TIE point is that om does not have to wait until a particular electron at the battery termiaal reaches the lamp for the lamp filament to mpond to the <xurent. When the switch is closed, the entire charge distribution within the condudor is set in motion almast instantaneously, much as water starts to flow in a long pipe ss soon as we open a tap.

4) a) drift velocity will be doubled,

b) drift velocity will be halved, i

c) drift velocity will remain unchanged.

5) i) By definition, E, the electric field, is related to the current density J, through the relationship

1 But a = - and therefore, P'

ii) E is related to V by b

Vb- Va m J E.d (1) a

Fmrn the Fig. 8.13, E is parallel to the axis of the cylindrical wire. If we evaluate (1) along a line in the direction of E and parallel to the cylinder axis, we obtain

Therefore, Vb is at a lower potential than Va.

6) a) Assuming radial flow of charge between rod and cylinder, we have at P

That is a 1ong.time. Yet we know that as soon as we close the proper switch, the charge flows through a circuit and lamps light up. We need not wait several minutes, not even seconds, to witness the effect of the current in a circuit, and .

there appears to be no observable dependence on the distance between the wall switch and the light fixture, a distance generally considerably greater than 10 cm.

J = - I andE - p.7 = & , where p is resistivity, with both J and 2rc rL

E in the direction of r. Then, by definition of the potential,

and so, noting the polarity of V,, we get .

solving for I, ,

I = ZLVr

p ln ( r z /n )

I b) From (a), J = - I ' and E - Vt 2zrL prln(rz/tr) P J m r l n ( d r 1 )

, . Vr p l n ( d r 1 ) c) Fmm Ohm's law, R - - I I 2 d '

20

UNIT 9 TIC FIELD

Structure

9.1 Introduction Objectives

9.2 Magnetic Field Source of Magnetic Field Definition o f Magnetic Field

9.3 Gauss's Law for Magnetism

9.4 Biot and Savart Law

9.5 Force between Two Parallel Conductors (Definition of Ampere)

9.6 Ampere's Law Applications of Ampere's Law Differential Form of Ampere's Law

9.7 Torque on a Current Loop

9.8 Summary

9.9 Terminal Questions

9.10 Solutions and Answers

In Block 1 of this course, you were introduced to the concept of an electric charge and studied some properties of charges at rest. You learnt that a static distribution of charge produces a static electric field. Similarly, steady flow of charge (i.e., a steady current) produces a static magnetic field, which is, infact, the topic of this unit. However, there . are some major differences between the two fields which you will discover in this unit.

In the science laboratory, during your school days, you must have been fascinated with magnets. Recall, when you tried to push two magnets together in a way they didn't want to go, you felt a mysterious force! In fact, magnetic fields or the effect of such fields have been known since ancient times when the effect of the naturally occurring permanent magnet (Fe304) was first observed. The north and south seeking properties of such materials played a large role in early navigation and exploration. Except for this application, magnetism was a little known phenomenoi~ until the 19th century, when Oersted discovered that an electric current in a wire deflects a compass needle. This discovery showed that electric current has something to do with the magnetic field because a compass needle gets deflected and finally points inathe north-south direction only when placed in a magnetic field.

In this Unit, we shall consider in detail the production of the magnetic fields due to steady currents, and the forces they exert on circuits carrying steady currents and on isolated moving charge. A good way of gaining a better understanding of the nature of fields is to know how they affect the charged particles on which they act. Hence, in the next unit, you will study the behaviour of charged particles in both electric and magnetic fields.

Objectives After studying this unit you should be able to :

r understand what is meant by the magnetic field, the right hand rule, Biot-Savart +

law, right hand method, Ampere's law,

r define the magnetic field at a point in tenns of the force on a steady current element and also on a moving charged particle,

r use the formula for the force on a steady current eleclent-or on charged particle due to a magnetic field to calculate the force on a certain simple current cirrrying '

circuits, and solve simple phblems,

BecQric Current and Magnetic Field

Fig. 9.1: ?When the magmet is Precly suspended, rr p d c u h r end of it points no& lWs end o t Uac magolet is def i id PO tbe north p B c

(b) Fig. 9.2: a) A mmpm needle

poinls In the direction d the magnetic field. b) Magnetic fie119 lhes of a magnet drawn using the Pact that a compass needle should line up along the Pield lines

show that the divergence of B vanishes,

+ use Biot-Savart law to describe and compute the magnetic field generated by a simple current flow,

s identify the nature of force (attraction or repulsion) on a given length of a long, straight cunent-carrying wire that is laid parallel to a similar currentcarrying wire,

e use Atnpere's law to calculate the magnetic field Erom steady current distributions having simple geometries,

e relate Ampere's law to its differential form via Stokes theorem,

compute the torque exerted by a steady magnetic field up011 closed current loops,

+ appreciate how the forces on current-carrying conductors, placed in a magnetic field, are used to understand the working of galvanometers and motors.

9.2 MAGNETIC FIELD

Wheneverwe speak of magnetic field, we speak in terms of bar magnets since this is the way the fields were first studied. You are already awareof the basic features of the magnetic field from your school days. For example, you know that the poles of a bar magnet experience forces when placed in a magnetic field. If a bar magnet is suspended by a delicate fibre as shown in Fig. 9.1, a particular end of the magnet will always point towards north. This end of the magnet is called the north pole of the magnet. The other end is the south pole. Do you recall that this arrangement is a simple compass? The north poles of two magnets repel each other. The south pole of a magnet is always attracted by the north pole of another magnet. If one tries to break off the north or south pole from a simple bar magnet, then this exercise proves to be futile. The broken magnet becomes two new bar magilets each having a north and a south pole. This shows that an isolated magnetic pole does not exist. a

In order to plot the magnetic field due to a bar magnet, we need only a compass needle. The direction in which the compass needle points is taken to be direction of the magnetic field. In class XII, you must have used this fact to plot the magnetic field in the vicinity of the bar magnet as shown in Fig. 9.2a. The magnetic field lines are drawn in such a way that a compass needle placed on the line will align itself tangentially to the line. Fig. 9.2b shows the typical magnetic field for the bar magnet. Notice that the field lines emerge from the north pole and enter the south pole. These are some qualitative features with which we are all familiar.

92.1 Source of Magnetic Field

As you know, the space near a ~ubbed glass rod (rubbed either by rubber or rabbit's fur) is characterised by an electric field which is denoted by E. Similarly, a magnetic field around a magnet may be represented by the symbol B. In electrostatics, the relation between the electric field E and the electric charge is represented as follows:

electric charge s E electric charge (9.1)

Tbat is, tlie electric charges set up an electric field and the field, in turn, exerts a force (electric in nature) on another electric charge that may be placed in that field. Now, by analogy, can you set up a similar relation for magnetism. Yes, the relation will be as follows:

magnetic charge 2 ]B ;;z magnetic charge (9.2)

You knowthat the two poles, i.e., north and south always, occur together. A single isolated pole is not known to exist. This means that there are no magnetic charges (also called magneticmonopoles). How does, then, the magnetic field arise? The answer to this question you will find in the following lines. Let us consider two wires, running parallel to one another, as shown in Fig. 9.3 a. As soon as the circuit is closed, the current in the two wires, flows in the same direction, and the wires are found to attract. If the direction of one of the currents is reversed, the wires repel each other. Thus the two sections of the wires in Fig.9.3 b, which are part of same circuit, tend to move apaft. If a sheet of metal is put between the two wires, the force with which wires attract or repel is not at all affected (Fig.9.3 c). How do. you explai~m this? Does electrostatic force account for the attraction of parallel ones? No, the

force acting is not an electrostatic or Coulomb force. This is because (i) there is no net charge on the conductor (the charge density of conduclion electrons just compensates for the positive charge on the lattice ions); (ii) the force is reversed in sign by revesir~g the direction of either current; (iii) the force ceases as soon as the circuit is broken; (iv) the force is not affected in the presence of a simple medium; (v) the attraction and repulsion of the electric currents is contrary to the attraction or repulsion of the electric charges.

Mngnelc Field

(0'

Fig. 9.3 : n) Parallel wires carrying currtnts in (be same direction w pulled logether. b) Pnralld wires carrying curnnls in opposik directions nrr pushed apart c) A sheet of metal bclween Lhe two wires docs not nlTect these foms.

'The experiments of Fig. 9.3 show that there is an additional force associated with a moving charge, which is different from the electrostatic force. This new force that comes into play when charges are moving is called the magnetic force. A charge sets up an electric field whether the charge is at rest or is moving. However, a charge sets up a magnetic field only if it is moving. You may ask a simple question. A bar magnet sets up a magnetic field in its vicinity, but where are the moving electric charges in a bar magnet? Actually the spinning and circulating electrons in the iron atoms of the magnetic material are responsible for its magnetism. You will learn more about it in Units 11 & 12 of this block. Hence. in magnetism, we can think in terms of the'following relation :

.PI

moving electric charge S B 3 moving electric charge (9.3)

As the moving charges constitute an electric current in a wire, Eq. (9.3) can be written Bs

electric current * B B electric current (9.4)

Eq. (9.3) or (9.4) tells us that (i) a moving charge or a current sets up a magnetic field and also (ii) if we place a moving charge or a wire carrying a current in a magnetic field. a force will' act on it. Now, let us define the magnetic field. But before doing this g try to answer the follotving SAQ.

SAQ 1

L

Fig. 9.41 A sLrrIgh1 wire CAI y h g cumnl and placed in n mnguetic field experiences n force. You have probably studied about an electric motor in your school, and you may be

knowing the principle on which it works. Briefly explain how an electric motor illustrates Eq. (9.4).

9.2.2 Definition o f Magnetic Field

. In Block 1, we defined the electric field E at a point in terms of the electric force FE that acted on a test charge q at rest at that point as follows :

FE = q E (9.5)

As suggested by Eq. (9.3), we can define the magnetic field in terms of the magnetic force exerted on a moving electric charge. It can also be defined in terms of the force on

23

Ekechic C u m t cmd Magnetic Field

a current. Since current is a flow of electric charge, the two definitions are related. First, let us state the definition in terms of force on a current-carrying wire.

a) Force on currents

Experiments show that a wire, carrying o current placed in a magnetic field, experiences a force. Fig. 9.4 shows a wire carrying a current I in a magnetic field produced by a magnet. Since the field lines come out of the north pole and enter the south pole, the field is directed from right to left. It is found that the wire experiences a force, which is proportional to both the current and the strength of the magnetic field. When the wire is placed parallel (or antiparallel) to the field lines, it experiences no force. But when the wire is placed perpendicular to the field, the force on the wire is the maximum. These two cases are shown in Fig. 9.5. This shows that the force on a wire is due entirely to the component of the field that is perpendicular to the wire. In other words, the force also depends on the relative orientation of the wire and the field lines. in Fig. 9.6, suppose, the angle between the field lines (represented by B) and the current carrying wire is 8. As said above, the force F on the wire of length L is dae entirely to the cornponelit of B that is perpendicular to the wire. This component represented as EL is given by (See Fig. 9.6) . .. .- . .

B-L = B sin 0,

I f there is surface area A placed perpdiculy to a uniform magnetic EieldB then the prodact o f magnctic field B and the surface area A is an important ph ysicaI quantity which i s called magnetic flux through the surface &A It i s dedotcd by 4 asfouows

ThouPit oEmagaetic flux is weber. Hence the unit of B is also weber per square metre. It is alao called Tesla 0.

F = maximum (inlo pagc)

Fm. 9.5 : The ronr oo a wire b due e d d y to (hc compownt &(be Bdd &at is pcrpendiculu to thc w i n

Fig. 9.6; The f o m on (he varks &s B sin 8, Lbat is, In proprtiw to'B~.

Further, the force on the wire of IengthL depends onL itself and the current I in the wire. We conclude that the force P on a length L of the wire is given by

F =. U B sin 0 (9.6)

Recall that the vector cross product A x B gives rise to a vector of magnitude AB sin 8 which is perpendiciilar to the plane containing A and B. Using this in Eq. (9.6) we get

F = I ( L x B ) (9.7)

Here L is a vector the magnitude of which is the length L of the wire and its direction is along the ahrent. I&. (9.6) or (9.7) shows that the B has the SI units of;N A- ' m- '. This unit is also given the name weber per square meter o r tesla (abbreviated as I?). One tesla h a strong magnetic field, so that a smaller unit called the gauss (G) is often used.

, I t a l a = la4 gauss (9.8)

Since gauss is not an SI unit, we should aiways convert it to tesIas before using it in equations. The quantity B has, several names. Its correct name is.magnetic induction. It is also designated as the magnetic field intensity, although another quantity, which you will study in Unit 12 of this Block, is also given this name.

The direction of the force on the wire is always perpendicular to the plane defined by B t Magnetic Field

and I. To find the direction of the force, we use right hand rule as shown in Fig. 9.7. According to right hand rule: if one's right hand is held flat with the fingers pointing in the direction of the field lines and the thumb pointing in the direction of the current, then the palm of the band will push in the direction of the force. I

Let us now apply Eq. (9.6) or (9.7) to a simple situation so that we can better see its f i meaning. - F

Fi. 9.7: Rigbl b n d rule.

Example 1

A horizontal wire is canying a current from east to west. What is the direction of the force on this current if we assume that at this location the magnetic field of eaPth points due north? If the wire carries a current of 20A, find the force per unit length on it due to the earth's magnetic field, which is about 1.0 G .

Solution

Using right hand rule, we find that when the thumb of the right hand points west and the fingers point north, the palm faces down, hence the force on this wire will be down (into the page).

Earth's magnetic field lin& are in a direction perpendicular to the wire. We hove

F - 'ILB sin 90"

Suppose the wire carrying current is not straight so that, at each point, its orientation relative to the field changes, or suppose the field changes in magnitude/direction over -

the length of the conductor, we can still useEq. (9.7) to find the force. For this purpose we imagine the wire to be broken up into small segments so that it is straight, and the field is essentially conitant over its length. See Fig. 9.8. Now Eq. (9.7) can be applied to each segment.

If the length of the small segment is dl then we can write for the small force dF on the segment as

(a) (b)

Fig. 9.8:(n) A curved wire in a nonunif~cm magnetic field (b) A s m d enough sqpment of tbe wire can be treated as rn stmight wire in a uniform Bdd

We can obtain the total force on this arbitrarily' shaped long current-carrying wire placed in a non-unifonn magnetic field B, by summing the expression for dF in Eq. (9.9)~verthe whole wire giving

Electric Current and Magnetic Field

If we let the length dl approach zero, this sum becomes an integral, and we call write the above expression as

F - I J ~ X B (9.1 0)

where, on the right hand side of the Eq. (9.10), we have the line integral taken over the length of the wire and I, being a constant, is taken out of the integral. In particular, if the magnetic field is uniform, which means that B is constant both in magnitude and direction at all points of the wire, then we can write Eq. (9.10) as

In this expressionJdl is the vector joining the initial point of thd wire to its final point. I

Further, if the current carrying wire is straight and of length L, the11 we have

This is the sanie as Eq. (9.7).

So far we considered the force on the current in a wire. An electric current is simply a group of charged particle sharing a colnnlon motion, so we should expect a charge to experience a force in the magnetic field. This gives another way of defining the magnetic fie'ld.

b) Force on a moving charge

The force which a magnetic field exerts on a moving positive charge can be obtained from Eq. (9.7). Recall from the discussion of Unit 8 that the velocity v of the charge q in a wire of cross-sectionA is related to I by Eq. (8.3) as follows :

I = qrdv

wser rz is the number of charge per unit volume. Substituting this expression for I into Eq. (9.9) gives

dF = ( d C ) A n q v x B (9.1 1)

Here (&)A represents the volume of the wire segment of length (dL). So (&)An is the number of moving charges in that portio~i of the wire for which we are writing the force. Hence the force F on a single charge is given dF/( dL )Ail, i.e.,

k

F = q v x B (9.12) I

The magnitude of the force is given by qvB sin 8. The direction of the force'on the moving charge can be obtained by right hand rule (Fig. 9.7 with I replaced by v ). Note, if the particle is negatively charged, the direction of F will be reversed.

SAQ 2 '

Of the three vectors in the equation F = q v x B, which pairs are alGays at right angles? Which may have ally angle between them. Read the following example so that you can understand how the force on a charged particle is calculated. It also illustrates the use of the right hand rule lor determining the

. direction of force.

Example 2

In a certain region a magnetic field of 0.10 T points vertically upward. Three protons enter the region, two horizontally and one vertically, as shown in Fig. 9.9. All the three are moving at 2.0 x lo3 ms-'. What is the force on each proton?

2.

F& 9.9 : Example 2 Solution

P ~ ~ t o n 2 is moving vertically, i.e., parallel to the field ( sin 8 = Q ) ia Eq. (9.12). Therefore, it experiences no force. Protons 1 and 3 are moving at right angles to the field, so sin 8 =, 1 in Eq. 9.12, and thus the forces on these two protons have the same magnitude given by

26

Since the protons carry a positive charge, the direction of the force is the direciion of the vectorv x 18. For proton 1, tnoving to the right, v x 81 is out of the page. For proton 3, moving to the left, the force is illto tlic page. This exi1111ple shows clearly thai the magnetic field alone does not detemiine the force. lde~~tical particles in the sanic field may experience different forccs, if their velocities are not identical. If ihc pallicles were electrons, the negative sign of the electron charge would have i~idicatcd a force opposite to tlie direction of v x B. Eq. (9.12) is equivalent to Eq. (9.7) so that either of them can be taken as the defining equation for B. In practice, we define B from Eq. (9.7), because it is much easier to measure the force acting on a wire than 011 a single ~noving charge.

In this section, you have learnt that a movink charge gives rise to a ~nagnetic field. Now, suppose there is a current carrying wire, and you are asked to calculate the magnetic field produced due to it at any point of spacc, then ccrtai~~ly you would like to have laws analogous to Coulomb's a~id Gauss's law. Let us first find out Gauss's law for magnetism.

9.3 GAUSS'S LAW FOR MAGNETISM

Suppose magnetic charges-monopoles-exist, then they would give rise to ~nag~ietic ,fields like the electric fields due to electric point charges. Then these fields-and those of magnetic charge distributions-would be described by laws analogous to Gauss's law. hi particular, we would find tliat the flux of the inng~ietic field through any closed surface would depend only on the enclosed ~nag~ietic charge. We may write Gauss's law for nag net is in as

- where the integral on the lefi is the flux of I3 over a closed surface enclosing the magnetic charge or monopoles denoted by g and po is some constant. But the very existence of the magnetic ~no~iopoles is uncertain. And even if they do exist, they seeiii to play no significant role in our world. hi the absence of the lnag~ictic monopoles, we must put g = 0 and then tlie magnetic flux through ally closed surlacc must be zero. We state this mathematically as Gauss's law for magnetism and write it as follows

Pig.9.LO; In Lhe absence of '

magnetic nionopolcs, the magnetic flux through a clnscd surface must be zero.(h) Thi! means there call be no point where magnetic field lhes begin or end, for a cl&ed surface

A consequence of Gauss's law for nlag~ietis~n is that iiiagnetic field lincs can never sumoundiug s idl a begin or end (Fig. 9.10). Unlike tlie electric field lilies, the ~izag~ietic field lines liavc to point would have non- forin closed loops. If we convert the above surface integral of Eq. (9.14) illto a volu~ne zero net flux (b) .

integral using the Divergence Theorem, we obtai~i Instend, n~egnctic licltl liues generally for111

[ v . B ~ v = 0 (9.15) closed lrrolls . .r

The integratio~i is ove'r the volu~ne etlcloscd by thc closcd surface of Eq. (9.14). Since the Eq. (9.16) holds for ai!y arbitrary volume of inlegmtion, we must have

This equation is true, even if B varies witli time hnd is, in fact, one of Maxwell's equations. Eq, (9.14) or Eq. (9.16) says that if the magnetic field exists, they l o ~ k different fonn the fields of the point charges. I11 the next section, we will find out the law analogous to Coulomb's law.

9.4 BIOT- AND SAVART LAW ' - Can we calculate the magnetic fields produced by a current? Can we show that a current loop has the magnetic field of a dipole? Interest in questions like these led the Fre~icli

Uecttic Cumnt and Magnetic Field

scientists Jean Baptiste Biot a~id Felix Savart to experimentally detennine the fornl of the magnetic field arising from a steady current. Known as Biot-Savart law, its result gives the magnetic field-at a point due to a small element of current.

In Unit 1, we learnt to calculate the electric field that a given distributio~~ of charges set up in the surrou~~ding space. Our approach was to divide the charge distribution into

Fig+.p.ll: (a) A ch.rge'daneot dq eshblirbts a diienntid decCric field dement dE at point P. (b) A curnut clement Id cshblIshes a diertntial magnetic IPekl W at point P. ' Ihc symbol x (the tail of an arrow) shows h a t the drmeut dB points into (he page along tbc ntgativt z - axis,

charge elements dqas in Fig. 9.1 la. We then calculated the field dE set up by a given charge elelllent at an arbitrary poi111 P. Finally, we calculated E at point P by integrating dE over the entire charge distribution.

The magnitude of dE is given as follows:

in which r is the distance from the charge element to the point P.

In the magnetic case, our approach will be the same. Fi$,.9.1 l b shows a wire of arbitrary shape carryinga current I. What is the magnetic fiel'd B at an arbitrary point P

1 near this wire? We first break up the wire into differential current elements Idl, 1 1 corresponding to the charge elements dq of Fig. 9.1 la. Here the vector dl is a

differential element of length, pointing along the tangent to the wire in'the direction of the current. Note that the differential charge element dq is a scalar, but the differential

- - current element Id is a vector.

Then the Biot-Savart law says that the magnitude of the magnetic h l d contributioii set ' up by a given current element at point P is give11 as follows :

po Zdlsin 8 a=--, 4~ r Z , .

Here pa is a constant, called the penneability co~lstaiit of free space. Its value is 4n; x TmA-I. This collstaiit plays a role in magnetic problems, lnGh like the role that the pennittivity constant ep plays in electrostatic problems.

The full expression for dB in vector form is

p, id lx ; a=-- ( Biot - Savart law ) ' ! 4~ r Z

Hew ; is a unit vector pointidg from dl towards P. Eq. (9.18) is the analogof Coulomb's law, and is called the law of Biot and Savart. TAe direction.of dB in Fig. 9.11b is that of the vector dl 'x ;, where i? is a unit vector that points from the current element to the point P at which yob Wish to know the field. The symbol x (representing the tail of an arrow) in Fig. 9.1 l b shows that dB at point P is directed into the plane of the page at right angles.

Coulomb's law gives the electric tield of a point c h a ~ e in tenns of the charge and the distance from the charge to the Beld point.The electric field varies as the invetse

square of the distance, and its direction lies along the line joining the charge with the '

field point. Analogously, the Biot-Savart law gives the magnetic field at a given point in terms of the current element (its source) and the distance to the field point. Like the electric field of a point charge, the magnetic field of an isolated current element varies . as the inverse square of the distance. But here the similarity ends. Unlike the electric charge in Coulomb's law, the current element Id1 has associated with it a directio~l as well as a magnitude. Hence the magnetic field of the current element is not symmetric about the element, but depends on the position of the field point relative to the directioii of the current element. This directional character is expressed by the cross product in Eq. (9.18). As shown in Fig. 9.1lb, the magnetic field is at right angles to both the ,121 Magactic ~ d d b c s

current element and the vector from the current element to the field point. Also, you geocnUy meircle r

have learnt in the previous section that the magnetic lines have no sources unlike the - current.

electric field lines which end or originate on electric charges, but are continuous and join back on themselves.

h t us see how Eq. (9.18) and Fig. 9.11b show this to be true. Let the point P move around the cumnt axis at a constant distance from the axis. From Eq. (9.18), the magnitude of dB is constant along this path, and at each point it has a direction tangent to the path. These are just the requirements for the lines to be concentric circles around the current. Hence, the magnetic field lines encircle the current as shown in Fig. 9.12. The direction in which the circular field lines point depends on the direction in which the current flows. If the direction of the current flow is reversed, the direction of the field line is also reversed as shown in Fig. 9.13.

( 4 (b) rig. 9.13 m e direction in which (he fidd lines pdat is debnnhcd by the direction in which tbe current

flows (a) When the curtent flows i n t ~ Lbe page, tbedidd lines t o m dockwise circles. (b) Whm the cumnt flows out , d ( h e page, the circles sn mnticlo~kwisc,

However, there is an easy way to remembe~ these directions. Just close the palm of your right hand and point your thumb in the direction of the current as shown in Fig. 9.13. In either case you will f h d that your fingers will naturally curl around in the directioti of the magnetic field lines. This simple tecliiiique for remembering the direction of the magnetic field is illustrated in Fig. 9.14, and is referred to asthe right-hand method. Do not confuse it with the right band rule for vectc cross products shown in Fig. 9.7.

Fig, 9.14 By using your right hand to 'gripp a cunmt- umyhg conduebr, you can llad out the dlrrdioa d a tbc magnetic fld& When your (humb points in the d i d m of current now, your Engem indicate

the dimdon of the mag&lic fId& -

4 Vig. 9-15 Cafculmting the mmgnetic Deld set up by s current tin e loug strnigbt wire. The f idd dB associated with $be cumcut element PiA poi& out of the pap, ns shown,

A horizontal vrlire crrrries a current from east to west. What is the direction of the magnetic field due to ellis current directly above and below the wire? Refer again to Fig. 9.1 1b. Since the inagnetic field obeys the superposition principle, the nct field at P clue to entire circuit, of which the wire is a part, will be the vector sum or line integral of the lields of individual current elements:

wlnere the integratiul~ is taken over the entire path through which the currellt I flows. Eel us apply Biot-Savart law to calculate a magnetic field for sinlple situation.

a) Ca8cmll;ltioaa oQB doe 80 a Ismg straight wise carrying la current

As show11 ita Fig. 9.15, suppose P is a poilie at which we are to calculate the inagnetic lield due to a long straight wire carryi~~g a current I. 7% distance between the point P and the wire is r. The: differential inagi~etic field set up at point P by the current eleiile~~t Id1 is given in ~lxigniiudc b y Eq. (9.191, i.e.,

116) Id1 sin 0 d~ = - (9.17) 4 7 ~ 1.2

The direction of d l 3 is given by the right hand nncthod. Here it points out of page. This is true irrespective of the positioi~ of (4 along the wire, so that at point P all the dB' s from all tlie current ellcme~lls Id! poillt in the same direction. Thus lo find the ~nagliitude of total lnagnetic ficld B at point P we integrate Eq. (9.17) obtainilig

n . . J ~ B = 4x

where 8 is the angle between and 1~61.

111 order to sulnup the co11trib~ttio11~ from all elelnents of [he long straight wire, we change Lhe variables fi-o1n8 and r to $ (See figure). Now, sill 0 = sin ( JX - ) = cos 41. (Frorn figure ) (9.20)

AC rd0 = cos q, I>rawAC J-PB, then - = - AB dl

[Angle 11etweenAC and AB beillg equal to angle between PA and PO j

. . . ril sin 0 dicosc) - Z -

I' 2 (From Eq. 9.20) I' -

B = - [ sin 4~ + sin $1 ] 43cR

For an infir~itely long straight wire

I The magnitude of B, thus, falls off invers~ly as the first power of the distance from an infinitely long wire. Eq. (9.22) shows that the lines of the magnetic field f o ~ m concentric circles around the wire. This expression for B is,analogous to the expressiot~

for E due to a long charged wire = - 42Ep ( ?), and thus shows its electrostatic

equivalent nature. -

30

A

b) Calculation of B along the axis oCa current loop Mawetic Field

Let us coiisider a circular loop of radius a aiid carrying a current I. As shown in Fig. 9.16, we choose a point P on its axis at a distance b froin its centre. The magnetic field dB at P due to a current element of length dl is give11 by

For all eleil~eiits around the loop, r is perpeildicular to I d , hence, the value of sin 8 in

Fig. 9.16 : Magnetic field along the axis d a current loop.

the cross-product liere is 1 and so I

p4lI dl 1

dB e-- 4~ r 2

Since r 1- dl lleilce dB is always perpendicular to the plane coilsistiiig of r and dl. Thus I

dB is perpeiidicular to r at the point P as show11 in Fig. 9.16. It can be resolvcd illto two compoilents, one dB sill cp along the axis aiid the other dB cos $ at right angles to the axis. Here 4 is the angle between rand the axis of the loop. You will iiolice that the compoilents of dB perpendicular to the axis will cancel, as inay easily bc seen by coilsidering the field due to an element opposite to dl. Therefore, the resultant I3 is in I the direction of the axis and will be given by summing only the components dS sin +. 1 Thus, B is given by i 1

B = S dB sill $

= !&?!dlsin$ - 41t r z I

F I *pl - - - 4~ r 2

As we are around the current loop, both + and rare coilstants, so they are taken outside the integral. The length integrated around the loop is b, so that

w,l sill $ 2na B =

4nr

If sin 4 and r are expressed in terms of the constants a and 0, we get

WI B e - a 2 2 3/2

(9.23) 2 ( a 2 + b )

When we choose the point P far from the loop ( b >> u ) , the expression for B is written as

B I -- 4~ r 3

(9.24)

Here we have writtenA = m2, the Area of the loop. This relation shows that the current loop generates its own magnetic field. Notice that the inagnetic field of the loop at large

distances is like the electric field of an electric dipole [ E - &(f) ]on i t sax i s

(See Sec. 3.6 of Unit 3.) This shows that the term (LA corresponds to electric dipole moment p of the electric dipole, Therefwe, the termL4 i s called the magnetic dipole ~ i g . 9.171 he current-canyiug . moment of the loop and is represented by p . This shows there is a similarity between a bar magnet which is.magnetic dipole, and a much like that ofthc

current loop. The similarity can also be seen by plotting the magnetic field around Ule short bar magnet shown m (c), current loop. When a compass is used to plot,the magnetic field, the result show11 in

Elecllic Current and Mngnelic Field

Fig. 9.17a and b is obtained. You should collvi~lce yourself that this is reaso~~able by applying the right halid method to a portio~i of the loop. Observe Fig. 9 . 1 7 ~ and note that rlle magnetic field of loop is much like that of a bar magnet. The current loop call be considered to have the north and south poles. We shall see in a later section that this is one aspect o l a very i~llportant similarity between bar magnets and current loops.

Alter going through this section I hope you can tell why the two wires, shown in Fig. 9.3, are attl-acted in one case while they are repelled in another case. If not, read the followi~~g section. It will also help you in defining the unit ampere which we have been using so far without defining it precisely.

But before defini~~g ampere do the following SAQ.

SAQ 4

Write otie analogy and one differe~~ce between Coulomb's law and Biot-Savart law.

9.5 FORCE BETWEEN TWO P LLEL CONDUCTORS @EFINITI[ON OF AMPERE)

In this section, we will find low much force docs one of the wires in Fig. 9.3 exert on the other. We assulile that the wires are Iiuear, parallel and very long. Here, one of the wires experie~ices a force, because it is in the magnetic field caused by the current in other wire.

Fig. 9.18 shows two long, parallel wires separated by a distance d and carryi~ig currents 11 and I2 in the same direction. The current in wire 2 prnduces a magnetic field Bz at all points around the wire. From Eq. (9.22) the magnitude of Bz at the site of wire 1 is given by

cro I2 Bz = - 2Rd

(9.25)

12

Fig. 9.1&Two parallel wims carrying currents in the same diitiw altrnct cach other.

The right hand method tells us that the direction of Bz at any p o i ~ ~ t on wire 1, is out of the page, as shown in F!g. 9.18.

Now, wire 1, which is carrying a current 11, finds itself immersed in an external magnetic lield B2. If L is the length of this wire, it will experience a lorce given by Eq. (9.7), whose magnitude is

What is the direction of this force? The right-hand rule says that FI points towards the wire 2. This means that wire 1 is attri. ted towards wire 2.

Similarly, for currents in the opposite direction, you should be able to sliow that the wires repel each other. The rule is that parallel current attract and antiparallel current repel.

The force between current-carrying conductors forms the basis for the definition of the ampere. The ampere is that constant current which, if maintaillea in two straight parallel

I conductors of infinite length, of negligible circular cross-section, and placed one meter apart in vacuum, would produce on each of these conductors a force equal to 2 x newtons per meter of length.

In other words, suppose we have two straight pamllel coilductors of infinite length, of negligible circular cross- section, and placed one meter apart in vacuum. When constant

current is made to flow in both the conductors, it is observed that each of these conductors experiences a force. The constant current which produces force equaI ttr 2 x newtons per meter of length of the conductor is known as ampere.

9.6 PEWE'S LAW

In electrostatics, we used Coulomb's law to calculate the electric field due to an arbitrary charge distributions. Do you remember that in Block 1, Unit 2, we used Gauss's law to solve electric field problems of appropriately high symmetry with ease and elegance. There we made simple observatio~a about the number of lines of force emerging from closed'surface and then formulated Gauss's law.

In magnetism, the situation is similar. We can calculate the magnetic field caused by any current distributions using Biot and Savart law. Now the question is can we make a statement analogous to Gauss's law that would help us to calculate a magnetic field with similar ease and elegance? Gauss's law for the electric field relates the amount of charge enclosed by a surface'to the number of lines of force emerging frgm that surface. To quantify "number of lines of force", we introduced the concept of flux. Is there an analogous concept that would prove useful in describing the magnetic field due to a current? Yes, a clue to its nature comes from the fact that the magnetic field lines are closed loops surrounding a current.

Consider the magnetic field of a long, current-carrying wire as shown in Fig. 9.19. The magnitude of this field decreases inversely with distance from the wire as evident from Eq. (9.22). Draw an arbitrary closed loop around the wire. The quantity that will prove useful is the length'of this loop, weighted'at each point by that component of the magnetic field, which is in the direction of the loop. We call this quantity the circulation around the loop. Let us calculate the circulation for a circular loop labelled 1 that coincides with a field line as shown in Fig. 9.19. At all points 011 the circle labelled 1, the magnitude of the magnetic field B1 is given by Eq. 9.22, i.e.,

where I is the current in the wire and r, is the radius of this circle. Since the circular loop is always in the same direction as the field (because it coincides with a field line), the circulatioll becomes simply the loop circumference multiplied by the field strength.

#

This shows that circulation does not depend on the loop radius r l . Let us calculate circulation around another circular loop of radius r2 labelld 2. This loop also coincides with a field line. Therefore,

This shows that the circulation aqund iny field line is the same. It also shows that the circulation around any loop coi~iciding with a field line is proportio~~al to the current in the wire which is encircled by that field line.

What will be the circulation around the closed loop that does not coincide with a field line? Fig. 9.20a shows the two field lines (shown thin). Let us consider, the closed loop (shown by bold line ig.Fig. 9.20b) that encircles the wire but does hot coincide'with a

. , - ..o. 0

(8) @)

Fig. 9.20: (a) M-dc field hes due to currentvrrying wire) (b)A dosed loop that docs noi coincide with a single afcld lint, The clrmlatioo around this loop is Lbe same as the drculntion around a field he.

Fig. 9.19: Magnetic field lines ' surrounding a

current-carrying wire.

Electric Current and Magnetic Field

Fig. 9.212 Ampen's l a w is applied to an arbitrary AmpcrInn loop that encloses two long straight wires but excludes a thM wire. Note the directions of the currents.

single field line. To calculate the circulation around this loop, we move around the loop, taking the product of the distance moved with the field compol~ent in tlie direction of motion. When we move along the straight portion cu anid bd of the loop, it contributes nothing to the ci~culation around the loop because when we move at right angles to the field there is no field component in the direction of motion.

What about tlie contribution to the circulation from the arc ab that lies on the inner field line? It will be the same as the contribution we would get if we moved along the arc cd which is not the part of our loop. This is because (1) the circulation aiound the inner and outer field lines is the same as Eq. 9.27 or Eq. 9.28 shows and (2) the arc ab occupies the same fraction of the inner field line as the arc cd does of the outer field line. Thus the circulation around our closed loop is just what wewould get while going around the outer field line. Therefore, according to Eq. (9.210, circulation has the value pal. This observation shows that the circulation around any closed loop eneinrling s steady straight cmrl-ent is proportiannl to the current I encircled by that loop and Is given by p d .

This statement is a simplified version ofAmperefs law. This law is true for any type of current and ally closed loop, as long as the encircled current is steady (never chonging in time). If the current is not in a single wire, but in a number of wires, we simply add all the currents to obtain the net current encircled by our loop. If there are currents flowing in opposite directions, then we give the opposite signs to opposite directions of the current. The algebraic sum of currents encircled by the loop is the net current that determines the circulation around the loop. For a counter-clockwise traversal of the loop, currents pointing out of the loop are take11 as positive, those pointing inward being negative. To remember this convention the right-hand method will help you. If the fingers of your right hand (the curly elenlent) represent the direction of traversal around the loop, then your extended right thumb (the straight element) represents the positive direction for currents ellcircled by the loop.

SAQ 5

Fig. 9.21 shows thecross-sections of three loiig straight wires that pierce the plarie of the page at right angles to it. The wires carry culTerlts i ~ , i2 and i s in the directions shown. Find the net current encircled by the Amperian loop.

In talking about circulation, we must specify the sense-clockwise or counter-clockwise -in which we traverse the loop. We adopt the convention that dirculation is positive if, when we curl the fingers of our right hand around the loop, our right thumb points in the general direction of the net current encircled by the loop. Let us evaluate the circulation around an irregular loop L in a magnetic field as shown in Fig. 9.22. 7'

(a (b)

Fig. 9.22: (a) Ao Lmgulnr loop in a magnetic tiel$ (b) a magnified view shows that the cmtribullon to the eireulnlion h an inhnitesimd segment d of the loop is just B.d.

Let us examine a small part of the loop so that it is essentially linear and the magnetic field is essentially constant in magnitude and direction over it. This small segment of the loop is represented by dl whose magnitude dl is the length of the segment and whose direction is the local direction of the loop. Then the contribuiion dC from d to the circulation C around the loop is the length of dl weighted by the componerlt of the magnetic field in the direction of dl, i.e.,

where 8 is the angle between the field and the direction of the Ioop segment dl. The circulation around the loop is just the sum bf contributio~ls from all the segrneuts dl, i.e., C = 2 dC. As the segments get arbitrarily small, this sum becomes an integral for the total circulation, i.e.,

You must have encountered line integrals similar to this one in Block 1 of this course when electric potential was defined. Here line integral means just a sum of many dot products of the field with segments dl of the loop. The circle on the integral sign reminds that you are dealing with a closed loop.

Therefore, mathematically, Ampere's law can be expressed as follows

This statement is true for any arbitrary loop provided the current I is steady and it is the net current encircled by the loop.

SAQ 6

Apply Ampere's law qualitatively to the three paths shown in Fig.9.23.

I11 electrostatics, we deterlnined the strength of the electric field due to various charge distributions using Gauss's Law. However, we could use Gauss's law only for certain symmetrical charge distributions by constructing suitable closed surfaces in the electric field. We consider Ampere's law as playing the same role in steady state magnetism as Gauss's law played in electrostatics. We shall see that Ampere's law can be used to determine the magnetic fields only due to symnletric current distributio~ls. For this purpose, we have to co~lstmct suitable closed loops called Amperinn loops in the

magnetic field over which the line integral$l . dl is to be evaluated. We shall illustrate

this by few examples.

9.6.1 Applications of Ampere's Law

1) Magnetic field due to a long strsight.current-carrying wire

We shall determine the magnetic field at a distance r. from a long straight wire carrying a current I as shown in Fig. 9.24. The wire is cylindrically sym~ilctric, so that the magnitude of the magnetic field cannot depend on angular position around the wire. We also know that since there are no magnetic monopoles on the wire, the magnetic field lines cannot begin or end on the wire and go radially outwad. Hence the magnetic field lines must be closed loops. The only field lines that are both closed and exhibt cyclindrical symmetry are circles concentric with the wire. It is useful to &member the right hand method for determining the direction of B (for your benefit wc are giving it again in the margin remark).

Now we shall use Ampere's law to determine the magnitude of B at a distance r metres from the axis of a long straight wire of radius R ( r > R ) metres and carrying a current I amperes. Here we assume that r is small in compariso~~ with the length of the wire so that the wire can be considered to be infinitely long.

To evaluate the line'integral in Ampere's law, welmust find an amperian loop. Here field lines are themselves appropriate loops. We construct a circular path of radius r with its centre on the axis of the wire. From symmetry, the magnitude of the field on this circular path (or line) is constant. Here amperian loop coincides with a field line. So the field is everywhere in the same direction as the loop. Hence

so that B . k -

II I Fig. 9.23 :

Fig. 9.24 I Magnetic nelcl due to long straigl~t wire,

Right Hnnd methodif we grasp the w h with the right haud, tllc thumb pointing in the direction of Lhe current, thou the fingers will curl round the wire in the direction of D,

EZPdric Current and Mmgaetic Field

Example 3 A long cylindrical wire of radius R carries a steady current I which is uniformly distributed over its cross-sectional area. Determine the magnetic field at a distance r ( < R ) from the axis of the wire.

Solution

We first notice that the point at a distance r < R is inside the wire. In this case also, we assert that, by symmetry, B has a constant magnitude at all points on a path which is a circle of radius r with its centre on the axis of the wire and with its plane perpendicular to the axis of the wire. The direction of B at every point along this circle is along the tangent to the circle at that point. We choose this circle as the path of integration for the line integral in Ampere's law. Hence,

However, the current enclosed by this path is not I but the part of the current which passes through the m-section of area zr '. The mmnt = JL r x cumnt per unit area of ms-section

Hence,

B = - IQJ r 2 . n ~ ~

SAQ 7

Plot B as a function of r from the axis of the wire (of radius R ) to some distance outside it.

2) Magnetic field due to a solenoid

In Block 2, we have found that we can produce a uniform electric field between the two closely spaced, charged conducting plates of a capacitor. Is there an analogous device that will produce a uniform magnetic field? Yes, the device is solenoid. Let us see how it produces a uniform magnetic field. You know that when a current flows in a circular loop, the magnetic field is found to be directed as shown in Fig. 9.17. The field lines have been drawn using the right hand method. You observe that field lines circle the win. A solenoid can be thought of as a cylindrical stack of current-caving loops.

'I

Fig. 9.25; A ioosely wound,cail of win The magnetic lldd arising ltam c m n t io Ule rrirc b stroagcet witbin the coil. Ibe field is shown only in tbe phw of the pnge) &b am pdob when current emerges from that plane, rrcrsses where eumnt g a s into plane ofpage.

Fig. 9.25 shows a solenoid of four turns. Here the turn are loosely wound compared to the common solenoids. Close to any part of the wire are magnetic field lines encircling the wire. We show these field lines at the lop and bottom of the coil, where the wires cross the plane of the page. The net field anywhere is the vector sum of the fields of the

individual parts of the loop. You can see that inside the coil, the fields from elements of wire at the top and bottom have a component to the right, and so tend to reinfore. Above the top of the coil, the fields arising from elements at the top all have a component to the left, while fields from elements at the bottom have a component to the right, thereby weakening the net field. 'A similar weakening of the field occurs below the bottom of the coil. Hence the net field is strong and points to the right within the coil, and is weaker and points to the left outside the coil, as shown in the Fig. 9.25.

Suppose the coil is tightly wound and its length is longer than its diameter, as suggested in Fig. 9.26. In such situation the field is still stroiig illside the coil of the solenoid, and as the individual turns get arbitrarily close, the irregularities in the field disappear, giving straight field lines inside the solenoid.

What about the field lines outside the solenoid? The exterior field lii~es must connect the field lines emerging from the right of the solenoid to those going into the left because field lines cannot begin or end. The field lines close to the solenoid axis bend very gradually, and spre'ad far from the solenoid before they return to the other end.

To find the value of B inside a solenoid by use of Ampere's law we must note two points: (i) the magnetic field is directed lengthwise along the axis of a tightly wound, long solenoid; (ii) if the solenoid is long, the field lines emerging from the end of the solenoid will fan out widely as they come back around to enter the other end. This indicates that the magnetic field outside the solenoid is many times weaker than it is inside. Consequently, we approximate the situation and consider the field outside thc solenoid to be negligibly small. We will apply Ampere's law to calculate the field within the solenoid. Consider a closed linear path PQRS as shown in Fig. 9.27. For this path,

The integrals over QR and SP are zero as for part of these paths (outside the solenoid)

Fie. 9.272 Cross stclloo of ? long eolenold, showGg s rectangular rmpcdau loop. . B = 0 and for another part inside the solenoid B is perpendicular to 4. The integral over RS is zero as B - 0 outside the solenoid along this path. The only integral that is different from zero is overPQ. Hence

Q For this path, B is along the direction of the path, which means $ B . dl = B . d so that

v BL

where L is the length of the path PQ. 1f thi's path encloses N turns of wire of the

Magnetic Ficlld

Fig. 9.26 : A longer ceutrd section of n long, more tightly wound mlenoid.

Elechic Current nnd Mngnetic Field

solenoid each carrying a current I, then the right hand side of Ampere's law is l-4~ NI. Finally, we have

where FZ is the number of turns per unit le~~gtli of the solenoid.

The fornlula given by the Eq. (9.34) derived for an infinitely loug solenoid, holds quite well for actual solenoids, for points well inside the sole~ioid away from its ends. Note that ]B does not depend upon the positio~~ of the point within the solenoid as long as we are far away from the ends of the solenoid. Therefore, we coilclude that B is uniform over the cross-sectio~i of the solenoid. This proves to be a practical way to set up a known uniform rnagletic field for experimeiital purposes.

3) Magnetic field Inside a toroid

If a solenoid is bent illto the form of a circle so as to join its two'ends, one obtains a toroid as shown in Fig. 9.28. Tlie field lines of B inside the toroid, by symmetry, are circular and concentric wit11 the center of the toroid. Also the magnitude of field is coilstant along any field line. Choosing our alnperian loop to be a circle of radius r that coincides with a field line, we call readily calculate the circulation around this loop:

~ i g ; 9.28 : Symmetry requires that the field lines be circular. Also shown is an amperia Imp (bold) for use in calculathg the Ileld

Here we could evaluate the line integral because B is constarit on the a~nperian loop and the loop coincides with a field line. As a result the integral is just the field strength times the circumference 21dr of the loop. How much current is encircled by the loop ? If the toroid consists ofN tunis, and carries a current I, the11 a11 alnperiail loop inside the tomid coil encircles a total current NI. This is because each tun1 carries current in the same direction through the path we have chosen. Usi11g Ampere's law to relate this current to the circulation, gives

~ I ' B = p&TI,

so that

This result holds when our amperiaii loop is within the toroid itself. On the other hand, if our amperian loop is inside the inner edge of the toroidal coils, there is no current encircled, and the magnetic field is zero. If the amperian loop is outside the outer edge of the coils, it encircles equal but opposite currents, again giving zero field. From Eq. (9.35) we see that B is not constant over the cross-section of the toroid unlike the stmiglit solenoid.

Magnetic Field

There is another way to express Ampere's law. Eq. (9.31) gives the integral form of this law. Zn this section, we will transform the integral form into the differential form viz:

GurlB = w j (9.36)

where j is the current density.

Let us consider the line integral$^. d around a closed pitlie, as shown in Fig. 9.29a. Here we have replaced the magnetic field, B by a general vector field F. The closed path C call he visualized as the hou~idary of some surface S which spans it. In the integrand, dl is the element of path which is an infinitesimal vector locally tangent to C. - - Now the closed path C is divided into two, thus making two loops C1 and C2, (Fig.9.29b). Take the line integral around each of these in the same directional sense. The sum of the circulatioils around C1 and Cz will be the same as original circulation around C. The reason is that the extra co~itribution from the bridge dividing the original F surface area into two parts cancels out since the two line integrals over the bridge have (a)

the same but opposite signs. The same reasoning is applied, no matter how many sub-divisions are made. Further sub-division into many loops C1, Cz, . . ., Ci, . . .. (Fig. 9.29~) leaves the sum unchanged, i.e.,

$ P . ~ - $ F . d l ~ t $ F . & + . . . C c1 cz

where each tenn in the sum of circulations around subareas of the original area is indicated .by a particular value of the subscript i. If we conti~iue to subdivide the whole loop indefinitely, then in the limit, we amve at a quantity, characterislic of the field F in a local neighbourhood. When we sub-divide the loops, we not'only make loops with s~naller circulation, but also with smaller area. So il is-natural to consider the ratio of the loop circulation to the loop area, just as we considered ratio of flux to volume in Unit 2 of Block 1. The area ai of a bit of surface that spans a small loop Ci is a vector, hence, the surface has an orientatioii in space. As we make loops smaller and smaller in some neighbourhood, we get a loop oriented in any direction we choose. Let us choo%e some particular orientation for the loop as it is sub-divided finally. If the unit vector n denotes the normal to the loop then it has to remain fixed in direction even if the loop su~ounding a particularpoint P shri~iks down towards the zero size. The direction of n, according to the right hand screw method, is as shown in Fig. 9.30. The limit of tlie ratio of circulatio~l to loop area will be written as follows:

SGF . mi lim - (9.38)

Oi-. 0 ai

Fig. 9.29: a) The circulation or F around the curve C is . the Line iutegral d R , the tangential component of F, b) The circulation around Ihe whole loop i s lhe sun1 ut the drculaiions around the two loops. c) When the whole loop b divided Into a number d snlall loops, fhe cirmlntion around the whde Iwp is the sum of (he clrcul~Uons around the little loops.

The limit obtained in the Eq. (9.38) is a scalar quantity which is associated with the point P in the vector field F and with direction I?. If the directions are towards 2, y^ and then we get three different numbers. It turns out that these numbers can be coilsidered as components of a vector called curl F. That is, the number we get for the limit with d in a particular direction is the component, in that direction, of the vector curl F.

( curl F ) . n" = lim - (9.39) oi - 0 ai

FIE. 9.301 Right-hand-smw Let us write again the expression for the circulation around the original path: relalion between the

N N surface normal md tbe

$ , ~ . r n - ZJ F e d a 1 . i (9.40) direction drculption lo line which Antegrad the

i - 1 '' i- 1 Is Wrca

III the last step we merely multiplied and divided by a;, In the.right hand side suppose N is made enormous and all the a'i's shrink. Then according to Eq. (9.40) the quantity in parentheses becomes ( curl F).$i where fii is the unit vector normal to the ith.100~. So, we have on the right hand the sum of the product "loop area times the normal

39

connporlent of curl F ". The sum is over all loops that make up the entire surface S ~~ ,aer~ l ing C. This is equal to the surface integral of the vector curl F over S, Thus

F . d i

.i ai ] ai ( curl F . i i i- 1 i - 1

We thus find that

where dS is the area enclosed by the closed path. Now using Eq. (9.42) we can write Eq. (9.31) as follows:

Since I = SJ . dS we write Eq. (9.43) as

Since d§ is nonzero, the quantity within brackets must be zero. INrlB;] (9.44)

which is the differential form of Ampere's' law. The relation given by Eq. (9.42) is known as Stokes Theorem. I

We now use the definition of the curl (Eq. 9.39) to find an expression for the curl in terms of cartesian co-ordinates. We would learn how to calculate curl F when the vector function F ( x, y, z ) is explicitly given. Since the curl of a vector isitself a vector we

will find its expression by finding its three mutually perpendicular components. Let us start by a calculation of the circulation about a path of very simple shape that encloses a rectangular patch of surface in the xy-plane as shown in Fig. 9.31. That is, we am taking n̂ 3 2. m e patch is considered to be ve j small so that vector F. does not change much along any one side of the rectangle. In Fig. 9.32 we look down on to the rectangular patch from above. To calculate the circulation we start at the point ( x, y ) - t h e lower left corner of the figure--we go around in the direction indicated by arrows. Along the first side which is marked (I), the tangential component is Fx (1) and the distance is Ax. Hence the first part of the integral is Fx ( 1 ) A x. Along the second side, marked (2), we get Fy ( 2 ) A y , Along the third we get - Fx (. 3 ) A x and along the f~ur th - Fy ( 4 ) A y. Qe minus sign is introduced because the tangential component has to be in the direction oftravel. The wlqle line integral is then

Fig. 9.32: Looking down on the patch in Fig. 9.31.

Now let us consider the first and third terms. Together they are

As we have assumed the patch to be infinitesimally small, the difference is zero. This is true to the first approximation. But we can be more accurate by taking into account the rate of change of F,. When we do so, we write

(See also Unit 2 of Block 1 .)

If we had considered the next approximation, it would have involved t e r m in ( A y ) also. But since we will ultimately think of the limit as A y - 0, such terms can be neglected, Using Eq. (9.47) in (9.46) we find

Similarly, for the other two terms in the circulation, we can write

Thus, the line integral around the whole rectangle is

Now Ax A y is the magnitude of the area of the enclosed rectangle wlrich we have represented by a vector in the z - direction.

Hence the quantity ( ---Y- at - - z ) is the limit of the ratio :

line iitegral arouhd patch as the patch shrinks to zero si72.

Area of patch

If the rectangular patch had been oriented with its nonnal in the positive y-direction like shown in Fig. 9.33, we would have found the expression

for the limit of the corresponding ratio.

If the patch had been oriented with its normal in the positivex-direction as shown in Fig. (9.33), we would have obtained

Thus curl F is given by the following expressions:

Magnetic Field

Flectric Cumat and Magnetic Field

FiS. 9.33: For each aritntatlon, (he Urnil of tbc nth dcircdatioa/m gives a componzent of F d ahnt point. H a e the patches hnve been shown sepamted. In arturl dl the patches should dwter a m w ~ d the polnt where curl F is king detenui~~ed

Now that we have finished our discussio~l of the magnetic field generated by a steady current, we turn again to the study of forces on conductors carrying currents. The force on a currentcarrying wire is the basis for many practical devices, including electric

- motors that start automobiles and run refrigerators. In this section, you will know that the working of galvanometer, which is the most important current measuring

.instrument, depends on the action of the magnetic field in exerting a torque on a current loop. Many electromechanical devices make use of the fact that a current-carrying coil of wire is caused to rotate by a magnetic field. When a current loop is placed in a uniform magnetic field as shown in Fig. 9.34, it is acted upon by equal and opposite forces and having same line of action. Therefore, the total force on the current loop is zero. But, a toque does exist on such a coil which can make it rotate. This $ easily seeh in Fig. 9.34.

If you apply the right-hand rule to the wires of the loop shown in Fig. 9.34 you will notice @e following: The forces on the upper and lower sides of the loop (not shown in Figure) are parallel to the axis of rotation and are equal and opposite. They cannot cause any rotationeither. However, the forces hat act on the sides of the loop can indeed cause it to turn. The turning effect zero when the coil is in the position shown in Fig. 9.34a, a torque is seen to exist for the position shown in Fig. 9.34b. Even though the magnitude of forces FI and Fi are same in the two casesF the lever arm from the axis is zero in (a) while non-zero in @). Let us now find the quantitative relation for the torque.

Consider the rectangular loop PQRS carrying a current I and immersed in a uniform magnetic field B as shown in Fig. 9.35a. Let PQ RT = 1 and QR - SP - b. The vertical sides PQ and RS of the loop are perpendicular to the magnetic field. Therefore, the magnitude of the force on these sides is given by

Fi = F2 = 113.

These two forces are equal, parallel and oppositely directed, and hence they form a couple.

42

(a) Parapoaivc view (b) Top 6~

Fi~9.35: Torque on the coil. i

Suppose, at any instant, the axis of the loop NN' (axis normal to the plane of the loop) makes an angle 0 with the magnetic field as shown in Fig. 9,35b. Then at that instant the magnitude of the torque z' due to forces F1 and W2 is given by

z1 .. F1 ( - Fz. ) x perpendicular distance

But 1 x b = A (area ofthe loop)

:. z' - L4B sill 0

This toque acts on every turn of the coil. If there are N turns, the total toque-c is

The quantities in parentheses are grouped together because they are all properties of the coil viz., its number of turns, its area and the current it carries. Eq. (9.53) tells us that a current carrying coil placed in a magnetic field will tend to rotate. We can express the torque in vector notation if A is defined as a vector such that its magnitude is the area ab of the loop and its direction is along the perpendicular to the plane of the loop. The direction of A is given by the right hand rule: Wrap your fingers around the loop in the direction of the current; your thumb then points in the direction of A. 'Ihen we can write

Eq. (9.54) should remind you of Eq. (3.40) for the torque on an electric dipole in an electric field. Eq. (3.40) is

with p the electric dipole moment and E the electric field. Comparison yith Eq. (9.54) suggests that a current loop in a magnetic field behaves analogously to all electric dipole in an electric field. The quantity NIA is called the magnetic dipole moment of the current loop. That is

p = NIA (9.55)

Magnetic dipole moment is a vector quantity and for a current loop its direction is along the direction of A. Using Eq. (9.59, the torque on a current loop can be written as

t = V X B (9.56)

The torque tends to align the magnetic moment with the field.

SAQ 8

A circular loop of radius 5.0 cm consists of 10 turns of wire, A current of 3.0 A flows in the wire. What is the magnitude of the loop's magnetic moment? Suppose initially the

Electric Current and Magnetic Field

magnetic moment is aligned with a uniform magnetic field of 100 Gauss. Now the loop is turned 90" from its original orientation. How much torque is required to hold the loop in its new orientation?

Eq. (9.56) shows that if a current carrying loop is placed in a fixed magnetic field then the torque experienced by it will depend on the current flowing in the loop. This is the principle of the galvanometer.

9.8 SUMMARY

e A long straight wire carrying a currentZ through a uiliforln lnag~ietic field B experiences a force due to the action of the field on the moving electric charges that constitute the current. The force on a section of the wire of length 1 is given by

F = Z l x B

where I is a vector of magnitude 1, pointing in the direction in wliicll the current flows along the wire.

e A magnetic field B is said to exist in any region in which a rlloving charge experiences a force that depends on its charge, its velocity v and the magnetic field. If B and v make an angle 8, the force on the moving charge is given by

F = q v x B

or F = qvB sin 0

e The direction of force on currents and moving charges is given by the right hand rule as shown in Fig. 9.7. It states that if one's right hand is held flat with the fingers pointing in the direction of the field lines and the thumb pointing in the direction of the current, then the palrn of the hand will push in the direction of the force.

Q Gauss's law for magnetism relates the number of magnetic field lines emerging from a closed surface to the net magnetic monopole enclosed. Siilce magiletic monopoles do not exist, Gauss's law for magnetism says that the magnetic flux through any closed surface is zero

This shows that magnetic lines have no beginning or end rather they formclosed loops.

r A current gives rise to a magnetic field. The magnetic field set up by a current carryiiig conductor can be found from Biot-Savart law:

where dS is the contribution to the field from a current I flowing along an infinitesimal vector dl . is a constant whose value is 4n x 10- NA-~ and f is a unit vector from the current element Id1 towards the poini where the field is being calculated.

The direction of the magnetic field is given by means of right' hand method as illustrated in Fig. 9.13. 8

e The magnetic field at a point which is at a distance of r from a long straight wire carrying a cumrit I is given by

e The magnetic field at a point along the axis of a circular loop carrying current is given by

Magnetic Field

where a is the radius of the circular loop carrying current I and bis the distance of the point (along the axis of the loop) from the centre of the loop.

When the point is far from the loop then

2 where A = rn . This shows that current loop behaves like a magnetic dipole.

Two parallel wires carrying currents in the same (or opposite) d i~ct ion attract (or repel) each other. If these two wires are separated by a distance d in a vacuum, then the force (q of attraction (or repulsion) on a segment of length I of either wire is given by

where I1 and I 2 are the currents flowing in the two wires. The force between two current carrying wires is used to define the ampere which is the basis of the whole system of electrical units.

Another way of finding the magnetic fieId in symmetrical cases is by use of Amperes law. This law relates the circulation (or line integral) of the magnetic field around an a&itraly closed loop to the cumnt encircled by the loop, i.e.,

It is used to calculate the magnetic field in situ,ations with high symmetry.

r A solenoid is a long cylindrical coil having many turns of wire, Inside the solenoid, there is a uniform field given by

B = W J where n is the number of turns per unit length.

r The field inside a toroidal coil is given by

MI Be- 2m

where r is the distance from the aytre of the toroid andN is the total number of turns wound on the toroid.

Differential form of ampere's law is

curl B = p~ J

where J is the cumnt density at a given point.

Stoke's theorem states that :J curl F . dS = SF. dl. S

o A closed current loop in a magnetic field behaves like a magnetic dipole with magnetic dipole moment ( p )

p = NIA

whereN is the number of turns in the loop, I the loop current and Avector perpendicular to the plane of the loop with magnitude equal to the loop area. The torque (z) experienced by such current loop is given by

z = p x B

This torque is the operating principle in ammeters, voltmeters and electric motors.

1) In Madras, ti= horizontal component of the earth's field is 3.6 x ~ b m - ~ . ~ f a vertical wire carries a current of 30A upward there, what is the magnitude and direction of the force on lm of the wire?

2) Find the force on each segment of the wire shown in Fig. 9.36, if B = 0.15 T. Assume that the current in the wire is 15A. ( sin 65" = 0.9063 )'

C

P

P i

B P

F

3) Fivevery long, straight, insulated wires are closely bound together to form a small cable. Currents carried by the wires are ill 5 204 I2 = - 6 4 13 = 1 2 4 14 = - 74 Is = 18A (negative currents are opposite in direction to the positive). Find B at a distance of 10 cm from the cable.

4) In the Bohr model of the hydrogen ato~n the electron follows a circular path centeredon the nucleus. Its speed is v and the radius of the orbit is r.

a) Show that the effective current in the orbit is e v / a r.

b) Show that p = - ( e/2m ) L , where L = rnr v is the angular momentum of the electron in its orbit.

9,10 SOLUTIONS AND MSWERS

SAQs

1) In most motors current in a wire sets up a magnetic field, and tlie magnetic field, in turn, exerts a force on a second current-carrying wire causing the shaft to rotate.

2) The pairF and v, and F and Bare always at right angles. Vectors v and B may have any angle between them.

3) Refer to Fig. 9.37, applying the right hand method shows that directly above the I wire B points north; and directly below it points south.

4) Both are inverse square laws. In Coulomb's law electrical force acts along r on Fig. 9.37 stationary charge. In Biot-Sava~t law, magnetic force acts perpendicular to r .

5) Net current i = il - i2. Current i3 lies outside the loop and is not encircled by it helice it is not included in calculating i.

6) ~ o r ~ a t h ~ o . 2 $ B . d = 0s-ntcunedbzem.~or~~~o.1and3$~.d - mi. k

7) Inside the wire, the field increases linearly with distance from the w i k axis. O m L

- Inside Ouuidc

, r --,

b W 8 r ' I h c m r g n d k W d M ~ d ~ b a t e . ' I b e d m B F k l B n t t b c b d t b e w i P .

46 1

we reach the surface bf the wire the field begins to decrease inversely with distance. Fig. 9.38 shows the rough plots of the field strengths both inside and outside a wire.

8) As described by Eq. (9.55), the magnetic moment p, is given by

The magnitude of the torque needed to hold the new orientation is given by Eq. (9.56) as

= 2.4 x Nm.

Terminal Questions

1) The vertical component of B is parallel to the current and does not contribute to the force; therefore, - F - EBH = ( 3 0 A ) ( 1 m ) (3.6 x 10-5 ~ b m - ~ ) - 10.8 x I O - ~ N , west.

2) For each straight segment F = I L x B, where E is the directed line segment. In sections AB and DE, L and IB are parallel, so sin 0 3 0, and F = 0. In section BC, F = ILB - ( 5 A ) ( 0.16 m ) ( 0.15 T ) - 0.12 N, into page, In section CD,F = (5A)(0 .20M)(O. l5T)s in65" = 0.136N,outofpage.

3) By superposition the field is just the sum of the fields due to the individual currents. At r = 10 cm all the fields are either parallel or antiparallel as the currents are parallel or antiparallel. Then

4) a) Since charge e passes a point once every revolution, I e/T, whereT = (2nr) /v ,sothat I = (ev)/(2nr).(b)In magnitude, the dipole moment is

and because the electron is negatively charged, p is antiparallel to L.

Magnetic Field