full time dependence of the fields and of the charges and currents, found a formula for the energy distribution of the electromagnetic field. But… hadn’t we already met a formula for this energy density? Well, we had found two. We write for simplicity only the electric density. The general formula adds the electric and the magnetic energies Only one of them is valid in general: the other one (that with the charge density ρ ) is only valid in the electrostatic case. Examples of Poynting’s vector. E x.. j k B S The energy enters the wire from outside!!!

13 Advanced EM - Master in Physics 2011-2012 13 The charging of a capacitor What we want to explain in this slide is how are the fields and the potentials in the case of a parallel plate capacitor being charged with a constant current. NB: between the capacitor plates, J=0 !! J h R B E S E and d E /dt are directed along the vertical axis. To find out how is B directed in the gap, let’s write A is the integral of J, therefore both A x and A y are zero. A z only depends on r, and so the curl of A is directed along the θ axis. Then S, the Poynting’s vector, is directed along the axis orthogonal to E and B, therefore is directed towards the axis of the capacitor, and has value The electrostatic energy has uniform density between the plates, U=(1/8π)E 2, the total energy is R 2 πhU and its rate of change is (Rh/4)E(dE/dt). The S vector is directed radially from outside towards the capacitor gap: ALL THE ENERGY STORED IN THE CAPACITOR HAS ENTERED LATERALLY FROM THE OUTSIDE AIR!


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