Electromagnetic induction

Transcription

1 9 Electromagnetic induction 9. Faraday s experiment A word induction appeared first in description of electromagnetic phenomena in the context of electrostatics. An electrostatic induction is a phenomenon where electrically neutral bodies gain electric properties when they are close to some other electrically charged bodies. In such a case an external electrostatic field causes motion of electric charges in the conductor until an equipotential configuration is reached. It results in appearance of electrostatic field associated with new distribution of charges. Faraday thought that there must exist a counterpart of this phenomenon in the case of circuits with electric currents. In other words, he suspected appearance of induced electric current in circuits closed to another circuits with non-vanishing currents. He did not realized initially that the essential condition for electromagnetic induction is time dependence of currents. His first experiments with stationary currents failed. The breakthrough in his research came when he began to work with time varying currents and time varying magnetic fields. Let us consider a sequence of experiments containing two coils (primary and secondary circuit), electromotive force sources (district current) and a permanent magnet.. The primary circuit is connected with a battery. There flows a district current in the circuit. We change the distance between the circuits by changing the position of the primary circuit. The secondary circuit is at rest with respect to the laboratory reference frame. The sign of induced electromotive force in the secondary circuit depend on sign of velocity of the primary circuit. After changing direction of district current the sign of induced electromotive force changes to the opposite.. The primary circuit is connected with a source of district current. The circuit remains at 39

2 9. ELETOMAGNETI INDUTION P. Klimas (a) (b) (c) Figure 9.: (a) Two circuits: the primary (left) and the secondary (right). (b) variation of a current in the primary circuit. (c) variation of magnetic field using a permanent magnet. 40

3 9. onductors in motion in magnetic field rest with respect to the laboratory reference frame. We change the distance between the circuits by changing the position of the secondary circuit. The sign of induced electromotive force depends on the sign of velocity of the secondary circuit. omparing this result with result of previous experience, we conclude that the signs concord when coils move close. Similarly, the signs corresponding with drifting apart of circuits also concord. We conclude that only matters the relative motion of circuits. 3. Both circuits are at rest with respect to the laboratory reference frame. We switch on and switch off the source of district current. There appears an electromotive force in the secondary circuit in short time intervals. 4. We repeat experiments,,3 including a magnetic needle which indicates correlation between induced electromotive force and variation of the magnetic field produced by the primary circuit. 5. We substitute the primary circuit by a permanent magnet and repeat experiences,,3. We observe that the sign of the electromotive force depend on velocity of the magnet and its orientation. 9. onductors in motion in magnetic field 9.. onductive rod in magnetic field Let us consider a conducting rod in external magnetic field B. The rod moves with velocity v perpendicular to its longitudinal axis and perpendicular to the vector of magnetic induction B. There exists the Lorentz force in the laboratory reference frame that acts on electrons according to F = q v B. (9..) c The force causes motion of electric charges until new equilibrium is reached. The rod is electrically polarized in the state of equilibrium. The electric field that exist inside the rode is such that each electric charge remains in equilibrium i.e. qe + hq v i c B =0. (9..) Such a configuration of electric charges has non-vanishing electric dipole moment. There is electric field inside and outside of the rod associated with such distribution of electric charges. Here, by equilibrium we mean no macroscopic drift. learly, on the microscopic level there exist quick random motion of electrons that form the Fermi gas. 4

4 9. ELETOMAGNETI INDUTION P. Klimas (a) (b) Figure 9.: (a) A rode in the magnetic field in the laboratory reference frame. (b) Electric field of the rode in the laboratory reference frame. How to describe such a system in the reference frame S 0 of the rod? Since the velocity of the rode vanishes in its reference frame then the Lorentz force must vanish as well. However, dislocation of the charges is observed in both reference frames! We conclude that there must be some other force in S 0 that is responsible for polarization of the rod. Since electric charges at rest (a) (b) Figure 9.3: (a) A rode in the magnetic field in its own reference frame. (b) Electric field in the rest frame of the rode. can interact only with electric field then there must exist an electric field E 0 in the reference frames 0 Such electric field can be obtained as result of Lorentz transformation of the magnetic field B. The magnetic field B 0 in the reference frame of the rod has different magnitude to the field observed in the laboratory reference frame. The field E 0 is uniform because B is uniform in S. Note, that the problem in the reference frame of the rode reduces to electrostatic problem of perfect conductor in external electric field (a magnetic field is irrelevant in this situation). It follows, that total electric field must vanish inside the conductive rode (in S 0 ). In other words, an external electric field E 0 inside the rode must cancels out with the field that originates in charge For v? B the electric field transform as E 0 = E + v c B where E = 0. Magnetic field transform as B 0 = B v c E where E = 0. For small velocities B0 B and E 0 v c B. 4

5 9. onductors in motion in magnetic field density of the rod. The electric field outside the rod is nonuniform in S onductive loop in an external magnetic field One can substitute the rod by a coil in form of closed rectangular loop with sides a and b. We consider loop which moves with velocity v parallel to sides of length a. When magnetic field is uniform then the only effect associated with motion of the coil is charge distribution (polarization of the coil). The sides of the coil having length a are equipotential so there is no (a) (b) Figure 9.4: A rectangular coil in uniform magnetic field. Figure 9.5: A rectangular coil in non-uniform magnetic field. induced current in the loop for the case of uniform magnetic field. On the contrary, there appears induced current for non uniform magnetic field. Let F be the force which acts on a given electric charge in the coil in motion. Existence of the current indicates that electric charges are transported by this force along the loop. The work 43

6 9. ELETOMAGNETI INDUTION P. Klimas done by the force on a closed path = reads I I v F dl = q c B dl = q 4X I k= k v c B dl k. (9..3) Integrals on 3 and 4 vanish because v k dl 3 and v k dl 4. We shall assume that b is short enough, which enable us aproximatie integrals Z v c B dl q v c B b, Z v c B dl q v c B b. Magnitudes of the magnetic field B and B can be taken as average values of the field at the respective paths. An integral on a closed loop is then I F dl = q v c (B B )b. (9..4) This expression represent a work realized on electric charges associated with transport of a single charge q along the closed loop. In physical system a transport along a closed path requires a long time. After a short period values of magnetic induction B and B would change so such interpretation cannot be taken literally. What really happens is that the force F dislocate many electrons on short distances. The sum of works done on the individual electrons is equal to H F dl. We can introduce electromotive force (emf) equal to E := I F dl. (9..5) q 9..3 elation with magnetic flux An electromotive force induced in the rectangular coil is given by expression E = v c (B B ). (9..6) Let us consider the loop in two very close instants of time t and t + magnetic fluxes (t + t) and (t) reads t. The difference of (t + t) (t) = (t) B bv t + B bv t (t)+o( t ) = v(b B )b t + O( t ). (9..7) 44

7 9. onductors in motion in magnetic field Figure 9.6: Variation of magnetic flux during an infinitesimal dislocation of the coil. where B and B are approximately constant for infinitesimal displacement of the loop. omparing (9..6) and (9..7) we find that (t + t) (t) E = lim = d t!0 t c. (9..8) Although we have shown (3..4) for a particular case, in fact, it is valid for any shape and any motion of the coil. In order to prove this statement we shall employ the result Z I da B = A dl (9..9) S which implies that the magnetic flux is the same for all open surfaces with a common = 0 i.e. S da B = 0 S da B. Figure 9.7: An arbitrary motion of the coil in magnetic field. We shall consider an arbitrary motion of a coil. Let and be two closed loops that correspond with position of the coil at t and t + t. Since the magnetic flux does not depend on particular choice of the surface then the flux H A dl can be represented by fluxes through S 45

8 9. ELETOMAGNETI INDUTION P. Klimas (with border ) and the flux through the surface connecting and. It follows that Z Z Z (t + t) = B da = B da + B da S+ S S {z } S {z } (t) where is the variation of the flux is exclusively associated with the flux through the side. We choose the surface S as a narrow strip with area element da := (v t) dl. For very small time interval t the surface integral can be given as a product of the line integral and t I I = B [(v t) dl]+o( t )= t dl (v B)+O( t ) = t I dl F + O( t )= t E + O( q t ). (9..0) Expression (9..0) gives (3..4) in the limit We shall discuss this law in detail in section (9.3.). t! 0. This result is known as Faraday s law. 9.3 Faraday s law 9.3. Lenz s law An electromotive force of induction causes a flow of the electric current in the coil. A direction of this flow is determined by the sign in Faraday s law. The minus sign means that the current flows in such a direction that variation of the flux associated with magnetic field of the induced current is opposite to variation of the flux that causes the induced current. This statement is known as Lenz s law. For instance, if a conducting thin ring moves in direction of a permanent magnet then increasing of the magnetic flux through any surface having the ring as its border leads to induction of the current in the ring. The magnetic field of this current reduces the flux through the surface spanned on the ring. A similar situation occurs for a permanent magnet that freely falls in a cooper tube. In such a case a material of the tube is exposed on time-varying magnetic field. It causes induction of Eddy currents which are responsible for generation of an extra magnetic field. An interaction of the magnet with this field reduces the velocity of the magnet. Electrons in the tube have vertical velocity in the reference frame of the magnet so they experience the Lorentz force. The magnetic field is static in this reference frame. The situation chances when passing to reference The velocity associates with thermal motion does exist, however, it is irrelevant for macroscopic number of electrons. The averge Lorentz force acting on electrons in thermal motion vanishes. 46

9 9.3 Faraday s law Figure 9.8: A ring falling down in presence of electromagnet which is a source of nonuniform magnetic field. (a) (b) Figure 9.9: A permanent magnet (a) at rest and (b) in motion. frame of the tube. A magnetic field in this frame is not static any longer. Moreover, there exist an electric field in this reference frame. The electric field cannot be electrostatic i.e r E 6= Faraday s law Substituting the electromotive force of induction E and the magnetic flux of the field B given by I E := Z E dl and := S B da 47

10 9. ELETOMAGNETI INDUTION P. Klimas into Faraday s law we get the following equation I E dl + Z d B da =0. (9.3.) c S We shall consider situations when the circuit is not deformed when moving. It follows that curve does not depend on time and then variation of the magnetic flux has origin in variation of the magnetic field itself. In such a case one gets d S B da = tb da. Application of the Stoke s theorem to (9.3.) yelds Z r E da =0. (9.3.) S This equation holds for any surface S. It is possible if and only if r E Equation (9.3.3) is a local version of the Faraday s law. =0. (9.3.3) Non-vanishing of the integral H E dl means that lines of the electric field form closed loops. Such a field cannot be generated by any stationary distribution of electric charge. Note, that equations (9.3.) and (9.3.3) do not determine completely the electric field because E and E +r' digger by gradient of a function ' so they are equivalent in the context of equations (9.3.) and (9.3.3). It follows from H r' dl 0 for integral formulation of Faraday s law and from r r' 0 for its local version Vector potential Nonexistence of magnetic monopoles is expressed by equation r B(t, x) =0. This equation became an identity for B(t, x) =r A(t, x). (9.3.4) In terms of the vector potential the local form of Faraday s law reads apple r E =0. (9.3.5) Equation (9.3.5) is satisfied as an identity if the expression inside the bracket is proportional to a gradient of any scalar function of t and x. Taking into account the static limit (electrostatics), we put explicitly the sign minus in front of the arbitrary function function i.e. electric field is given by expression '(t, x). The E = 48

11 9.3 Faraday s law The same result can be obtained analyzing the integral form of the Farday s law. Plugging (9.3.4) to the integral (9.3.) and applying the Stokes theorem we get I apple E dl =0. (9.3.6) Since is a fixed curve then d H A dl = ta dl. The most general solution of (9.3.6) is given by the function r'(t, x). Similarly to the case of static vector potential A(x) the equation (9.3.4) for time dependent vector potential has the symmetry A(t, x)! A 0 (t, x) =A(t, x)+r (t, x) (9.3.7) which results in B 0 (t, x) =B(t, x). The electric field would not depend on gauge transformation for appropriate choice of the function ' 0. Let us consider the set of transformations '! ' 0 and (9.3.7). One gets E 0 = r' 0 t[a + r ] = r[' 0 + t ] ta. (9.3.8) The fields E 0 and E are equal for ' 0 + t = '. It leads to set of transformations '! ' 0 = ' t (9.3.9) A! A 0 = A + r (9.3.0) which preserve electric and magnetic field. The gauge symmetry is a fundamental symmetry of electromagnetism. Moreover, it allows to introduce fundamental interactions as a gauge fields which helps to preserve local symmetry of action for matter fields Example: Infinitely long solenoid We shall analyze the electromagnetic field of an infinitely long solenoid with radius a and n turns per unit of length. The magnetic field does not vanish only inside the solenoid. Applying the Ampere s law to a small rectangular circuit which surrounds few coils one gets that H B dl = 4 c I tot. The magnetic field must be of the form B = B(r) (a the Ampere s law we get r)ẑ. Plugging this formula to B = 4 ni (a r)ẑ. (9.3.) c 49

12 9. ELETOMAGNETI INDUTION P. Klimas Figure 9.0: Electric field induced outside of the solenoid. The vector potential of the solenoid can be determined from the identity I Z A dl = B da (9.3.) where the surface S can be chosen as a disc perpendicular to axis z with radius r and the center i at the longituidinal axis of the solenoid. It follows that dl = ˆ rd S and da = ẑ da. A vector potential should be a function which has only azimuthal component and depends on radial coordinate A = A(r) ˆ (9.3.3) Plugging this result to (9.3.) we get r A(r) = r B for r<aand r A(r) = a B for r>a, then ( A(r) = rb ˆ for r<a, a r B ˆ for r<a, where B 4 c ni. Note that ˆ = ẑ ˆr so B ˆ = r B r. It follows that A(r) = Note, that r A =0for r>a. ( B r for r<a a B r r for r<a. When the electric current density is a slow-varying function of time, then our result serves for determination of electric field inside and outside of the solenoid. Note, that for quickly varying currents the retardation effect became relevant so the expression for A loses its validity. In such a case the vector potential can be determined as the integral of current density and fundamental solution for d Alembert equation. For quasi-stationary currents ( A(t, r) = B(t) r ˆ for r<a a B(t) ˆ r for r<a, 50

13 9.4 Inductance ni(t). In absence of electrostatic sources the electric field is given by expres- where B(t) 4 c sion what gives E(t, r) E(t, r) =( c a r ˆ for r<a r omponents (r E)ˆx and (r E)ŷ vanish immediately whereas the component (r E)ẑ reads (r E)ẑ = h r h r (h E ˆ) (h r E ˆr ) = r(reˆ). (9.3.5) The electric field inside the solenoid r < a has the form E ˆ r = tb(t) what leads to equation (r E)ẑ On the other hand, outside of the solenoid (r E)ẑ = 0 because E ˆ = a tb(t). The electric field outside of the solenoid does exist, however, its curl vanishes. It follows from the fact that curl of the vector potential vanishes in this region. An integral along the closed loop which surrounds the solenoid does not vanish I E dl = 9.4 Inductance 9.4. Mutual inductance Z d = = 0 a c n di. (9.3.6) We consider two electric circuits (coils) that correspond with closed loops and given by position vectors r and r. A magnetic field associated with current I in reads B (r) = I I dl r r c r r 3. (9.4.7) A flux of this field through any surface S with border reads Z apple Z I = B da = da S c S dl r r r r 3 {z } const I. (9.4.8) The constant expression depends only on geometry of circuits and its mutual position. Let us assume that current I is a slowly varying function of time. In such a case its magnetic field 5

14 9. ELETOMAGNETI INDUTION P. Klimas Figure 9.: oils and. B (t, r) can be obtained in framework of quasi-stationary approximation, what leads to a magnetic flux (t) = const I (t). An electromotive force induced in the circuit reads E = M di M const c = ci (9.4.9) where M is coefficient of mutual inductance. It has physical unit in SI system is Henry H = Vs/A. Note, that in Gaussian system of units [M ]= s cm Example As example we consider two concentric and coplanar circular coils with radii and, where. Each coil has respectively N and N turns. A magnetic field of the coil is non-uniform. The assumption allows to approximate the field B at the surface (disc) S by a uniform field that value correspond with B (t, r = 0). This field has the value B (t, 0) = N I (t) c Z 0 d ˆ ˆr = N I (t) c ẑ, (9.4.30) so the flux is a product of this value and N -th multiplicity of area of a disc c N N I (t). (9.4.3) The electromotive force of induction is then E = c d di = M where M = c N N. (9.4.3) 5

15 9.4 Inductance eciprocity theorem A coil with the electric current intensity I (t) induces an electromotive force E = M di in the coil. Similarly, the coil with the current I (t) induces an electromotive force Figure 9.: Two arbitrary circuits. di E = M in the coil. Let A and A be two vector potentials that originate in currents I and I. The magnetic fluxes through the surfaces S and S spanned on the a = a, a =,, read = = I dr A (r )= I I I dr dr (9.4.33) c r r I dr A (r )= I I I dr dr c r r. (9.4.34) Note, that expressions containing integrals are equal. The only difference is an order of integration which cannot change the result. One gets /I = /I. It follows that coefficients of mutual inductance are equal: M = c I = c I = M (9.4.35) i.e. there exists only one such coefficient for each two coils. It means that the current intensity I (t) = I(t) in the circuit induces electromotive force in the circuit whose value is exactly the same as value of the force induced in due to variation of the current I (t) =I(t) in. learly, this result has nothing to do with a geometric symmetry of circuits. In fact, such a symmetry does not exist in majority of cases. In the case of two coplanar coils there is no symmetry $. Indeed, we have M = c N N = M. which is not simple replacing by and vice versa. 53

16 9. ELETOMAGNETI INDUTION P. Klimas 9.4. Self-inductance Any variation of the electric current in the coil leads to induction of the electromotive force in the proper coil because the induced electric field acts on electrons inside the material of the conductor that coil is made of. The proportionality coefficient between magnetic flux and the current in the circuit is called self-inductance coefficient (t) =cli(t) (9.4.36) which depends only on geometry of the coil. The induced electromotive force reads E = c d = LdI. (9.4.37) Example As a simple example we consider a toroidal coil having internal radius and external radius. The coil has N turns and its hight has value h. We assume that inside the coil there is a paramagnetic material with permeability µ. The magnetic strength can have only an azimuthal component H = H(t, r) ˆ. Taking into account the geometric symmetry of the problem we choose integration paths as a family of circles coplanar with z = const. A total intensity if the Figure 9.3: A toroidal inductor. current reads 8 >< 0 for r<, I tot = NI for <r<, >: 0 for r>. (9.4.38) The Ampere s law H H dl = 4 c I tot gives magnetic intensity inside the coil H(t, r) = N c I(t) r for <r< (9.4.39) 54

17 9.4 Inductance and H(t, r) =0outside of the coil. A total magnetic flux through N turns of the coil reads =N Z h 0 Z dz dr µ N c The coefficient of self-inductance takes the value apple I(t) µn = c r c h ln I(t). L = µn c h ln. (9.4.40) A general expression for coefficient of self-inductance of the coil having N turns is given in terms of magnetic flux and current intensity L = N ci = N c SH B da H dl c 4 which can be also written in terms two closed line integrals L = 4 N c Note, that capacity of capacitor was obtained as = Q V = H S D da 4 L E dl. H H A dl (9.4.4) H dl Electric circuits with self-inductance The simplest electric circuit containing inductance consists of a battery, a coil and a resistor. An intensity of the current in such a circuit depends on time. An electromotive force is a sum of Figure 9.4: A circuits containing an inductor. 55

18 9. ELETOMAGNETI INDUTION P. Klimas electromotive force of the battery E 0 and EMF due to electromagnetic induction E ind = From Ohms law E 0 + E ind = I(t) we get L di. di + L I = E 0 L. (9.4.4) This equation has a solution I(t) = E 0 +Ae L t, where free constant A can be determined from initial condition I(0) = 0 which gives A = E 0 I(t) = E 0 e L t. (9.4.43) Let us now consider an electric circuit with a stationary current I 0. At t = t the battery is removed. The current flows only through resistor and the inductor. The current must satisfy equation di + I =0 (9.4.44) L which has solution I(t) =Ae L t where we determine A from condition I(t )=I 0 giving A = I 0 e L t I(t) =I 0 e L (t t ). (9.4.45) 9.5 ircuits containing inductance 9.5. Example: electric circuit in vicinity of linear, infinitely long conducting wire We shall consider a linear wire of infinite length with slowly varying electric current I 0 (t). A rectangular conductive loop with resistance is localized in the plane = const (in cylindrical coordinates), see Fig.9.5. The magnetic field of quasi-stationary current in rectangular wire has the form B = I 0 (t) ˆ (9.5.46) c r and the flux through the surface spanned on the rectangular circuit reads Z = S B da = I 0(t) c Z b 0 Z r+a dr 0 apple b dz r r 0 = c ln r + a r It follows that coefficient M := M = M takes the value I 0 (t). (9.5.47) M(r) := b c ln r + a. (9.5.48) r 56

19 9.5 ircuits containing inductance Figure 9.5: A rectangular coil an an infinitely long electric wire. A value of this coefficient is a function of time when a distance between the wire and a rectangular circuit changes according to r = r(t). We shall denote the coefficient of self-inductance of the rectangular coil by L. A total electromotive force is a sum E = E self + E mutual = equation di(t) L di d c which result in + L I(t) = d L (M(r)I 0(t)) f(t) (9.5.49) which HS is an explicitly given function that depends on external mechanical force and external electromotive force. A solution of this equation I(t) can be represented by the integral Z I(t) =Ae L t + 0 D(t t 0 )f(t 0 ) (9.5.50) where A is a free constant and D(t t 0 ) is fundamental solution i.e. it is a solution of distributional equation (D 0 (t)+ L D(t),'(t)) = ( (t),'(t)). (9.5.5) A fundamental solution can be written in the form D(t) = (t)z(t) where Z 0 + L Z =0. h L = 4N c (a + b)+ p a + b a ln( a+p a +b b ) b ln( b+p a +b a )+aln a d where d is a radius of the wire and N stands for number of turns. i b + b ln d 57

20 9. ELETOMAGNETI INDUTION P. Klimas Plugging this solution to equation (9.5.5) we find Z so it must be Z(0) =. It follows that (D, ' 0 )+ (D, ') =(, ') L Z Z(t)' 0 (t) + Z(t)'(t) = '(0) 0 L 0 Z Z(0)'(0) + Z 0 + L Z '(t) = '(0) 0 {z } 0 D(t) = (t)e L t. (9.5.5) Let us consider that there is no current in a rectangular circuit at t =0i.e. I(0) = 0 A = Z 0 D( t 0 )f(t 0 ) (9.5.53) ase: dm =0, di 0 6= 0. We shall consider the case when the coil remains at rest and the current intensity I 0 (t) is explcit function of time. An induced current satisfies equation di + L I = M L di 0 f(t). (9.5.54) Let us consider an electric current I 0 (t) in a form of an impulse with a finite time of duration I 0 (t) = i 0 cos T t for 0 <t<t (9.5.55) and I 0 (t) =0for t<0 and t>t, where i 0 = const. An electric current intensity I(t) in the rectangular wire reads The function f(t) has the form 8 >< I n () for t<0, I(t) =Ae L t + I n () for 0 <t<t, >: I n (3) for t>t. f(t) =f 0 sin T t ( (t) (t T )) where f 0 := M L (9.5.56) T i 0. (9.5.57) 58

21 9.5 ircuits containing inductance A non-homogeneous part of the solution is of the form I n (t) = = Z Z 0 D(t t 0 )f(t 0 ) 0 (t t 0 )e L (t t0) f(t 0 ) = e L t Z t 0 e L t0 f(t 0 ) Z t = f 0 e L t 0 e L t0 sin = f 0 e L t i Z t T t0 ( (t 0 ) (t 0 T )) 0 e ( +i L T )t0 e ( L i )t0 T ( (t 0 ) (t 0 T )) (9.5.58) The integral I () n (t) =0for t<0 because (t 0 ) (t 0 T )=0for t 0 <t. For 0 <t<twe have (t 0 ) (t 0 T )=in the interval 0 <t 0 <tso I n () (t) =f 0 e L t i = f 0 e L t i = = L L f 0 e L t + T f 0 e L t + T Z t 0 e ( +i L T )t0 e ( L i )t0 T 0 e ( +i L T )t e( L + T i apple L i apple L e L t sin L T i T t i T )t L T i e ( +i L T )t L + T i e ( L T e L t cos T t + T. i T )t The solution of non-homogeneous equation reads I () n (t) = L f 0 + T apple T e L t cos T t + L sin T t. (9.5.59) At t = T this current takes the value I () n (T )= L f 0 + T T e L T. (9.5.60) Finally, we compute the integral I (3) n defined in the interval t>t. There is no contribution 59

22 9. ELETOMAGNETI INDUTION P. Klimas from the integral on the interval t 0 >tso It gives I n (3) (t) =f 0 e L t Z T 0 e ( +i L T )t0 e ( L i )t0 T i 0 = L f 0 e L t + T = f 0 e (T t) L L + T apple L e L T sin T T T e L T. T e L T cos T T + T I (3) n (t) =I () n (T ) e L (T t). (9.5.6) This result shows that there is only an exponentially decreasing component of the current for t>t. ase: di 0 =0, dm 6= 0. An induced current obes equation di + L I = I 0 dm L f(t) (9.5.6) where dm = dm dr dr dm dr v(t). Derivative of M with respect to r reads dm dr = b apple c r + a r. (9.5.63) A solution of non-homogeneous equation is given by I n (t) = Z D(t t 0 )f(t 0 ) (9.5.64) where f(t) = I apple 0b c L r(t)+a r(t) v(t). (9.5.65) It gives I n (t) = Z t = I 0b c L e L t 0 e L (t t0) f(t 0 ) Z t apple 0 e L t0 v(t 0 ) r(t 0 )+a r(t 0 ) then I n (t) = I 0b c L e L t W (r(t)) (9.5.66) 60

23 9.5 ircuits containing inductance where W (r(t)) = Z t apple 0 e L t0 v(t 0 ) r(t 0 )+a r(t 0 ). (9.5.67) In order to simplify our considerations we shall study an uniform motion given by r = r 0 + vt where v = const. We shall assume that a motion of the circuit was initiated at t =0. It follows that t = r r 0 v. After change of the variable of integration one gets Z r apple W (r) = dr 0 e Lv (r0 r 0 ) r 0 r 0 + a r 0 = e Lv (r 0+a) Z r Lv (r0 +a) dr 0 e r 0 r 0 + a e Lv r 0 We define new variable u := Lv (r0 + a) for the first integral, so dr0 take u := Lv r0 for the second integral. The integral W (r) reads W (r) =e Z Lv (r 0+a) = e Lv (r 0+a) e Lv r 0 Z Lv (r+a) Lv (r 0+a) Z du e u u du e u Lv (r 0+a) u du e u Lv r u 0 Z Each integral can be cast in the form of exponential integral e Lv r 0 Z du e u Lv r u Z r Lv r0 dr 0 e r 0 r 0. (9.5.68) r 0 +a = du u Z Lv r du e u Lv r u 0 du e u Lv (r+a) u. By analogy, we. (9.5.69) Ei(z) := Z z du e u u (9.5.70) what gives W (r) =e e Lv r 0 apple Lv (r 0+a) apple Ei Ei Lv Lv r (r + a) Ei Lv r 0 Ei Lv (r 0 + a). (9.5.7) 9.5. oupled circuits Inductors in series Two inductors in series can be coupled in two non-equivalent ways. Depending on the mutual orientation of the inductors a magnetic field of the second/first inductor can reinforce or diminish the magnetic flux in the first/second inductor. 6

24 9. ELETOMAGNETI INDUTION P. Klimas Figure 9.6: Two inductors in series. When fluxes sum constructively then induced electromotive forces in coils of inductor a =, read di E a = L a M di (9.5.7) where M M = M and I I = I because circuits are in series. Another possibility leads to subtracting of magnetic fluxes giving E a = L a di + M di. (9.5.73) Assuming that conductors have resistances and and there is another source of electro- (a) (b) Figure 9.7: (a) Positive and (b) negative magnetic coupling of inductors. motive force (a battery, etc) E 0 we get equation E 0 + E + E = I + I which results in (L + L ± M) di +( + )I = E 0. (9.5.74) 6

25 9.5 ircuits containing inductance The circuit can be replaced by an equivalent circuit with total resistance eff = + and total inductance L eff = L + L ± M. When mutual inductance is very small (inductors are far from each other) a total inductance is a sum of self-inductances L eff = L + L. In the case of connection in series the circuit can be reduced to a circuit containing a single inductor and a single resistor Inductors in parallel Similarly to connection in series there are two ways how magnetic coupling between conductors can be realized. It is taken into account by the choice of the sign in expression ±M. When circuits are connected in parallel, then electromotive forces of induction associated with each inductor read E = L di (±M) di E = L di (±M) di (9.5.75) We shall assume that there is a source of electromotive force represented by E 0. The system of Figure 9.8: Two inductors in parallel. circuits is described by two equations E 0 + E = I and E 0 + E = I, which can be cast in the form where L ±M ±M L {z } A di di I I = E 0 E 0 (9.5.76) det A = L L M 6=0. (9.5.77) 63

26 9. ELETOMAGNETI INDUTION P. Klimas Multiplying by an inverse matrix A = det A L M M L (9.5.78) one gets di di {z } di L ±M I det A ±M {z L I } {z } I = E 0 det A L M L M. (9.5.79) Before studying the complete solution we shall consider an idealized case = 0. In such a case equations depend only on derivatives of currents I and di di = E 0 det A L M L M I. (9.5.80) Taking a sum of last two equations we find that I tot = I + I obeys the equation di tot = L + L M L L M E 0. When magnetic coupling between circuits is very small M 0, then di tot = + E 0 ) = +. L L L eff L L A general solution of the original problem (without aproximations) is the sum of general solution of a homogeneous equation and particular solution of a non-homogeneous equation. When external electromotive force does not depend on time E 0 = const then particular solution of non-homogeneous equations is constant as well. From (9.5.76) we get that the solution is given by expression I p I p = E0 E 0 We have to find general solution of the set of homogeneous equations (9.5.8) di I =0 (9.5.8) where is a matrix defined in (9.5.79). We shall assume the following form of solution I(t) =v () e t + v () e t where v (a) v (a) = v (a) a =,. (9.5.83) 64

27 9.5 ircuits containing inductance Plugging this ansatz to equation (9.5.8) we get an algebraic equation ( v () v () )e t +( v () v () )e t =0. which is satisfied for any t if ( a)v (a) =0. (9.5.84) It follows that parameters a and v (a) are, respectively, eigenvalues and eigenvectors of matrix. Eigenvalues are determined from equation det( ) =0with = L det A ±M det A ±M L det A which gives We find (det A) +(L + L ) + =0. (9.5.85) =(L L ) +4M (9.5.86) so eigenvalues of read = p (L + L ) (L L M ) = p +(L + L ) (L L M ) (9.5.87) so det A = + p L L, det A = p L L. Plugging this eigenvalues to (9.5.84) and defining where := L + det A = L det A = hp i (L L ) (9.5.88) and := L + det A = L det A = hp +(L L )i (9.5.89) we get ±M ±M v () v () = 0 0 (9.5.90) 65

28 9. ELETOMAGNETI INDUTION P. Klimas and ±M ±M v () v () = 0 0 (9.5.9) It follows from (9.5.86) that the product of and reads = 4 (L L ) = M. (9.5.9) As expected, each set of equations (9.5.90) and (9.5.9) is linearly dependent det ±M ±M =0=det ±M ±M From the first equation of (9.5.90) we have v () ±M v () =0. Then, choosing v () := we get v () = ± M. Similarly, the second equation in (9.5.9) gives ±M v () + v () = 0. Taking v () := we get v () = M. The eigenvectors read v () = ± M where sign ± refers to cumulative (+) or anti-cumulative ( v () = M. (9.5.93) ) magnetic coupling of conductors. Note, that another form of eigenvectors is possible by redefinition of constants and. Taking = ± M and = M we get ṽ () = ± M ṽ () = M. (9.5.94) Eigenvectors (9.5.93) or (9.5.94) are not orthogonal because an associated matrix is not symmetric. A solution of the problem has general form I (t) I (t) = ± M e t + M e t + E0 E 0 (9.5.95) where and are free constants. These constants can be fixed by initial condition. We shall assume following initial conditions I (0) = I 0 and I (0) = I 0. The solution given by(9.5.95) at t =0takes the form M ± M {z } U = I 0 E 0 I 0 E 0 (9.5.96) 66

29 9.5 ircuits containing inductance where det U =+ we find that = =. Multiplying the equation (9.5.96) by U ± = M apple E 0 I 0 apple M M ± M I 0 E 0 + I 0 E 0 I 0 E 0, (9.5.97). (9.5.98) Limit of weak magnetic coupling M! 0. The leading terms of eigenvalues read q / = L L ± L L + 4 M (L L ) L L M apple ± +... (9.5.99) L L L L and so / = L L s + apple L L (L L )+ 4 M (L L ) (L L ) It follows from this expression that for L >L we have M! 0 M! in the limit M! 0. On the other hand, for L <L one gets M! M! 0. L L M + O(M 4 ). learly, the eigenvectors v () and v () are more convenient for the first case whereas the choice ṽ () and ṽ () is better for the second case. The solutions take the form I (t) = e L t 0 + e L E0 t + I (t) 0 for L >L and I (t) 0 = e L t + I (t) 0 e L E 0 E 0 t + E0 (9.5.00). (9.5.0) for L <L. The solutions are totally equivalent. The only difference are free constants $, $. 67