The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 27 of 39

502. Let us suppose that we have a system of circuits, which we shall denote by the numbers 1, 2, .... Let us suppose that when a unit current flows through 1, all the other circuits being devoid of currents, a magnetic field is produced such that the numbers of tubes of induction which cross circuits 1, 2, 3, ... are

■"11 J -"12 > -"13 > ••••

501-503] Coefficients of Induction 443

Similarly, when a unit current flows through 2, let the numbers of tubes of induction be

-"21 > -"22 > -"23 ) • • • •

The theorem of § 446 shews at once that

'12

cose

L21 = \ — dsds, etc (428).

r

If currents i1, i2, ... flow through the circuits simultaneously, and if the

numbers of tubes of induction which cut the circuits are Nu N2, N3, ..., we

have

Nx = Luix + L12i2 + L13i3 +... }

r ■•• ~hJiU )•

N2 = L2lix + Z^ + ^23*3 + • •■> etc. J The energy of the system of currents is

= iXh (Zut'i + L12i2 +...),

= i £n Vs + £12^2 + %L22i22 + (430).

Coefficients of Induction.

503. The coefficient Ln is commonly called the coefficient of self-induc- tion (or, more briefly, the self-inductance) of circuit 1, while L12 is called the coefficient of mutual induction of the two circuits 1 and 2. The value of L12 for any pair of circuits can be calculated from formula (428).

As an example, consider the important case of two circular wires, radii a, a' in parallel planes, the line joining their centres being perpendicular to the planes and of length d, b. Formula (428) gives

^f2* aa' cos (d-e')ddde'

JV2

o J o [a2 + a'2 + b2- 2aa' cos (6 - 6')f

= 2tt

2n aa' cos ijr dty

o [a2 + a'2 + b2 - 2aa cos \jr]% '

Aiflff

Put °°=(« + ay + b» -(—+)

and we readily find

t t t >\k fhlT 2sin2(£-l ,, Ll2 = ^ir (aa )* c = deb

Jo (l-c2sin2<^ Y = 4>ir(aa'^l(^-c)K(o)-^E(c) ,

where K (c), E (c) are the complete elliptic functions to modulus c.

When the circles nearly coincide, b is small and a and a' are nearly equal. Thus c is nearly equal to unity, and E(c) approximates to unity. Put

c' - (1 - c2)K

444 The Magnetic Field produced by Electric Currents [oh. xiii so that c' is small, then

Kic)^' £— r-/1* & j,

-'o (l-c2sin2</>)* Jo (cos2</> + c'2sin2(/>)i

of which the approximate value is found to be log (4/c').

If r is the nearest distance apart of the two circles, we have, when r is small, c = r/2a, so that

K(c) = \og(8a/r),

L12 = 4<7ra(\og—-2\ (430a).

and

504. It might be expected that we could obtain the value of Ln in any problem by making the two circuits 1 and 2 coincide, but this proves not to be the case; the value of the integral in equation (428), where the integral is taken twice round the same circuit, is always infinite. As an instance, we may notice that on putting r = 0 in the formula just obtained, we find L12 = oo .

We can readily see why this must be. When there is only one current flowing, we have

iZfch" = g£ (//(«* + P2 + Y2) dxdydz,

each side of this equation representing the energy of the current. Near to

the wire, at a small distance r from it, the magnetic force is 2i\r so that

a2 + /S2 + 72 = 4di/r2. Thus the energy contained within a thin ring formed of

coaxal cylinders of radii ru r2, bent so as to follow the wire conveying the

current, will be

/* fff 419

„ rdrddds, ottJ J J r2

where the integration with respect to r is from rx to r2, that with respect to

6 is from 0 to 27r, and that with respect to s is along the wire. Integrating,

we find energy

fii2 log {r^n)

per unit length, and on taking rx = 0, the energy is seen to be infinite.

Suppose that the wire has a circular cross-section of radius a, and that the current is uniformly distributed over this cross-section. A circle of radius r inside the wire will enclose a current ir2/a2, so that the magnetic force at distance r from the centre will be 2ir/a2, and

(aj

On integrating this from r = 0 to r = a we find that there is magnetic energy inside the wire of amount £/A'2 per unit length, where p! is the magnetic permeability of the material of the wire. Hence the total energy per unit length inside a cylinder of radius r2 enclosing the wire is

^'t'2 + /u'2log(r2/a) (4306).

503-505] Coefficients of Induction 445

Even when a is finite this still becomes infinite when r2 is made infinite — i.e. when the magnetic field extends to infinity. Thus the self-induction per unit length of a straight wire in free space is infinite except when the magnetic field is limited by the presence of other conductors.

Suppose that the return current is carried by .a concentric cylinder of radius b surrounding the wire. The total flow of current through a circle of radius greater than b is zero, so that there will be no magnetic force outside the cylindrical conductor, and the magnetic field will be limited by the cylinder r = b. The energy per unit length is now given by formula (4306) with r2 put equal to b, so that the coefficient of self-induction per unit length is

L=%fi+2filog(b/a) (430c),

and this is finite for all finite values of b and a.

505. The energy of the magnetic field produced by a current i in a wire will always be the sum of the energies of the magnetic field in the wire and of the magnetic field outside the wire. If the current is uniformly distributed in the wire, the former energy will always be ^fi'i2 as in § 504. Thus L, the self-induction of a wire of length I, will always be of the form

L = Wl + L' (430d),

where the term \xl arises from the field inside the wire, and L' arises from the field outside the wire.

When the circuit lies entirely in one plane and the radius of cross-section of the wire is small a simple value can be obtained for L'. Let 8 denote the curve formed by the centres of the cross-sections of the wire, and let S' denote the curve formed by the inner edge of the wire in the plane in which the circuit lies. Then it will be easily verified that the magnetic force at any point inside S' is the same as if the whole current % flowed along the curve 8. Hence the number of tubes of induction which flow through S' when the current flows in the wire is the same as if a current i flowed in 8, and so is equal to i times L\2 where L'K is the coefficient of mutual induction between 8 and 8'. Thus in formula (430 d), L' will be the coefficient of mutual induc- tion between the circuits S and S'.

As an example, let us find the coefficient of self-induction in a wire of length lira whose cross-section is a circle of radius r, bent into a circle of radius a. The curve S is a circle of radius a, the curve 8' is a concentric circle of radius a — r. By formula (430 a),

Z' = 4™ (log — -2

8a

so that L = ira/jL + 47ra f log 2 j .

446 The Magnetic Field produced by Electric Currents [ch. xiii

As a second example, let us find the coefficient of self-induction of a rect- angular circuit of sides a, b made of wire of circular cross-section of radius r. In this case the circuit S will be a rectangle of sides a, b, while the circuit S' is a coplanar concentric rectangle of sides a — r,b — r. We evaluate L' the coefficient of mutual induction of S and S' from formula (428). There is no contribution from pairs of elements on sides perpendicular to one another, since for these cose = 0; the whole value of L' is contributed by parallel pairs of elements.

For two parallel lines of lengths I, V at distance h apart, we find

"dsds' [& ft1' dxdx

hi J -V'[(ac'-ocY + h*$

hi -hi

• i 00 ""~ Jb

smn-1 -

dx af=-W

h

= (I + V) sinh-1 l-±l-{l- V) sinh-1 ^ - [4A2 + (l + Ijf

• [4A*+(J_Z')»]1.

On making I — I' small, and replacing sinh-1 by its logarithmic value, this

becomes

21 log * + (ft + A')* _ 2 (Ji + &■)* + 2A.

By repeated use of this formula we find

U = - 8 (a + b) + 8 (a2 + 62)^ - 4a log [a + (a2 + 62)*]

• 46 log [6 + (a2 + t2)-] + 4 (a + 6) log — ,

and the coefficient of self-induction is now given by

L = (a + b)p, + L'.

505 a. Formula (430 c), expressing the self-induction per unit length of a

circular wire with a concentric return, can be put in the form L = h/uf + L',

where

L' = 2fi log (b/a).

If K is the electrostatic capacity per unit length of the condenser formed by the wire and its surrounding cylinder, we have, from § 82,

K

K =

2 log (b/a) '

where k is the inductive capacity of the insulating material surrounding the wire. Thus

L'=^ (430*).

It is not a mere accident that this simple relation holds. Suppose we solve the electrostatic problem by the method of conjugate functions (§ 312).

505, 505 a] Examples 447

The appropriate transformation is readily found to be (cf. § 318)

U+iV= Cons. + 2 log r + 1i0,

where x = r cos 6, y = r sin 6. In this transformation U may be taken to be the electrostatic potential due to unit charge per unit length, and V will clearly be the magnetic potential due to unit current. It follows at once that the value of X2 + Y2 at any point when there is unit charge per unit length is the same as the value of a2 + /32 at the same point when there is unit current flowing, and relation (430 e) is at once seen to be true.

The argument can be applied equally well to any conjugate-function trans- formation whatever. Thus relation (430 a) is seen to be universally true for any straight conductor accompanied by a parallel return.

EXAMPLES.

1. A  current  i  flows  in  a  very  long  straight  wire.     Find  the  forces  and  couples  it 
   

exerts upon a small magnet.

Shew that if the centre of the small magnet is fixed at a distance c from the wire, it has two free small oscillations about its position of equilibrium, of equal period

'-A/

where Mk2 is the moment of inertia, and /u the magnetic moment, of the magnet.

2. Two parallel straight infinite wires convey equal currents of strength i in opposite directions, their distance apart being 2a. A magnetic particle of strength p and moment of inertia ink"1 is free to turn about a pivot at its centre, distant c from each of the wires. Shew that the time of a small oscillation is that of a pendulum of length I given by

4:ialjx = mgk'icl.

3. Two equal magnetic poles are observed to repel each other with a force of 40 dynes when at a decimetre apart. A current is then sent through 100 metres of thin wire wound into a circular ring eight decimetres in diameter and the force on one of the poles placed at the centre is 25 dynes. Find the strength of the current in amperes.

4. Regarding the earth as a uniformly and rigidly magnetised sphere of radius a,

and denoting the intensity of the magnetic field on the equator by H, shew that a wire

surrounding the earth along the parallel of south latitude A, and carrying a current i

from west to east, would experience a resultant force towards the south pole of the

heavens of amount

QnaiH sin X cos2 X.

5. Shew that at any point along a line of force, the vector potential due to a current in a circle is inversely proportional to the distance between the centre of the circle and the foot of the perpendicular from the point on to the plane of the circle. Hence trace the lines of constant vector potential.

6. A current i flows in a circuit in the shape of an ellipse of area A and length I. Shew that the force at the centre is nil/ A.

448 The Magnetic Field produced by Electric Currents [ch. xiii

7. A current i flows round a circle of radius a, and a current i' flows in a very long straight wire in the same plane. Shew that the mutual attraction is 47m' (sec a - 1), where a is the angle subtended by the circle at the nearest point of the straight wire.

8. If, in the last question, the circle is placed perpendicular to the straight wire with its centre at distance c from it, shew that there is a couple tending to set the two wires in the same plane, of moment 2irii'a2lc or 2nii'c, according as c> or <a.

9. A long straight current intersects at right angles a diameter of a circular current, and the plane of the circle makes an acute angle a with the plane through this diameter and the straight current. Shew that the coefficient of mutual induction is

4it {c sec a -(c2 sec2 a -a2) 2} or 47rctan ( - - - ) ,

according as the straight current passes within or without the circle, a being the radius of the circle, and c the distance of the straight current from its centre.

10. Prove that the coefficient of mutual induction between a pair of infinitely long straight wires and a circular one of radius a in the same plane and with its centre at a distance b (> a) from each of the straight wires, is

87r(6-V62-a2).

11. A circuit contains a straight wire of length 2a conveying a current. A second straight wire, infinite in both directions, makes an angle a with the first, and their common perpendicular is of length c and meets the first wire in its middle point. Prove that the additional electromagnetic forces on the first straight wire, due to the presence of a current in the second wire, constitute a wrench of pitch

„ / . , a sin a\ / . _ , a sin a 2 a sin a - c tan ~ 1 / sin 2a tan ~ 1 .

/ . _ asincA / ( a sin a - c tan 1 J / s

12. Two circular wires of radii a, b have a common centre, and are free to turn on an insulating axis which is a diameter of both. Shew that when the wires carry currents i, i', a couple of magnitude

*?('-5)

is required to hold them with their planes at right angles, it being assumed that b\a is so small that its fifth power may be neglected.

13. Two circular circuits are in planes at right angles to the line joining their centres. Shew that the coefficient of induction

= 27r(a2-c2) P

cos 26 d6

/a2 sin2 6 + c2 cos2 6

where a, c are the longest and shortest lines which can be drawn from one circuit to the other. Find the force between the circuits.

14. Two currents i, V flow round two squares each of side a, placed with their edges parallel to one another and at right angles to the distance c between their centres. Shew that they attract with a force

,j/2a2 + c2 , a2 + 2c2

ow

.., |W2a2 + c2 , a2 + 2c2 ) 1 a2 + c- cVa2 + c2J

V«2 + <

15. A current i flows in a rectangular circuit whose sides are of lengths 2a, 2b, and the circuit is free to rotate about an axis through its centre parallel to the sides of length 2a. Another current i' flows in a long straight wire parallel to the axis and at a distance

Examples 449

d from it. Prove that the couple required to keep the plane of the rectangle inclined at an angle <£ to the plane through its centre and the straight current is

%ii'abd(b2 + cP)sm(f> bi + di-2b2d2cos2cf>'

16. Two circular wires lie with their planes parallel on the same sphere, and carry opposite currents inversely proportional to the areas of the circuits. A small magnet has its centre fixed at the centre of the sphere, and moves freely ahout it. Shew that it will be in equilibrium when its axis either is at right angles to the planes of the circuits, or makes an angle tan-1^ with them.

17. An infinitely long straight wire conveys a current and lies in front of and parallel to an infinite block of soft iron bounded by a plane face. Find the magnetic potential at all points, and the force which tends to displace the wire.

18. A small sphere of radius b is placed in the neighbourhood of a circuit, which when carrying a current of unit strength would produce magnetic force H at the point where the centre of the sphere is placed. Shew that, if k is the coefficient of induced magnetization for the sphere, the presence of the sphere increases the coefficient of self- induction of the wire by an amount approximately equal to

87rb(3 + 27rK)E2

19. A circular wire of radius a is concentric with a spherical shell of soft iron of radii 6 and c If a steady unit current flow round the wire, shew that the presence of the iron increases the number of lines of induction through the wire by

2ff2a4 (c3 _ &3) (^ _ !) (M + 2)

63 {(2M + 1) Qx + 2) Cs - 2 0* - 1)2 62} approximately, where a is small compared with 6 and c.

20. A right circular cylindrical cavity is made in an infinite mass of iron of perme- ability n In this cavity a wire runs parallel to the axis of the cylinder carrying a steady current of strength /. Prove that the wire is attracted towards the nearest part of the surface of the cavity with a force per unit length equal to

2(,*-l)/2

0*+i)rf '

where d is the distance of the wire from its electrostatic image in the cylinder.

21. A steady current C flows along one wire and back along another one, inside a long cylindrical tube of soft iron of permeability p, whose internal and external radii are ax and a2, the wires being parallel to the axis of the cylinder and at equal distance a on opposite sides of it. Shew that the magnetic potential outside the tube will be

F=^ sin 6+ -| sin 38+ ^sin50 + ...,

Hence shew that a tube of soft iron, of 150 cm. radius and 5 cm. thickness, for which the effective value of n is 1200 c.G.s., will reduce the magnetic field at a distance, due to the current, to less than one-twentieth of its natural strength.

9Q

450 The Magnetic Field produced by Electric Currents [oh. xtii

22. A wire is wound in a spiral of angle a on the surface of an insulating cylinder of radius a, so that it makes n complete turns on the cylinder. A current i flows through the wire. Prove that the resultant magnetic force at the centre of the cylinder is

27rm

a(l + 7r2»2tan2a)i along the axis.

23. A current of strength i flows along an infinitely long straight wire, and returns in a parallel wire. These wires are insulated and touch along generators the surface of an infinite uniform circular cylinder of material whose coefficient of induction is h. Prove that the cylinder becomes magnetized as a lamellar magnet whose strength is 2irkiJ{l + 2Trk).

24. A fine wire covered with insulating material is wound in the form of a circular disc, the ends being at the centre and the circumference. A current is sent through the wire such that / is the quantity of electricity that flows per unit time across unit length of any radius of the disc. Shew that the magnetic force at any point on the axis of the disc is

27r/{cosh-1 (sec a) — sin a},

where a is the angle subtended at the point by any radius of the disc.

25. Coils of wire in the form of circles of latitude are wound upon a sphere and produce a magnetic potential ArnPn at internal points when a current is sent through them. Find the mode of winding and the potential at external points.

26. A tangent galvanometer is to have five turns of copper wire, and is to be made so that the tangent of the angle of deflection is to be equal to the number of amperes flowing in the coil. If the earth's horizontal force is -18 dynes, shew that the radius of the coil must be about 17 "45 cms.

27. A given current sent through a tangent galvanometer deflects the magnet through an angle 6. The plane of the coil is slowly rotated round the vertical axis through the centre of the magnet. Prove that if 6 > \iv, the magnet will describe complete revolu- tions, but if 6< j7r, the magnot will oscillate through an angle sin-1(tan#) on each side of the meridian.

28. Prove that, if a slight error is made in reading the angle of deflection of a tangent galvanometer, the percentage error in the deduced value of the current is a minimum if the angle of deflection is j7r.

29. The circumference of a sine galvanometer is 1 metre : the earth's horizontal magnetic force is "18 c.G.s. units. Shew that the greatest current which can be measured by the galvanometer is 4-56 amperes approximately.

30. The poles of a battery (of electromotive force 2-9 volts and internal resistance 4 ohms) are joined to those of a tangent galvanometer whose coil has 20 turns of wire and is of mean radius 10 cms. : shew that the deflection of the galvanometer is approximately 45°. The horizontal intensity of the earth's magnetic force is 1*8 and the resistance of the galvanometer is 16 ohms.

31. A tangent galvanometer is incorrectly fixed, so that equal and opposite currents give angular readings a and /3 measured in the same sense. Shew that the plane of the coil, supposed vertical, makes an angle e with its proper position such that

2 tan e = tan a + tan /3.

32. If there be an error a in the determination of the magnetic meridian, find the true strength of a current which is i as ascertained by means of a sine galvanometer.

Examples 451

33. In a tangent galvanometer, the sensibility is measured by the ratio of the incre- ment of deflection to the increment of current, estimated per unit current. Shew that

the galvanometer will be most sensitive when the deflection is — , and that in measuring

the current given by a generator whose electromotive force is E, and internal resistance B, the galvanometer will be most sensitive if there be placed across the terminals a shunt of resistance

BRr E-H(R+rY

where r is the resistance of the galvanometer, and H is the constant of the instrument. What is the meaning of the result if the denominator vanishes or is negative ?

34. A tangent galvanometer consists of two equal circles of radius 3 cms. placed on a common axis 8 cms. apart. A steady current sent in opposite directions through the two circles deflects a small needle placed on the axis midway between the two circles through an angle a. Shew that if the earth's horizontal magnetic force be R in c.G.s. units, then the strength of the current in c.G.s. units will be 125.£ftana/367r.

35. A galvanometer coil of n turns is in the form of an anchor-ring described by the revolution of a circle of radius b about an axis in its plane distant a from its centre. Shew that the constant of the galvanometer

~ a J

K

cn2 wdn2 u du (£ = &/«)

■■(8nj3k2a)[{l +k2) E-(l-tf) K].

29—2

CHAPTEE XIV

INDUCTION OF CURRENTS IN LINEAR CIRCUITS

Physical Principles.

506. It has been seen that, on moving a magnetic pole about in the presence of electric currents, there is a certain amount of work done on the pole by the forces of the field. If the conservation of energy is to be true of a field of this kind, the work done on the magnetic pole must be represented by the disappearance of an equal amount of energy in some other part of the field. If all the currents in the field remain steady, there is only one store of energy from which this amount of work can be drawn, namely the energy of the batteries which maintain the currents, so that these batteries must, during the motion of the magnetic poles, give up more than sufficient energy to maintain the currents, the excess amount of energy representing work performed on the poles. Or again, if the batteries supply energy at a uniform rate, part of this energy must be used in performing work on the moving poles, so that the currents maintained in the circuits will be less than they would be if the moving poles were at rest.

Let us suppose that we have an imaginary arrangement by which addi- tional electromotive forces can be inserted into, or removed from, each circuit as required, and let us suppose that this arrangement is manipulated so as to keep each current constant.

Consider first the case of a single movable pole of strength m and a single circuit in which the current is maintained at a uniform strength i. If a> is the solid angle subtended by the circuit at the position of the pole at any instant, the potential energy of the pole in the field of the current is miw, so that in an infinitesimal interval dt of the motion of the pole, the work per- formed on the pole by the forces of the field is mi -=- dt. The current which has flowed in this time is idt, so that the extra work done by the additional

batteries is the same as that of an additional electromotive force m -r- .

at

506, 507] Physical Principles 453

Thus the motion of the pole must have set up an additional electromotive force in the circuit of amount — m-57 , to counteract which the additional

electromotive forces are needed. The electromotive force — m -7- which

at

appears to be set up by the motion of the magnets is called the electromotive

force due to induction.

The number of tubes of induction which start from the pole of strength m is 47rm, and of these a number mm pass through the circuit. Thus if n is the number of tubes of induction which pass through the circuit at any instant,

the electromotive force may be expressed in the form — -=- .

So also if we have any number of magnetic poles, or any magnetic system

of any kind, we find, by addition of effects such as that just considered, that

dN there will be an electromotive force 7— arising from the motion of the

whole system, where N is the total number of tubes of induction which cut the circuit.

It will be noticed that the argument we have given supplies no reason for taking N to be the number of tubes of induction rather than tubes of force. But if the number of tubes crossing the circuit is to depend only on the boundary of the circuit we must take tubes of induction and not tubes of force, for the induction is a solenoidal vector while the force, in general, is not.

dN 507. The electromotive force of induction — -5— has been supposed to

be measured in the same direction as the current, and on comparing this with the law of signs previously given in § 483, we obtain the relation between the directions of the electromotive force round the circuit, and of the lines of induction across the circuit. The magnitude and direction of the electromotive force are given in the two following laws:

Neumann's Law. Whenever the number of tubes of magnetic induction which are enclosed by a circuit is changing, there is an electromotive force acting round the circuit, in addition to the electromotive force of any batteries which may be in the circuit, the amount of this additional electromotive force being equal to the rate of diminution of the number of tubes of induction enclosed by the circuit.

Lenz's Law. The 'positive direction of the electromotive force f j—j and

the direction in which a tube of force must pass through the circuit in order to be counted as positive, are related in the same way as the forward motion and rotation of a right-handed screw.

454 Induction of Currents in Linear Circuits [ch. xiv

If there is no battery in the circuit, the total electromotive force will be

dN — , and the current originated by this electromotive force is spoken of as

CLZ

an " induced " current.

508. In order that the phenomena of induced currents may be consistent with the conservation of energy, it must obviously be a matter of indifference whether we cause the magnetic lines of induction to move across the circuit, or cause the circuit to move across the lines of induction. Thus Neumann's Law must apply equally to a circuit at rest and a circuit in motion. So also if the circuit is flexible, and is twisted about so as to change the number of lines of induction which pass through it, there will be an induced current of which the amount will be given by Neumann's Law.

509. For instance if a metal ring is spun about a diameter, the number of lines of induction from the earth's field which pass through it will change continuously, so that currents will flow in it. Furthermore, energy will be consumed by these currents so that work must be expended to keep the ring in rotation. Again the wheels and axles of two cars in motion on the same line of rails, together with the rails themselves, may be regarded as forming a closed circuit of continually changing dimensions in the earth's magnetic field. Thus there will be currents flowing in the circuit, and there will be electromagnetic forces tending to retard or accelerate the motions of the cars.

510. If, as we have been led to believe, electromagnetic phenomena are the effect of the action of the medium itself, and not of action at a distance, it is clear that the induced current must depend on the motion of the lines of force, and cannot depend on the manner in which these lines of force are pro- duced. Thus induction must occur just the same whether the magnetic field originates in actual magnets or in electric currents in other parts of the field. This consequence of the hypothesis that the action is propagated through the medium is confirmed by experiment — indeed in Faraday's original investiga- tions on induction, the field was produced by a second current.

511. Let us suppose that we have two circuits 1, 2, of which 1 contains a battery and a key by which the circuit can be closed and broken, while circuit 2 remains permanently closed, and contains a galvanometer but no battery. On closing the circuit 1, a current flows through circuit 1, setting up a magnetic field. Some of the tubes of induction of this field pass through

circuit 2, so that the number of these tubes Key

changes as the current establishes itself in a ery

circuit 1, and the galvanometer in 2 will • "

accordingly shew a current. When the current in 1 has reached its steady

507-513] General Equations 455

value, as given by Ohm's Law, the number of tubes through circuit 2 will no longer vary with the time, so that there will be no electromotive force in circuit 2, and the galvanometer will shew no current. If we break the circuit 1, there is again a change in the number of tubes of induction passing through the second circuit, so that the galvanometer will again shew a momentary current.

General Equations of Induction in Linear Circuits.

512. Let us suppose that we have any number of circuits 1, 2,

Let their resistances be Ru R2, ..., let them contain batteries of electro- motive forces Elf E2, ..., and let the currents flowing in them at any instant

DO tj, 12) • • • .

The numbers of tubes of induction N1} J¥2, ... which cross these circuits are given by (cf. equations (429))

J^j = Zuij. + L12i2 + L13i3 + . . ., etc.

In circuit 1 there is an electromotive force Ex due to the batteries, and an

dN

electromotive force r-1 due to induction. Thus the total electromotive

at

force at any instant is E-^ — -7—1, and this, by Ohm's Law, must be equal to

Riix. Thus we have the equation

Ei--r (LnH + Ll2i2 + L13i3 + ...) = Rih (431).

Similarly for the second circuit,

E2 - jt (Lnh. + L,2i2 + L23iz +...) = R2i2 (432),

and so on for the other circuits.

Equations (431), (432), ... may be regarded as differential equations from which we can derive the currents i1} i2, ... in terms of the time and the initial conditions. We shall consider various special cases of this problem.

Induction in a Single Circuit.

513. If there is only a single circuit, of resistance R and self-induction L, equation (431) becomes

E-jt(Li1) = Ri1 (433).

Let us use this equation first to find the effect of closing a circuit pre- viously broken. Suppose that before the time t = 0 the circuit has been open, but that at this instant it is suddenly closed with a key, so that the current is free to flow under the action of the electromotive force E.

456

Induction of Currents in Linear Circuits [ch. xiv

The first step will be to determine the conditions immediately after the

d circuit is closed. Since -n(Lii) is, by equation (433), a finite quantity, it

follows that Lix must increase or decrease continuously, so that immediately after closing the circuit the value of Lix must be zero.

To find the way in which ix increases, we have now to solve equation (433), in which E, L and R are all constants, subject to the initial condition that ix = 0 when t = 0. Writing the equation in the form

we see that the general solution is

-5*

E-Ri1 = Ce L

where G is a constant, and in order that ix may vanish when t — 0, we must have G = E, so that the solution is

;,=|(i-e-z<)

.(434).

It will be seen

Fm. 131.

The graph of ix as a function of t is shewn in fig. 131. that the current rises gradually to its final value EjR given by Ohm's Law, this rise being rapid if L is small, but slow if L is great. Thus we may say that the increase in the current is retarded by its self-induction. We can see why this should be. The energy of the current i± is ^Lif, and this is large when L is large. This energy represents work per- formed by the electric forces: when the current is i1} the rate at which these forces perform work is E, a quantity which does not depend on L. Thus when L is large, a great time is required for the electric forces to establish the great amount of energy Li*.

A simple analogy may make the effect of this self-induction clearer. Let the flow of the current be represented by the turning of a mill-wheel, the action of the electric forces being represented by the falling of the water by which the mill-wheel is turned. A large value of L means large energy for a finite current, and must therefore be represented by supposing the mill-wheel to have a large moment of inertia. Clearly a wheel with a small moment of inertia will increase its speed up to its maximum speed with great rapidity, while for a wheel with a large moment of inertia the speed will only increase slowly.

Alternating Current.

514. Let us next suppose that the electromotive force in the circuit is not produced by batteries, but by moving the circuit, or part of the circuit, in a magnetic field. If N is the number of tubes of induction of the

513, 514] Induction in a Single Circuit 457

external magnetic field which are enclosed by the circuit at any instant, the equation is

-jt(Li1 + N) = Ri1 (435).