9 electromagnetic induction - paweł klimas9 electromagnetic induction 9.1 faraday’s experiment a...

9

Electromagnetic induction

9.1 Faraday’s experimentA word “induction” appeared first in description of electromagnetic phenomena in the contextof 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 sucha case an external electrostatic field causes motion of electric charges in the conductor until anequipotential configuration is reached. It results in appearance of electrostatic field associatedwith new distribution of charges.

Faraday thought that there must exist a counterpart of this phenomenon in the case of circuitswith electric currents. In other words, he suspected appearance of induced electric current incircuits closed to another circuits with non-vanishing currents. He did not realized initially thatthe essential condition for electromagnetic induction is time dependence of currents. His firstexperiments with stationary currents failed. The breakthrough in his research came when hebegan to work with time varying currents and time varying magnetic fields.

Let us consider a sequence of experiments containing two coils (primary and secondarycircuit), electromotive force sources (district current) and a permanent magnet.

1. The primary circuit is connected with a battery. There flows a district current in thecircuit. We change the distance between the circuits by changing the position of theprimary circuit. The secondary circuit is at rest with respect to the laboratory referenceframe. The sign of induced electromotive force in the secondary circuit depend on signof velocity of the primary circuit. After changing direction of district current the sign ofinduced electromotive force changes to the opposite.

2. The primary circuit is connected with a source of district current. The circuit remains at

239

9. ELECTROMAGNETIC INDUCTION P. Klimas

(a)

(b)

(c)

Figure 9.1: (a) Two circuits: the primary (left) and the secondary (right). (b) variation of acurrent in the primary circuit. (c) variation of magnetic field using a permanent magnet.

240

9.2 Conductors in motion in magnetic field

rest with respect to the laboratory reference frame. We change the distance between thecircuits by changing the position of the secondary circuit. The sign of induced electromo-tive force depends on the sign of velocity of the secondary circuit. Comparing this resultwith result of previous experience, we conclude that the signs concord when coils moveclose. Similarly, the signs corresponding with drifting apart of circuits also concord. Weconclude that only matters the relative motion of circuits.

3. Both circuits are at rest with respect to the laboratory reference frame. We switch onand switch off the source of district current. There appears an electromotive force in thesecondary circuit in short time intervals.

4. We repeat experiments 1,2,3 including a magnetic needle which indicates correlationbetween induced electromotive force and variation of the magnetic field produced by theprimary circuit.

5. We substitute the primary circuit by a permanent magnet and repeat experiences 1,2,3.We observe that the sign of the electromotive force depend on velocity of the magnet andits orientation.


9.2.1 Conductive rod in magnetic fieldLet 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 accordingto

F = qv

c⇥B. (9.2.1)

The force causes motion of electric charges until new equilibrium is reached. The rod is elec-trically polarized in the state of equilibrium. The electric field that exist inside the rode is suchthat each electric charge remains in equilibrium1 i.e.

qE +

hqv

c⇥B

i= 0. (9.2.2)

Such a configuration of electric charges has non-vanishing electric dipole moment. There iselectric field inside and outside of the rod associated with such distribution of electric charges.

1Here, by equilibrium we mean “no macroscopic drift”. Clearly, on the microscopic level there existquick random motion of electrons that form the Fermi gas.

241


(a) (b)

Figure 9.2: (a) A rode in the magnetic field in the laboratory reference frame. (b) Electricfield of the rode in the laboratory reference frame.

How to describe such a system in the reference frame S0 of the rod? Since the velocity ofthe 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 besome other force in S0 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 inthe rest frame of the rode.

can interact only with electric field then there must exist an electric field E0 1 in the referenceframeS0 Such electric field can be obtained as result of Lorentz transformation of the magneticfield B. The magnetic field B0 in the reference frame of the rod has different magnitude to thefield observed in the laboratory reference frame. The field E0 is uniform because B is uniformin S. Note, that the problem in the reference frame of the rode reduces to electrostatic problemof perfect conductor in external electric field (a magnetic field is irrelevant in this situation). Itfollows, that total electric field must vanish inside the conductive rode (in S0). In other words, anexternal electric field E0 inside the rode must cancels out with the field that originates in charge

1For v ? B the electric field transform as E0= �

�E +

vc ⇥B

�where E = 0. Magnetic field

transform as B0= �

�B � v

c ⇥E�

where E = 0. For small velocities B0 ⇡ B and E0 ⇡ vc ⇥B.

242


density of the rod. The electric field outside the rod is nonuniform in S0.

9.2.2 Conductive 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 magneticfield 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 appearsinduced current for non uniform magnetic field.

Let F be the force which acts on a given electric charge in the coil in motion. Existence ofthe current indicates that electric charges are transported by this force along the loop. The work

243


done by the force on a closed path C = C1

+ C2

+ C3

+ C4

readsICF · dl = q

IC

⇣vc⇥B

⌘· dl

= q4X

k=1

IC

k

⇣vc⇥B

⌘· dlk. (9.2.3)

Integrals on C3

and C4

vanish because v k dl3

and v k dl4

. We shall assume that b is shortenough, which enable us aproximatie integralsZ

C1

⇣vc⇥B

⌘· dl

1

⇡ qv

cB

1

b,ZC2

⇣vc⇥B

⌘· dl

2

⇡ �qv

cB

2

b.

Magnitudes of the magnetic field B1

and B2

can be taken as average values of the field at therespective paths. An integral on a closed loop is thenI

CF · dl = q

v

c(B

1

�B2

)b. (9.2.4)

This expression represent a work realized on electric charges associated with transport of asingle charge q along the closed loop. In physical system a transport along a closed path requiresa long time. After a short period values of magnetic induction B

1

and B2

would change so suchinterpretation cannot be taken literally. What really happens is that the force F dislocate manyelectrons on short distances. The sum of works done on the individual electrons is equal toHC F · dl. We can introduce electromotive force (emf) equal to

E :=

1

q

ICF · dl. (9.2.5)

9.2.3 Relation with magnetic fluxAn electromotive force induced in the rectangular coil is given by expression

E =

v

c(B

1

�B2

). (9.2.6)

Let us consider the loop in two very close instants of time t and t + �t. The difference ofmagnetic fluxes �(t+�t) and �(t) reads

�(t+�t)� �(t) = �(t)�B1

b v�t+B2

b v�t� �(t) + O(�t2)

= �v(B1

�B2

)b�t+ O(�t2). (9.2.7)

244


Figure 9.6: Variation of magnetic flux during an infinitesimal dislocation of the coil.

where B1

and B2

are approximately constant for infinitesimal displacement of the loop. Com-paring (9.2.6) and (9.2.7) we find that

E = lim

�t!0

�(t+�t)� �(t)

�t= �1

c

d�

dt. (9.2.8)

Although we have shown (13.1.4) for a particular case, in fact, it is valid for any shape and anymotion of the coil.

In order to prove this statement we shall employ the resultZSda ·B =

ICA · dl (9.2.9)

which implies that the magnetic flux is the same for all open surfaces with a common border@S = C = @S0 i.e.

RS0 da ·B =

RS da ·B.

Figure 9.7: An arbitrary motion of the coil in magnetic field.

We shall consider an arbitrary motion of a coil. Let C1

and C2

be two closed loops thatcorrespond with position of the coil at t and t+�t. Since the magnetic flux does not depend onparticular choice of the surface then the flux

HC2

A · dl2

can be represented by fluxes through S

245


(with border C1

) and the flux through the surface connecting C1

and C2

. It follows that

�(t+�t) =

ZS+�S

B · da =

ZSB · da| {z }�(t)

+

Z�S

B · da| {z }��

where �� is the variation of the flux is exclusively associated with the flux through the side. Wechoose the surface �S as a narrow strip with area element da := (v�t) ⇥ dl. For very smalltime interval �t the surface integral �� can be given as a product of the line integral and �t

�� =

ICB · [(v�t)⇥ dl] + O(�t2) = ��t

ICdl · (v ⇥B) + O(�t2)

= ��t1

q

ICdl · F + O(�t2) = ��tE+ O(�t2). (9.2.10)

Expression (9.2.10) gives (13.1.4) in the limit �t ! 0. This result is known as Faraday’s law.We shall discuss this law in detail in section (9.3.2).

9.3 Faraday’s law

9.3.1 Lenz’s lawAn electromotive force of induction causes a flow of the electric current in the coil. A directionof this flow is determined by the sign in Faraday’s law. The minus sign means that the currentflows in such a direction that variation of the flux associated with magnetic field of the inducedcurrent is opposite to variation of the flux that causes the induced current. This statement isknown as Lenz’s law.

For instance, if a conducting thin ring moves in direction of a permanent magnet then in-creasing of the magnetic flux through any surface having the ring as its border leads to inductionof the current in the ring. The magnetic field of this current reduces the flux through the surfacespanned on the ring.

A similar situation occurs for a permanent magnet that freely falls in a cooper tube. In sucha case a material of the tube is exposed on time-varying magnetic field. It causes induction ofEddy currents which are responsible for generation of an extra magnetic field. An interactionof the magnet with this field reduces the velocity of the magnet. Electrons in the tube havevertical velocity1 in the reference frame of the magnet so they experience the Lorentz force. Themagnetic field is static in this reference frame. The situation chances when passing to reference

1The velocity associates with thermal motion does exist, however, it is irrelevant for macroscopicnumber of electrons. The averge Lorentz force acting on electrons in thermal motion vanishes.

246

9.3 Faraday’s law

Figure 9.8: A ring falling down in presence of electromagnet which is a source of non-uniform 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 existan electric field in this reference frame. The electric field cannot be electrostatic i.e r⇥E 6= 0.

9.3.2 Faraday’s law

Substituting the electromotive force of induction E and the magnetic flux of the field B givenby

E :=

ICE · dl and � :=

ZSB · da

247


into Faraday’s law we get the following equationICE · dl+ 1

c

d

dt

ZSB · da = 0. (9.3.11)

We shall consider situations when the circuit C is not deformed when moving. It follows thatcurve C does not depend on time and then variation of the magnetic flux has origin in variationof the magnetic field itself. In such a case one gets d

dt

RS B · da =

RS @tB · da. Application of

the Stoke’s theorem to (9.3.11) yeldsZS

✓r⇥E +

1

c

@B

@t

◆· da = 0. (9.3.12)

This equation holds for any surface S. It is possible if and only if

r⇥E +

1

c

@B

@t= 0. (9.3.13)

Equation (9.3.13) is a local version of the Faraday’s law.Non-vanishing of the integral

HC 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 (9.3.11) and (9.3.13) 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(9.3.11) and (9.3.13). It follows from

HC r' · dl ⌘ 0 for integral formulation of Faraday’s law

and from r⇥r' ⌘ 0 for its local version.

9.3.3 Vector potentialNonexistence of magnetic monopoles is expressed by equation r ·B(t,x) = 0. This equationbecame an identity for

B(t,x) = r⇥A(t,x). (9.3.14)

In terms of the vector potential the local form of Faraday’s law reads

r⇥E +

1

c

@A

@t

�= 0. (9.3.15)

Equation (9.3.15) is satisfied as an identity if the expression inside the bracket is proportional toa 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. �'(t,x). Theelectric field is given by expression

E = �r'� 1

c

@A

@t.

248

9.3 Faraday’s law

The same result can be obtained analyzing the integral form of the Farday’s law. Plugging(9.3.14) to the integral (9.3.11) and applying the Stokes theorem we getI

C

E +

1

c

@A

@t

�· dl = 0. (9.3.16)

Since C is a fixed curve then ddt

HC A · dl =

HC @tA · dl. The most general solution of (9.3.16)

is given by the function �r'(t,x).Similarly to the case of static vector potential A(x) the equation (9.3.14) for time dependent

vector potential has the symmetry

A(t,x) ! A0(t,x) = A(t,x) +r�(t,x) (9.3.17)

which results in B0(t,x) = B(t,x). The electric field would not depend on gauge transforma-

tion for appropriate choice of the function '0. Let us consider the set of transformations ' ! '0

and (9.3.17). One gets

E0= �r'0 � 1

c@t[A+r�]

= �r['0+

1

c@t�]�

1

c@tA. (9.3.18)

The fields E0 and E are equal for '0+

1

c@t� = '. It leads to set of transformations

' ! '0= '� 1

c@t� (9.3.19)

A ! A0= A+r� (9.3.20)

which preserve electric and magnetic field. The gauge symmetry is a fundamental symmetry ofelectromagnetism. Moreover, it allows to introduce fundamental interactions as a gauge fieldswhich helps to preserve local symmetry of action for matter fields.

9.3.4 Example: Infinitely long solenoid

We shall analyze the electromagnetic field of an infinitely long solenoid with radius a and n turnsper unit of length. The magnetic field does not vanish only inside the solenoid. Applying theAmpere’s law to a small rectangular circuit which surrounds few coils one gets that

HC B · dl =

4⇡c Itot. The magnetic field must be of the form B = B(r) ✓(a � r)z. Plugging this formula to

the Ampere’s law we get

B =

4⇡

cnI✓(a� r)z. (9.3.21)

249


Figure 9.10: Electric field induced outside of the solenoid.

The vector potential of the solenoid can be determined from the identityICA · dl =

ZSB · da (9.3.22)

where the surface S can be chosen as a disc perpendicular to axis z with radius r and the centeri at the longituidinal axis of the solenoid. It follows that dl = � rd� and da = z da. Avector potential should be a function which has only azimuthal component and depends onradial coordinate

A = A(r)� (9.3.23)

Plugging this result to (9.3.22) we get 2⇡r A(r) = ⇡r2B for r < a and 2⇡r A(r) = ⇡a2B forr > a, then

A(r) =

(1

2

rB� for r < a,1

2

a2

r B� for r < a,

where B ⌘ 4⇡c nI . Note that � = z ⇥ r so B� =

1

rB ⇥ r. It follows that

A(r) =

(1

2

B ⇥ r for r < a1

2

a2

r2B ⇥ r for r < a.

Note, that r⇥A = 0 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 quicklyvarying 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 andfundamental solution for d’Alembert equation. For quasi-stationary currents

A(t, r) =

(1

2

B(t) r � for r < a1

2

a2B(t) 1

r � for r < a,

250

9.4 Inductance

where B(t) ⌘ 4⇡c nI(t). In absence of electrostatic sources the electric field is given by expres-

sionE(t, r) = �1

c

@A

@t(9.3.24)

what gives

E(t, r) =

(� 1

2c@B(t)@t r � for r < a

�a2

2c@B(t)@t

1

r � for r < a.

Components (r ⇥ E)

x and (r ⇥ E)

y vanish immediately whereas the component (r ⇥ E)

z

reads

(r⇥E)

z=

1

hrh�

h@r(h�E

ˆ�)� @�(hrE

r)

i=

1

r@r(r E

ˆ�). (9.3.25)

The electric field inside the solenoid r < a has the form Eˆ�= � r

2c@tB(t) what leads toequation (r ⇥ E)

z= �1

c@B@t . On the other hand, outside of the solenoid (r ⇥ E)

z= 0

because Eˆ�= � a2

2rc@tB(t). The electric field outside of the solenoid does exist, however, itscurl 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 vanishI

CE · dl = �a2

2c

@B(t)

@t

Z2⇡

0

d� = �⇡a2

c

@B(t)

@t= �

✓2⇡a

c

◆2

ndI

dt. (9.3.26)

9.4 Inductance

9.4.1 Mutual inductanceWe consider two electric circuits (coils) that correspond with closed loops C

1

and C2

given byposition vectors r

1

and r2

. A magnetic field associated with current I1

in C1

reads

B1

(r) =I1

c

IC1

dl1

⇥ r � r1

|r � r1

|3 . (9.4.27)

A flux of this field through any surface S2

with border C2

reads

�

21

=

ZS2

B1

· da2

=

1

c

ZS2

da2

·IC1

dl1

⇥ r � r1

|r � r1

|3

�| {z }

const

I1

. (9.4.28)

The constant expression depends only on geometry of circuits and its mutual position. Let usassume that current I

1

is a slowly varying function of time. In such a case its magnetic field

251


Figure 9.11: Coils C1

and C2

.

B1

(t, r) can be obtained in framework of quasi-stationary approximation, what leads to amagnetic flux �

21

(t) = const I1

(t). An electromotive force induced in the circuit C2

reads

E21

= �M21

dI1

dtM

21

⌘ const

c=

�

21

c I1

(9.4.29)

where M21

is coefficient of mutual inductance. It has physical unit in SI system is Henry1H = V s/A. Note, that in Gaussian system of units [M

21

] =

s2

cm

.

9.4.1.1 Example

As example we consider two concentric and coplanar circular coils with radii R1

and R2

, whereR

2

⌧ R1

. Each coil has respectively N1

and N2

turns. A magnetic field of the coil C1

isnon-uniform. The assumption R

2

⌧ R1

allows to approximate the field B1

at the surface (disc)S2

by a uniform field that value correspond with B1

(t, r2

= 0). This field has the value

B1

(t,0) =N

1

I1

(t)

c

Z2⇡

0

d�R1

�⇥✓� r

R2

1

◆= N

1

I1

(t)2⇡

cR1

z, (9.4.30)

so the flux is a product of this value and N2

-th multiplicity of area of a disc ⇡R2

2

�

21

⇡ 2⇡2

c

R2

2

R1

N1

N2

I1

(t). (9.4.31)

The electromotive force of induction is then E21

= �1

cd�21dt = �M

21

dI1dt where

M21

=

2⇡2

c2R2

2

R1

N1

N2

. (9.4.32)

252

9.4 Inductance

9.4.1.2 Reciprocity theorem

A coil with the electric current intensity I1

(t) induces an electromotive force E21

= �M21

dI1dt

in the coil C2

. Similarly, the coil C2

with the current I2

(t) induces an electromotive force

Figure 9.12: Two arbitrary circuits.

E12

= �M12

dI2dt in the coil C

1

. Let A1

and A2

be two vector potentials that originate incurrents I

1

and I2

. The magnetic fluxes through the surfaces S1

and S2

spanned on the coils,@Sa = Ca, a = 1, 2, read

�

21

=

IC2

dr2

·A1

(r2

) =

I1

c

IC2

dr2

·IC1

dr1

|r2

� r1

| (9.4.33)

�

12

=

IC1

dr1

·A2

(r1

) =

I2

c

IC1

dr1

·IC2

dr2

|r1

� r2

| . (9.4.34)

Note, that expressions containing integrals are equal. The only difference is an order of integra-tion which cannot change the result. One gets �

21

/I1

= �

12

/I2

. It follows that coefficients ofmutual inductance are equal:

M21

=

1

c

�

21

I1

=

1

c

�

12

I2

= M12

(9.4.35)

i.e. there exists only one such coefficient for each two coils. It means that the current intensityI1

(t) = I(t) in the circuit C1

induces electromotive force in the circuit C2

whose value isexactly the same as value of the force induced in C

1

due to variation of the current I2

(t) = I(t)

in C2

. Clearly, this result has nothing to do with a geometric symmetry of circuits. In fact, sucha symmetry does not exist in majority of cases. In the case of two coplanar coils there is nosymmetry 1 $ 2. Indeed, we have

M21

=

2⇡2

c2R2

2

R1

N1

N2

= M12

.

which is not simple replacing 1 by 2 and vice versa.

253


9.4.2 Self-inductanceAny variation of the electric current in the coil leads to induction of the electromotive forcein the proper coil because the induced electric field acts on electrons inside the material of theconductor that coil is made of. The proportionality coefficient between magnetic flux and thecurrent in the circuit is called self-inductance coefficient

�(t) = c L I(t) (9.4.36)

which depends only on geometry of the coil. The induced electromotive force reads

E = �1

c

d�

dt= �L

dI

dt. (9.4.37)

9.4.2.1 Example

As a simple example we consider a toroidal coil having internal radius R1

and external radiusR

2

. The coil has N turns and its hight has value h. We assume that inside the coil there is aparamagnetic material with permeability µ. The magnetic strength can have only an azimuthalcomponent H = H(t, r)�. Taking into account the geometric symmetry of the problem wechoose integration paths as a family of circles coplanar with z = const. A total intensity if the

Figure 9.13: A toroidal inductor.

current reads

Itot =

8><>:0 for r < R

1

,

NI for R1

< r < R2

,

0 for r > R2

.

(9.4.38)

The Ampere’s lawHC H · dl = 4⇡

c Itot gives magnetic intensity inside the coil

H(t, r) =2N

c

I(t)

rfor R

1

< r < R2

(9.4.39)

254

9.4 Inductance

and H(t, r) = 0 outside of the coil. A total magnetic flux through N turns of the coil reads

� = N

Z h

0

dz

Z R2

R1

dr

✓µ2N

c

I(t)

r

◆= c

2µN2

c2h ln

R2

R1

�I(t).

The coefficient of self-inductance takes the value

L =

2µN2

c2h ln

R2

R1

. (9.4.40)

A general expression for coefficient of self-inductance of the coil having N turns is given interms of magnetic flux and current intensity

L =

N�

cI=

N

c

RS B · da

c4⇡

HC H · dl

which can be also written in terms two closed line integrals

L =

4⇡N

c2

HC A · dlHC H · dl

. (9.4.41)

Note, that capacity of capacitor was obtained as

C =

Q

V=

1

4⇡

HS D · daRLE · dl

.

9.4.2.2 Electric circuits with self-inductance

The simplest electric circuit containing inductance consists of a battery, a coil and a resistor. Anintensity of the current in such a circuit depends on time. An electromotive force is a sum of

Figure 9.14: A circuits containing an inductor.

255


electromotive force of the battery E0

and EMF due to electromagnetic induction Eind = �LdIdt .

From Ohms law E0

+ Eind = RI(t) we get

dI

dt+

R

LI =

E0

L. (9.4.42)

This equation has a solution I(t) = E0R +Ae�

R

L

t, where free constant A can be determined frominitial condition I(0) = 0 which gives A = �E

0

R

I(t) =E0

R

⇣1� e�

R

L

t⌘. (9.4.43)

Let us now consider an electric circuit with a stationary current I0

. At t = t1

the batteryis removed. The current flows only through resistor and the inductor. The current must satisfyequation

dI

dt+

R

LI = 0 (9.4.44)

which has solution I(t) = Ae�R

L

t where we determine A from condition I(t1

) = I0

givingA = I

0

eR

L

t1

I(t) = I0

e�R

L

(t�t1). (9.4.45)

9.5 Circuits containing inductance

9.5.1 Example: electric circuit in vicinity of linear, infinitely longconducting wire

We shall consider a linear wire of infinite length with slowly varying electric current I0

(t). Arectangular conductive loop with resistance R is localized in the plane � = const (in cylindricalcoordinates), see Fig.9.15. The magnetic field of quasi-stationary current in rectangular wire hasthe form

B =

2

c

I0

(t)

r� (9.5.46)

and the flux through the surface spanned on the rectangular circuit reads

� =

ZSB · da =

2I0

(t)

c

Z b

0

dz

Z r+a

r

dr0

r0=

2b

cln

r + a

r

�I0

(t). (9.5.47)

It follows that coefficient M := M12

= M21

takes the value

M(r) :=2b

c2ln

r + a

r. (9.5.48)

256


Figure 9.15: 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 rectan-gular circuit changes according to r = r(t). We shall denote the coefficient of self-inductanceof the rectangular coil by L1.

A total electromotive force is a sum E = Eself + Emutual = �LdIdt �

1

cd�dt which result in

equation

dI(t)

dt+

R

LI(t) = � 1

L

d

dt(M(r)I

0

(t)) ⌘ f(t) (9.5.49)

which RHS is an explicitly given function that depends on external mechanical force and externalelectromotive force.

A solution of this equation I(t) can be represented by the integral

I(t) = Ae�R

L

t+

Z 1

�1dt0D(t� t0)f(t0) (9.5.50)

where A is a free constant and D(t � t0) is fundamental solution i.e. it is a solution of distri-butional equation

(D0(t) + R

LD(t),'(t)) = (�(t),'(t)). (9.5.51)

A fundamental solution can be written in the form D(t) = ✓(t)Z(t) where Z 0+

RLZ = 0.

1L =

4N2

c2

h�2(a+ b) + 2

pa2 + b2 � a ln(a+

pa2+b2

b )� b ln( b+pa2+b2

a ) + a ln 2ad + b ln 2b

d

iwhere d is a radius of the wire and N stands for number of turns.

257


Plugging this solution to equation (9.5.51) we find

�(D,'0) +

R

L(D,') = (�,')

�Z 1

0

Z(t)'0(t)dt+

R

L

Z 1

0

Z(t)'(t)dt = '(0)

Z(0)'(0) +

Z 1

0

✓Z 0

+

R

LZ

◆| {z }

0

'(t)dt = '(0)

so it must be Z(0) = 1. It follows that

D(t) = ✓(t)e�R

L

t. (9.5.52)

Let us consider that there is no current in a rectangular circuit at t = 0 i.e. I(0) = 0

A = �Z 1

�1dt0D(�t0)f(t0) (9.5.53)

Case: dMdt = 0, dI0

dt 6= 0.We shall consider the case when the coil remains at rest and the current intensity I

0

(t) is explcitfunction of time. An induced current satisfies equation

dI

dt+

R

LI = �M

L

dI0

dt⌘ f(t). (9.5.54)

Let us consider an electric current I0

(t) in a form of an impulse with a finite time of duration

I0

(t) =i0

2

✓1� cos

✓2⇡

Tt

◆◆for 0 < t < T (9.5.55)

and I0

(t) = 0 for t < 0 and t > T , where i0

= const. An electric current intensity I(t) in therectangular wire reads

I(t) = Ae�R

L

t+

8><>:I(1)n for t < 0,

I(2)n for 0 < t < T,

I(3)n for t > T.

(9.5.56)

The function f(t) has the form

f(t) = f0

sin

✓2⇡

Tt

◆(✓(t)� ✓(t� T )) where f

0

:= �M

L

⇡

Ti0

. (9.5.57)

258


A non-homogeneous part of the solution is of the form

In(t) =

Z 1

�1dt0D(t� t0)f(t0)

=

Z 1

�1dt0✓(t� t0)e�

R

L

(t�t0)f(t0)

= e�R

L

t

Z t

�1dt0e

R

L

t0f(t0)

= f0

e�R

L

t

Z t

�1dt0e

R

L

t0sin

✓2⇡

Tt0◆(✓(t0)� ✓(t0 � T ))

= f0

e�R

L

t 1

2i

Z t

�1dt0

⇣e(

R

L

+i 2⇡T

)t0 � e(R

L

�i 2⇡T

)t0⌘(✓(t0)� ✓(t0 � T )) (9.5.58)

The integral I(1)n (t) = 0 for t < 0 because ✓(t0)� ✓(t0 � T ) = 0 for t0 < t.

For 0 < t < T we have ✓(t0)� ✓(t0 � T ) = 1 in the interval 0 < t0 < t so

I(2)n (t) = f0

e�R

L

t 1

2i

Z t

0

dt0⇣e(

R

L

+i 2⇡T

)t0 � e(R

L

�i 2⇡T

)t0⌘

= f0

e�R

L

t 1

2i

"e(

R

L

+i 2⇡T

)t � 1

RL +

2⇡T i

� e(R

L

�i 2⇡T

)t � 1

RL � 2⇡

T i

#

=

f0

e�R

L

t�RL

�2

+

�2⇡T

�2

1

2i

✓R

L� 2⇡

Ti

◆⇣e(

R

L

+i 2⇡T

)t � 1

⌘�

✓R

L+

2⇡

Ti

◆⇣e(

R

L

�i 2⇡T

)t � 1

⌘�

=

f0

e�R

L

t�RL

�2

+

�2⇡T

�2

R

Le

R

L

tsin

✓2⇡

Tt

◆� 2⇡

Te

R

L

tcos

✓2⇡

Tt

◆+

2⇡

T

�.

The solution of non-homogeneous equation reads

I(2)n (t) =f0�

RL

�2

+

�2⇡T

�2

2⇡

T

✓e�

R

L

t � cos

✓2⇡

Tt

◆◆+

R

Lsin

✓2⇡

Tt

◆�. (9.5.59)

At t = T this current takes the value

I(2)n (T ) =f0�

RL

�2

+

�2⇡T

�2

2⇡

T

⇣e�

R

L

T � 1

⌘. (9.5.60)

Finally, we compute the integral I(3)n defined in the interval t > T . There is no contribution

259


from the integral on the interval t0 > t so

I(3)n (t) = f0

e�R

L

t 1

2i

Z T

0

dt0⇣e(

R

L

+i 2⇡T

)t0 � e(R

L

�i 2⇡T

)t0⌘

=

f0

e�R

L

t�RL

�2

+

�2⇡T

�2

R

Le

R

L

Tsin

✓2⇡

TT

◆� 2⇡

Te

R

L

Tcos

✓2⇡

TT

◆+

2⇡

T

�

=

f0

eR

L

(T�t)�RL

�2

+

�2⇡T

�2

2⇡

T

⇣e�

R

L

T � 1

⌘.

It gives

I(3)n (t) = I(2)n (T ) eR

L

(T�t). (9.5.61)

This result shows that there is only an exponentially decreasing component of the current fort > T .

Case: dI0dt = 0, dM

dt 6= 0.An induced current obes equation

dI

dt+

R

LI = �I

0

L

dM

dt⌘ f(t) (9.5.62)

where dMdt =

dMdr

drdt ⌘

dMdr v(t). Derivative of M with respect to r reads

dM

dr=

2b

c2

1

r + a� 1

r

�. (9.5.63)

A solution of non-homogeneous equation is given by

In(t) =

Z 1

�1D(t� t0)f(t0) (9.5.64)

wheref(t) = �2I

0

b

c2L

1

r(t) + a� 1

r(t)

�v(t). (9.5.65)

It gives

In(t) =

Z t

�1dt0 e�

R

L

(t�t0)f(t0)

= �2I0

b

c2Le�

R

L

t

Z t

�1dt0 e

R

L

t0v(t0)

1

r(t0) + a� 1

r(t0)

�then

In(t) = �2I0

b

c2Le�

R

L

tW (r(t)) (9.5.66)

260


where

W (r(t)) =

Z t

�1dt0 e

R

L

t0v(t0)

1

r(t0) + a� 1

r(t0)

�. (9.5.67)

In order to simplify our considerations we shall study an uniform motion given by r = r0

+ vt

where v = const. We shall assume that a motion of the circuit was initiated at t = 0. It followsthat t = r�r0

v . After change of the variable of integration one gets

W (r) =

Z r

r0

dr0eR

Lv

(r0�r0)

1

r0 + a� 1

r0

�= e�

R

Lv

(r0+a)

Z r

r0

dr0e

R

Lv

(r0+a)

r0 + a� e�

R

Lv

r0

Z r

r0

dr0e

R

Lv

r0

r0. (9.5.68)

We define new variable u := � RLv (r

0+ a) for the first integral, so dr0

r0+a =

duu . By analogy, we

take u := � RLv r

0 for the second integral. The integral W (r) reads

W (r) = e�R

Lv

(r0+a)

Z � R

Lv

(r+a)

� R

Lv

(r0+a)du

e�u

u� e�

R

Lv

r0

Z � R

Lv

r

� R

Lv

r0

due�u

u

= e�R

Lv

(r0+a)

"Z 1

� R

Lv

(r0+a)du

e�u

u�

Z 1

� R

Lv

(r+a)du

e�u

u

#

� e�R

Lv

r0

"Z 1

� R

Lv

r0

due�u

u�

Z 1

� R

Lv

rdu

e�u

u

#. (9.5.69)

Each integral can be cast in the form of exponential integral

Ei(z) := �Z 1

�zdu

e�u

u(9.5.70)

what gives

W (r) = e�R

Lv

(r0+a)

Ei

✓R

Lv(r + a)

◆� Ei

✓R

Lv(r

0

+ a)

◆�� e�

R

Lv

r0

Ei

✓R

Lvr

◆� Ei

✓R

Lvr0

◆�. (9.5.71)

9.5.2 Coupled circuits

9.5.2.1 Inductors in series

Two inductors in series can be coupled in two non-equivalent ways. Depending on the mutualorientation of the inductors a magnetic field of the second/first inductor can reinforce or diminishthe magnetic flux in the first/second inductor.

261


Figure 9.16: Two inductors in series.

When fluxes sum constructively then induced electromotive forces in coils of inductor a =

1, 2 read

Ea = �LadI

dt�M

dI

dt(9.5.72)

where M ⌘ M12

= M21

and I ⌘ I1

= I2

because circuits are in series. Another possibilityleads to subtracting of magnetic fluxes giving

Ea = �LadI

dt+M

dI

dt. (9.5.73)

Assuming that conductors have resistances R1

and R2

and there is another source of electro-

(a) (b)

Figure 9.17: (a) Positive and (b) negative magnetic coupling of inductors.

motive force (a battery, etc) E0

we get equation E0

+ E1

+ E2

= R1

I + R2

I which results in

(L1

+ L2

± 2M)

dI

dt+ (R

1

+R2

)I = E0

. (9.5.74)

262


The circuit can be replaced by an equivalent circuit with total resistance Reff = R1

+ R2

andtotal inductance Leff = L

1

+ L2

± 2M . When mutual inductance is very small (inductors arefar from each other) a total inductance is a sum of self-inductances Leff = L

1

+L2

. In the caseof connection in series the circuit can be reduced to a circuit containing a single inductor and asingle resistor.

9.5.2.2 Inductors in parallel

Similarly to connection in series there are two ways how magnetic coupling between conductorscan be realized. It is taken into account by the choice of the sign in expression ±M . Whencircuits are connected in parallel, then electromotive forces of induction associated with eachinductor read

E1

= �L1

dI1

dt� (±M)

dI2

dtE2

= �L2

dI2

dt� (±M)

dI1

dt(9.5.75)

We shall assume that there is a source of electromotive force represented by E0

. The system of

Figure 9.18: Two inductors in parallel.

circuits is described by two equations E0

+ E1

= R1

I1

and E0

+ E2

= R2

I2

, which can be castin the form "

L1

±M

±M L2

#| {z }

A

"dI1dtdI2dt

#+

"R

1

0

0 R2

#"I1

I2

#=

"E0

E0

#(9.5.76)

where

detA = L1

L2

�M2 6= 0. (9.5.77)

263


Multiplying by an inverse matrix

A�1

=

1

detA

"L2

⌥M

⌥M L1

#(9.5.78)

one gets"dI1dtdI2dt

#| {z }

dI

dt

� 1

detA

"�L

2

R1

±MR2

±MR1

�L1

R2

#| {z }

C

"I1

I2

#| {z }

I

=

E0

detA

"L2

⌥M

L1

⌥M

#. (9.5.79)

Before studying the complete solution we shall consider an idealized case R1

= R2

⇡ 0.In such a case equations depend only on derivatives of currents ˙I

1

and ˙I2"

dI1dtdI2dt

#=

E0

detA

"L2

⌥M

L1

⌥M

#. (9.5.80)

Taking a sum of last two equations we find that Itot = I1

+ I2

obeys the equation

dItotdt

=

L1

+ L2

⌥ 2M

L1

L2

�M2

E0

.

When magnetic coupling between circuits is very small M ⇡ 0, then

dItotdt

=

✓1

L1

+

1

L2

◆E0

) 1

Leff=

1

L1

+

1

L2

.

A general solution of the original problem (without aproximations) is the sum of generalsolution 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 solu-tion of non-homogeneous equations is constant as well. From (9.5.76) we get that the solutionis given by expression "

Ip1

Ip2

#=

"E0R1E0R2

#(9.5.81)

We have to find general solution of the set of homogeneous equations

dI

dt� CI = 0 (9.5.82)

where C is a matrix defined in (9.5.79). We shall assume the following form of solution

I(t) = v(1)e�1t+ v(2)e�2t

where v(a) =

"v(a)1

v(a)2

#a = 1, 2. (9.5.83)

264


Plugging this ansatz to equation (9.5.82) we get an algebraic equation

(�1

v(1) � Cv(1))e�1t+ (�

2

v(2) � Cv(2))e�2t= 0.

which is satisfied for any t if(C � �a1)v

(a)= 0. (9.5.84)

It follows that parameters �a and v(a) are, respectively, eigenvalues and eigenvectors of matrixC. Eigenvalues are determined from equation det(C � �1) = 0 with

C � �1 =

1

detA

"�L

2

R1

� � detA ±MR2

±MR1

�L1

R2

� � detA

#

which gives(detA)�2

+ (L2

R1

+ L1

R2

)�+R1

R2

= 0. (9.5.85)

We find� = (L

2

R1

� L1

R2

)

2

+ 4M2R1

R2

(9.5.86)

so eigenvalues of C read

�1

=

p�� (L

2

R1

+ L1

R2

)

2(L1

L2

�M2

)

�2

= �p�+ (L

2

R1

+ L1

R2

)

2(L1

L2

�M2

)

(9.5.87)

so

�1

detA =

1

2

⇣+

p�� L

1

R2

� L2

R1

⌘,

�2

detA =

1

2

⇣�p�� L

1

R2

� L2

R1

⌘.

Plugging this eigenvalues to (9.5.84) and defining where

�1

:= L2

R1

+ �1

detA = �L1

R2

� �2

detA

=

1

2

hp�� (L

1

R2

� L2

R1

)

i(9.5.88)

and

�2

:= L1

R2

+ �1

detA = �L2

R1

� �2

detA

=

1

2

hp�+ (L

1

R2

� L2

R1

)

i(9.5.89)

we get "��

1

±MR2

±MR1

��2

#"v(1)1

v(1)2

#=

"0

0

#(9.5.90)

265


and "�2

±MR2

±MR1

�1

#"v(2)1

v(2)2

#=

"0

0

#(9.5.91)

It follows from (9.5.86) that the product of �1

and �2

reads

�1

�2

=

1

4

⇥�� (L

1

R2

� L2

R1

)

2

⇤= M2R

1

R2

. (9.5.92)

As expected, each set of equations (9.5.90) and (9.5.91) is linearly dependent

det

"��

1

±MR2

±MR1

��2

#= 0 = det

"�2

±MR2

±MR1

�1

#.

From the first equation of (9.5.90) we have ��1

v(1)1

±MR2

v(1)2

= 0. Then, choosing v(1)1

:= ↵1

we get v(1)2

= ± �1MR2

↵1

. Similarly, the second equation in (9.5.91) gives ±MR1

v(2)1

+�1

v(2)2

=

0. Taking v(2)2

:= ↵2

we get v(2)1

= ⌥ �1MR1

. The eigenvectors read

v(1) = ↵1

"1

±�1MR2

#v(2) = ↵

2

"⌥�1MR1

1

#(9.5.93)

where sign “±” refers to cumulative (+) or anti-cumulative (�) magnetic coupling of conduc-tors. Note, that another form of eigenvectors is possible by redefinition of constants ↵

1

and ↵2

.Taking ↵

1

= ± �2MR1

↵1

and ↵2

= ⌥ �2MR2

↵1

we get

v(1) = ↵1

"±�2MR1

1

#v(2) = ↵

2

"1

⌥�2MR2

#. (9.5.94)

Eigenvectors (9.5.93) or (9.5.94) are not orthogonal because an associated matrix is not sym-metric. A solution of the problem has general form"

I1

(t)

I2

(t)

#= ↵

1

"1

±�1MR2

#e�1t

+ ↵2

"⌥�1MR1

1

#e�2t

+

"E0R1E0R2

#(9.5.95)

where ↵1

and ↵2

are free constants. These constants can be fixed by initial condition. We shallassume following initial conditions I

1

(0) = I01

and I2

(0) = I02

. The solution given by(9.5.95)at t = 0 takes the form "

1

⌥�1MR1

±�1MR2

1

#| {z }

U

"↵1

↵2

#=

"I01

� E0R1

I02

� E0R2

#(9.5.96)

266


where detU = 1 +

�1�2

. Multiplying the equation (9.5.96) by

U�1

=

�2

�1

+ �2

"1

±�1MR1

⌥�1MR2

1

#we find that

↵1

=

�2

�1

+ �2

✓I01

� E0

R1

◆± �

1

MR1

✓I02

� E0

R2

◆�, (9.5.97)

↵2

=

�2

�1

+ �2

⌥ �

1

MR2

✓I01

� E0

R1

◆+

✓I02

� E0

R2

◆�. (9.5.98)

Limit of weak magnetic coupling M ! 0. The leading terms of eigenvalues read

�1/2 =

1

2

�L1

R2

� L2

R1

± |L1

R2

� L2

R1

|q1 +

4R1R2M2

(L1R2�L2R1)2

L1

L2

�M2

⇡ 1

2

�R

2

L2

� R1

L1

±��R2

L2

� R1

L1

��+ . . . (9.5.99)

and so

�1/2 =

1

2

"|L

1

R2

� L2

R1

|

s1 +

4R1

R2

M2

(L1

R2

� L2

R1

)

2

⌥ (L1

R2

� L2

R1

)

#

⇡ 1

2

|L

1

R2

� L2

R1

|⌥ (L1

R2

� L2

R1

) +

2R1

R2

|L1

R2

� L2

R1

|M2

+ O(M4

)

�.

It follows from this expression that for L1

R2

> L2

R1

we have�1

M! 0

�2

M! 1

in the limit M ! 0. On the other hand, for L1

R2

< L2

R1

one gets�1

M! 1 �

2

M! 0.

Clearly, the eigenvectors v(1) and v(2) are more convenient for the first case whereas the choicev(1) and v(2) is better for the second case. The solutions take the form"

I1

(t)

I2

(t)

#= ↵

1

"1

0

#e�L1

R1t+ ↵

2

"0

1

#e�L2

R2t+

"E0R1E0R2

#(9.5.100)

for L1

R2

> L2

R1

and"I1

(t)

I2

(t)

#= ↵

1

"0

1

#e�L2

R2t+ ↵

2

"1

0

#e�L1

R1t+

"E0R1E0R2

#. (9.5.101)

for L1

R2

< L2

R1

. The solutions are totally equivalent. The only difference are free constants↵1

$ ↵2

, ↵2

$ ↵1

.

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9 electromagnetic induction - paweł klimas9 electromagnetic induction 9.1 faraday’s experiment a...

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