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Motivating the Magnetic Field Concept

Motivating the Magnetic Field Concept
UNIT III UNIT III Static Magnetic Fields Static Magnetic Fields B.Hemalatha APECE 1Motivating the Magnetic Field Concept: Motivating the Magnetic Field Concept: Forces Between Currents Forces Between Currents Magnetic forces arise whenever we have charges in motion. Forces between current carrying wires present familiar examples that we can use to determine what a magnetic force field should look like: Here are the easilyobserved facts: B.Hemalatha APECE 2BiotSavart Law BiotSavart Law • The current flowing in any path, can be considered as many infinitesimal current elements, each of length dl, flowing in a magnetic field dB, at any point P ˆ Idlr 0 dB 2 4r B.Hemalatha APECE 3Idl sin 0 • The magnitude of dB is: dB 2 4r where θ is the angle between dl and r ˆ Idlr 0 BdB • The total magnetic field at point P is: 2  2r • This is equivalent to Coulomb’s Law written in differential form: 1dq dE 2 4r 0 B.Hemalatha APECE 4Magnetic Field due to a Current Carrying Wire BiotSavart Law Hans Christian Oersted, 1820 B.Hemalatha APECE 5• Magnetic fields are caused by currents. • Hans Christian Oersted in 1820’s showed that a current carrying wire deflects a compass. No Current in the Wire Current in the Wire B.Hemalatha APECE 6Right Hand Curl Rule Right Hand Curl Rule B.Hemalatha APECE 7Magnetic Fields of Long CurrentCarrying Wires B =  I o 2r I I = current through the wire (Amps) r = distance from the wire (m)  = permeability of free space o 7 = 4 x 10 T m / A B = magnetic field strength (Tesla) B.Hemalatha APECE 8Magnetic Field of a Current Carrying Wire B.Hemalatha APECE 9What if the currentcarrying wire is not straight Use What if the currentcarrying wire is not straight Use the BiotSavart Law: the BiotSavart Law: Assume a small segment of wire ds causing a field dB: ˆ  dsr  0 dBI 2  4  r Note: dB is perpendicular to ds and r B.Hemalatha APECE 10BiotSavart Law allows us to calculate the BiotSavart Law allows us to calculate the Magnetic Field Vector Magnetic Field Vector • To find the total field, sum up the contributions from all the current elements I ds ˆ  dsr  0ii BI  2  4  r i • The integral is over the entire current distribution ˆ μIdsr o B 2  4πr B.Hemalatha APECE 11Note on BiotSavart Law Note on BiotSavart Law • The law is also valid for a current consisting of charges flowing through space • ds represents the length of a small segment of space in which the charges flow. • Example: electron beam in a TV set B.Hemalatha APECE 12Comparison of Magnetic to Electric Field Magnetic Field Electric Field 2 2 • B proportional to r • E proportional to r • Vector • Vector • Perpendicular to F , ds, r • Same direction as F B E • Magnetic field lines have no • Electric field lines begin on beginning and no end; they positive charges and end on form continuous circles negative charges • BiotSavart Law • Coulomb’s Law • Ampere’s Law (where there • Gauss’s Law (where there is is symmetry symmetry) B.Hemalatha APECE 13Derivation of B for a Long, Straight CurrentCarrying Derivation of B for a Long, Straight CurrentCarrying Wire Wire Integrating over all the current elements gives ˆ ˆ dsrdx sin θ k  θ μI 2 o Bsinθ dθ  θ 1 4πa μI o cosθcosθ  1 2 4πa B.Hemalatha APECE 14If the conductor is an infinitely long, straight If the conductor is an infinitely long, straight wire, = 0 and =  wire, = 0 and =    • The field becomes: a μI o B 2πa B.Hemalatha APECE 15B.Hemalatha APECE 16B for a Curved Wire Segment B for a Curved Wire Segment • Find the field at point O due to the wire segment A’ACC’: B=0 due to AA’ and CC’ Due to the circular arc: ˆ μIdsr o B 2  4πr μI o Bθ 4πR q=s/R, will be in radians B.Hemalatha APECE 17B at the Center of a Circular Loop of Wire B at the Center of a Circular Loop of Wire μIμI oo Bθ 2π 4πR 4πR Consider the previous result, with  = 2 μI o B 2R B.Hemalatha APECE 18Note The overall shape of the magnetic field of the circular loop Is similar to the magnetic field of a bar magnet. B.Hemalatha APECE 19B along the axis of a Circular Current Loop B along the axis of a Circular Current Loop • Find B at point P ˆ μIdsr o B 2  4πr 2 μIR o If x=0, B same as at center of a loop B x 3 2 2 2 2xR  B.Hemalatha APECE 20If x is at a very large distance away from the loop. xR: 2 2 μIRμIR oo B x 3 3 2 2 2 2x 2xR  B.Hemalatha APECE 21Magnetic Force Between Two Parallel Conductors The field B due to the current 2 in wire 2 exerts a force on wire 1 of F = I ℓ B 1 1 2 μI o 2 B 2 2πa μII o 1 2 F 1 2πa B.Hemalatha APECE 22Magnetic Field at Center of a Solenoid B =  NI o L N: Number of turns L : Length n=N/L L B.Hemalatha APECE 23Direction of Force Between Two Parallel Conductors If the currents are in the: –same direction the wires attract each other. –opposite directions the wires repel each other. B.Hemalatha APECE 24Magnetic Force Between Two Parallel Conductors, F B FμII Bo 1 2 Force per unit length:   2πa B.Hemalatha APECE 25Definition of the Ampere • When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 7 m is 2 x 10 N/m, the current in each wire is defined to be 1 A B.Hemalatha APECE 26Definition of the Coulomb • The SI unit of charge, the coulomb, is defined in terms of the ampere • When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C B.Hemalatha APECE 27BiotSavart Law: ˆ  dsr  0 Field produced by current carrying wires dBI 2  4  r I 0 B – Distance a from long straight wire 2a I 0 B – Centre of a wire loop radius R 2R NI 0 B – Centre of a tight Wire Coil with N turns 2R FII 0 1 2 • Force between two wires  l 2a B.Hemalatha APECE 28Differences between Ampere’s law the Biot Savart Law • The difference between Ampere’s Law and the Biot BdlI Savart Law is that in Ampere’s Law ( 0 e n c l), the  magnetic field is not necessarily due only to the current enclosed by the path of integration, as Ampere suggests • In the BiotSavant Law, dB is due entirely to the current element I·dl. To find the total B, it is necessary to include all currents B.Hemalatha APECE 29Magnetic Field Due to Current in a Straight Wire • To find the magnetic field near an infinitely long, straight wire, carrying a current I, the Biot Savant Law gives us: y Idysin 0 B 2  4r y • The solution of this integral yields: I 0 B 2R • This is the same as Ampere’s Law B.Hemalatha APECE 30The Magnetic Field OnAxis of a Current Loop • To find the magnetic field OnAxis, of a Current Loop, applying the BiotSavant Law yields: Idl sin  0 ˆ dB , 90becausedlr 2 4R Idl 0 dB 2 4R II 0 0 Bdl2R 2 2  4R 4R I 0 B 2R B.Hemalatha APECE 31Magnetic Field of a Loop at a Point OnAxis • B = 0 is from the symmetry of the y problem • B is calculated using the Biot Savant Law: z Savart Law 2  2RI 0 B z 3 2 2 2 4 zR • For zR then the above relationship simplifies to: 2  2RI 0 B z 3 4 z B.Hemalatha APECE 32ˆ r Magnetic Field of a Wire Segment  • Again: , the BiotSavant Law ˆ  90becausedlr gives: Idl 0 dB 2 4R ˆ r • Solving the Integral yields: dB I 1 0 B 8 R B.Hemalatha APECE 33Magnetic fields of long wires • Example for two wires. Find B at P , P , and P . 1 2 3 B.Hemalatha APECE 34Magnetic fields of long wires • P :  I/2d) +  I/2(4d) =  I/8d 1 0 0 0 • P : + I/2d) +  I/2(d) = + I/d 0 0 0 2 • P : +  I/2d)  I/2(d) =  I/3d 0 0 0 3 B.Hemalatha APECE 35Ampere’s law (general statement) • General statement of Ampere’s law B.Hemalatha APECE 36Ampere’s law (special case) • Ampere’s law for a circular path around a long straight conductor. B.Hemalatha APECE 37Applications of Ampere’s Law   Ampère’s law in differential form  B(r) J(r) 0   ( B) da B d J da I Ampère’s law in integral form 0 0 enc  Example 1  Find the magnetic field of an infinite uniform surface ˆ K Kx current , flowing over the xy plane B.Hemalatha APECE 38Example 2 Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R and carrying a steady current I.     Bds B NI Bds Bds B ds B   0 side1 side1 Where N is the number of turns in the length N B I nI 0 0  B.Hemalatha APECE 39Example 3 Find the magnetic field of a toroidal coil, consisting of a circular ring around which a long wire is wrapped.   Bds B ds B(2r) NI  0  NI 0 B 2r B.Hemalatha APECE 40Applications of Ampere’s Law Problem 1 A steady current I flows down a long cylindrical wire of radius a. Find the magnetic field, both inside and outside the wire, if (a). The current is uniformly distributed over the outside surface of the wire. (b). The current is distributed in such a way that J is proportional to s, the distance from the axis. a I (a)  B d 2sB I 0 enc   0 for s a   B  I  0 ˆ  for s a  2s B.Hemalatha APECE 41Applications of Ampere’s Law a 3 2ka 3I (b) J ks I Jda ks(2s)ds k  3 3 2a 0 3Is J ks B d 2sB I 3 0 enc  2a For s a s 3  3Is 3 I 1 Is 3 0 0 B d 2sB I 2sds s 0 enc 0 3 3 3  2a a 3 a 0 2  Is 0 B 3 2a For s a a  3Is 3 I 1 3 0 B d 2sB I 2sds a I 0 enc 0 0  3 3 2a a 3 0  I 0 B 2s B.Hemalatha APECE 42Comparison of Magnetostatics and Electrostatics The divergence and curl of the electrostatic field are  1  Gauss’s law  E    0    E 0  The divergence and curl of the magnetostatic field are    B 0   Ampère’s law  B J  0 “The electric force is stronger than the magnetic force. Only when both the source charge and the test charge are moving at velocities comparable to the speed of light, the magnetic force approaches the electric force.” B.Hemalatha APECE 43Magnetic Vector Potential   V: electric scalar potential  E 0 EV You can add to V any function whose gradient is zero (V f )VfV   A : magnetic vector potential  B 0 B A  A You can add to any function whose curl is zero   (A) A A   2  B ( A)( A) A J B A 0  We will prove that   A  0 : B.Hemalatha APECE 44Magnetic Vector Potential  Suppose that our original vector potential is not divergenceless A 0   A 0 0  Because we can add to any function whose curl is zero A 0   2 A A  A A A 0 0 0 If a function  can be found that satisfies  2  A A 0 0  2  V Mathematically identical to Poisson’s equation  0 In particular, if  goes to zero at infinity, then the solution is 1 V d'  4 0 B.Hemalatha APECE 45 By the same token, if goes to zero at infinity, then  A 0  1 A 0  d'  4 It is always possible to make the vector potential divergenceless   A 0   2 2 so  B( A) A A J 0 This again is a Poisson’s equation  Assuming goes to zero at infinity, then J     J(r') 0 A(r) d'  4 For line and surface currents        K(r')  I(r') 0 0 A(r) da' A(r) d'   4 4 B.Hemalatha APECE 46Magnetic Vector Potential Example 1 A spherical shell, of radius R, carrying a uniform surface charge , is set spinning at angular velocity . Find the vector potential it  produces at point . r  r  The integration is easier  if we let lie on the z r  r' axis, so that  is titled at an angle . B.Hemalatha APECE 47Magnetic Vector Potential       K(r') 2 2 0 where A(r) da' Kv R r 2Rr cos'  4 ˆ ˆ ˆ x y z  v r'sin 0cos R sin'cos' R sin'sin' R cos'  R(cos sin'sin')x ˆ (cos sin'cos' sin cos')y ˆ (sin sin'sin')z ˆ 2 2  ˆ Because We just consider vRsin cos' y sin'd' cos'd' 0  0 0  (Rsin cos') 2 0 ˆ A(r) R sin'd'd'y  2 2 4 R r 2Rr cos'  3  Rsin sin'cos' 0 ˆ  ( d')y  2 2 2 R r 2Rr cos' 0 B.Hemalatha APECE 48Magnetic Vector Potential Letting u  c o s  ' , the integral becomes  1 sin'cos' u I d' du  2 2 2 2 R r 2Rr cos' R r 2Rru 01 1 u 1 2 2 2 2 3/ 2   R r 2Rru (R r 2Rru) 2 2 Rr 3R r 1 1 2 2 (R r Rru) 2 2   R r 2Rru 2 2 3R r 1 1 2 2 2 2  (R r Rr) R r (R r Rr)(R r) 2 2 3R r  2r I If the point lies inside the sphere, Then R r.  r 2 3R  2R If the point lies outside the sphere, Then R r.  I r 2 3r B.Hemalatha APECE 49Magnetic Vector Potential  3 3   Rsin sin'cos' Rsin 0 0 ˆ ˆ A(r) ( d')y Iy  2 2 2 2 R r 2Rr cos' 0  2r I If the point lies inside the sphere, Then R r.  r 2 3R  2R If the point lies outside the sphere, Then R r.  I r 2 3r  ˆ ( r)r siny Noting that For the point inside the sphere 3 3   Rsin Rsin 2r R 0 0 0 ˆ ˆ A(r) Iy y ( r) 2 2 2 3R 3 For the point outside the sphere 3 3 4   Rsin Rsin 2R R 0 0 0 ˆ ˆ A(r) Iy y ( r) 2 3 2 2 3r 3r B.Hemalatha APECE 50Magnetic Vector Potential   We revert to the original coordinates, in which coincides  with the z axis and the point is at (r,,) r For the point inside the sphere   R R 0 0 ˆ A(r) ( r) r sin 3 3   2 R 2 R 0 0 ˆ ˆ ˆ B A(r) (cosr sin) z 3 3 The magnetic field inside this spherical shell is uniform For the point outside the sphere 4 4   R R 0 0 ˆ A(r) ( r) sin 3 2 3r 3r B.Hemalatha APECE 51Magnetic Vector Potential Example 2 Find the vector potential of an infinite solenoid with n turns per unit length, radius R, and current I     I(r') 0 We cannot use because the current itself extends A(r) d'  4 to infinity.   A d ( A) da B da Notice that   ˆ B nIz Since the magnetic field is uniform inside the solenoid : 0      nIs 2 0 ˆ A d A(2s) B da nI(s ) A For s R 0  2 For an amperian outside the solenoid 2    nIR 2 0 ˆ For s R A d A(2s) B da nI(R ) A 0  2s B.Hemalatha APECE 52Motivating the Magnetic Field Concept: Forces Between Currents Magnetic forces arise whenever we have charges in motion. Forces between currentcarrying wires present familiar examples that we can use to determine what a magnetic force field should look like: Here are the easilyobserved facts: B.Hemalatha APECE 53Magnetic Field The geometry of the magnetic field is set up to correctly model forces between currents that allow for any relative orientation. The magnetic field intensity, H, circulates around its source, I , 1 in a direction most easily determined by the righthand rule: Right thumb in the direction of the current, fingers curl in the direction of H Note that in the third case (perpendicular currents), I is in the same direction as H, so that their 2 cross product (and the resulting force) is zero. The actual force computation involves a different field quantity, B, which is related to H through B =  H in free space. This will be taken up in  a later lecture. Our immediate concern is how to find H from any given current distribution. B.Hemalatha APECE 54Magnetic Field Arising From a Circulating Current At point P, the magnetic field associated with the differential current element IdL is To determine the total field arising from the closed circuit path, we sum the contributions from the current elements that make up the entire loop, or The contribution to the field at P from any portion of the current will be just the above integral evaluated over just that portion. B.Hemalatha APECE 55Two and ThreeDimensional Currents On a surface that carries uniform surface current density K A/m, the current within width b is ..and so the differential current quantity that appears in the BiotSavart law becomes: The magnetic field arising from a current sheet is thus found from the twodimensional form of the BiotSavart law: In a similar way, a volume current will be made up of threedimensional current elements, and so the BiotSavart law for this case becomes: B.Hemalatha APECE 56Example of the BiotSavart Law In this example, we evaluate the magnetic field intensity on the y axis (equivalently in the xy plane) arising from a filament current of infinite length in on the z axis. Using the drawing, we identify: and so.. so that: B.Hemalatha APECE 57Example: continued We now have: Integrate this over the entire wire: ..after carrying out the cross product B.Hemalatha APECE 58Example: concluded Evaluating the integral: Current is into the page. Magnetic field streamlines are concentric circles, whose magnitudes finally: decrease as the inverse distance from the z axis B.Hemalatha APECE 59Field Arising from a Finite Current Segment In this case, the field is to be found in the xy plane at Point 2. The BiotSavart integral is taken over the wire length: ..after a few additional steps (see Problem 7.8), we find: B.Hemalatha APECE 60Another Example: Magnetic Field from a Current Loop Consider a circular current loop of radius a in the xy plane, which carries steady current I. We wish to find the magnetic field strength anywhere on the z axis. We will use the BiotSavart Law: where: B.Hemalatha APECE 61Example: Continued Substituting the previous expressions, the BiotSavart Law becomes: carry out the cross products to find: but we must include the angle dependence in the radial unit vector: with this substitution, the radial component will integrate to zero, meaning that all radial components will cancel on the z axis. B.Hemalatha APECE 62Example: Continued Now, only the z component remains, and the integral evaluates easily: Note the form of the numerator: the product of the current and the loop area. We define this as the magnetic moment: B.Hemalatha APECE 63Ampere’s Circuital Law Ampere’s Circuital Law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. In the figure at right, the integral of H about closed paths a and b gives the total current I, while the integral over path c gives only that portion of the current that lies within c B.Hemalatha APECE 64Ampere’s Law Applied to a Long Wire Symmetry suggests that H will be circular, constant valued at constant radius, and centered on the current (z) axis. Choosing path a, and integrating H around the circle of radius  gives the enclosed current, I:  so that: as before. B.Hemalatha APECE 65Coaxial Transmission Line In the coax line, we have two concentric solid conductors that carry equal and opposite currents, I. The line is assumed to be infinitely long, and the circular symmetry suggests that H will be entirely  directed, and will vary only with radius . Our objective is to find the magnetic field for all values of  B.Hemalatha APECE 66Field Between Conductors The inner conductor can be thought of as made up of a bundle of filament currents, each of which produces the field of a long wire. Consider two such filaments, located at the same radius from the z axis,  , but which lie at symmetric   coordinates,  and Their field contributions  superpose to give a net H component as shown.  The same happens for every pair of symmetrically located filaments, which taken as a whole, make up the entire center conductor. a  b B.Hemalatha APECE 67Field Within the Inner Conductor With current uniformly distributed inside the conductors, the H can be assumed circular everywhere. Inside the inner conductor, and at radius we again have: But now, the current enclosed is so that or finally: B.Hemalatha APECE 68Field Outside Both Conductors Outside the transmission line, where  c, no current is enclosed by the integration path, and so 0 As the current is uniformly distributed, and since we have circular symmetry, the field would have to be constant over the circular integration path, and so it must be true that: B.Hemalatha APECE 69Field Inside the Outer Conductor Inside the outer conductor, the enclosed current consists of that within the inner conductor plus that portion of the outer conductor current existing at radii less than  Ampere’s Circuital Law becomes ..and so finally: B.Hemalatha APECE 70Magnetic Field Strength as a Function of Radius in the Coax Line Combining the previous results, and assigning dimensions as shown in the inset below, we find: B.Hemalatha APECE 71Magnetic Field Arising from a Current Sheet For a uniform plane current in the y direction, we expect an xdirected H field from symmetry. Applying Ampere’s circuital law to the path we find: or In other words, the magnetic field is discontinuous across the current sheet by the magnitude of the surface current density. B.Hemalatha APECE 72Magnetic Field Arising from a Current Sheet If instead, the upper path is elevated to the line between 3 and 3’ , the same current is enclosed and we would have from which we conclude that So the field is constant in each region (above and below the current plane) By symmetry, the field above the sheet must be the same in magnitude as the field below the sheet. Therefore, we may state that B.Hemalatha APECE 73Magnetic Field Arising from a Current Sheet The actual field configuration is shown below, in which magnetic field above the current sheet is equal in magnitude, but in the direction opposite to the field below the sheet. The field in either region is found by the cross product: where a is the unit vector that is normal N to the current sheet, and that points into the region in which the magnetic field is to be evaluated. B.Hemalatha APECE 74Magnetic Field Arising from Two Current Sheets Here are two parallel currents, equal and opposite, as you would find in a parallelplate transmission line. If the sheets are much wider than their spacing, then the magnetic field will be contained in the region between plates, and will be nearly zero outside. These fields cancel for current H (z d/2 ) x2 sheets of infinite width. H (z d/2 ) x1 These fields are equal and K = K a 1 y y add to give H (d /2 z d/2 ) x1 H (d /2 z d/2 ) x2 H = K x a (d/2 z d/2 ) N K = K a 2 y y where K is either K or K 1 2 H (z d/2 ) x1 These fields cancel for current H (z d/2 ) sheets of infinite width. x2 B.Hemalatha APECE 75Current Loop Field Using the BiotSavart Law, we previously found the magnetic field on the z axis from a circular current loop: We will now use this result as a building block to construct the magnetic field on the axis of a solenoid formed by a stack of identical current loops, centered on the z axis. B.Hemalatha APECE 76OnAxis Field Within a Solenoid We consider the single current loop field as a differential contribution to the total field from a stack of N closelyspaced loops, each of which carries current I. The length of the stack (solenoid) is d, so therefore the density of turns will be N/d. Now the current in the turns within a differential length, dz, will be We consider this as our differential “loop current” z d/ 2 so that the previous result for H from a single loop: now becomes: d/2 in which z is measured from the center of the coil, where we wish to evaluate the field. B.Hemalatha APECE 77Solenoid Field, Continued The total field on the z axis at z = 0 will be the sum of the field contributions from all turns in the coil or the integral of dH over the length of the solenoid. z d/2 d/2 B.Hemalatha APECE 78Approximation for Long Solenoids We now have the onaxis field at the solenoid midpoint (z = 0): z d/2 Note that for long solenoids, for which , the result simplifies to: ( ) d/2 This result is valid at all onaxis positions deep within long coils at distances from each end of several radii. B.Hemalatha APECE 79Another Interpretation: Continuous Surface Current The solenoid of our previous example was assumed to have many tightlywound turns, with Several existing within a differential length, dz. We could model such a current configuration as a continuous surface current of density K = K a A/m. a  d/2 Therefore: d/2 In other words, the onaxis field magnitude near the center of a cylindrical current sheet, where current circulates around the z axis, and whose length is much greater than its radius, is just the surface current density. B.Hemalatha APECE 80Solenoid Field OffAxis To find the field within a solenoid, but off the z axis, we apply Ampere’s Circuital Law in the following way: The illustration below shows the solenoid crosssection, from a lengthwise cut through the z axis. Current in the windings flows in and out of the screen in the circular current path. Each turn carries current I. The magnetic field along the z axis is NI/d as we found earlier. B.Hemalatha APECE 81Toroid Magnetic Field A toroid is a doughnutshaped set of windings around a core material. The crosssection could be circular (as shown here, with radius a) or any other shape. Below, a slice of the toroid is shown, with current emerging from the screen around the inner periphery (in the positive z direction). The windings are modeled as N individual current loops, each of which carries current I. B.Hemalatha APECE 82Ampere’s Law as Applied to a Toroid Ampere’s Circuital Law can be applied to a toroid by taking a closed loop integral around the circular contour C at radius Magnetic field H is presumed to be circular, and a function of radius only at locations within the toroid that are not too close to the individual windings. Under this condition, we would assume: This approximation improves as the density of turns gets higher (using more turns with finer wire). Ampere’s Law now takes the form: so that…. Performing the same integrals over contours drawn in the regions or will lead to zero magnetic field there, because no current is enclosed in either case. B.Hemalatha APECE 83Surface Current Model of a Toroid Consider a sheet current molded into a doughnut shape, as shown. The current density at radius crosses the xy plane in the z direction and is given in magnitude by K a Ampere’s Law applied to a circular contour C inside the toroid (as in the previous example) will take the form: leading to… inside the toroid…. and the field is zero outside as before. B.Hemalatha APECE 84Approximation of H Along One Segment Along path 12, we may write: where: And therefore: B.Hemalatha APECE 85Stokes’ Theorem We now take our previous result, and take the limit as In the limit, this side In the limit, this side becomes the path integral becomes the integral of H over the outer perimeter of the curl of H over because all interior paths surface S cancel . The result is Stokes’ Theorem This is a valuable tool to have at our disposal, because it gives us two ways to evaluate the same thing B.Hemalatha APECE 86Obtaining Ampere’s Circuital Law in Integral Form, using Stokes’ Theorem Begin with the point form of Ampere’s Law for static fields: Integrate both sides over surface S: ..in which the far right hand side is found from the left hand side using Stokes’ Theorem. The closed path integral is taken around The perimeter of S. Again, note that we use the righthand convention in choosing the direction of the path integral. The center expression is just the net current through surface S, so we are left with the integral form of Ampere’s Law: B.Hemalatha APECE 87Magnetic Flux and Flux Density We are already familiar with the concept of electric flux: Coulombs in which the electric flux density in free space is: and where the free space permittivity is In a similar way, we can define the magnetic flux in units of Webers (Wb): Webers in which the magnetic flux density (or magnetic induction) in free space is: and where the free space permeability is This is a defined quantity, having to do with the definition of the ampere B.Hemalatha APECE 88A Key Property of B If the flux is evaluated through a closed surface, we have in the case of electric flux, Gauss’ Law: If the same were to be done with magnetic flux density, we would find: The implication is that (for our purposes) there are no magnetic charges specifically, no point sources of magnetic field exist. A hint of this has already been observed, in that magnetic field lines always close on themselves. B.Hemalatha APECE 89Another Maxwell Equation We may rewrite the closed surface integral of B using the divergence theorem, in which the right hand integral is taken over the volume surrounded by the closed surface: Because the result is zero, it follows that This result is known as Gauss’ Law for the magnetic field in point form. B.Hemalatha APECE 90Example: Magnetic Flux Within a Coaxial Line Consider a length d of coax, as shown here. The magnetic field strength between conductors is: and so: The magnetic flux is now the integral of B over the flat surface between radii a and b, and of length d along z: B d The result is: The coax line thus “stores” this amount of magnetic flux in the region between conductors. This will have importance when we discuss inductance in a later lecture. B.Hemalatha APECE 91Scalar Magnetic Potential We are already familiar with the relation between the scalar electric potential and electric field: So it is tempting to define a scalar magnetic potential such that: This rule must be consistent with Maxwell’s equations, so therefore: But the curl of the gradient of any function is identically zero Therefore, the scalar magnetic potential is valid only in regions where the current density is zero (such as in free space). So we define scalar magnetic potential with a condition: B.Hemalatha APECE 92Further Requirements on the Scalar Magnetic Potential The other Maxwell equation involving magnetic field must also be satisfied. This is: in free space Therefore: ..and so the scalar magnetic potential satisfies Laplace’s equation (again with the restriction that current density must be zero: B.Hemalatha APECE 93Example: Coaxial Transmission Line With the center conductor current flowing out of the screen, we have Thus: So we solve: .. and obtain: where the integration constant has been set to zero B.Hemalatha APECE 94Ambiguities in the Scalar Potential The scalar potential is now: where the potential is zero at ) the potential is At point P ( But wait As increases to we have returned to the same physical location, and the potential has a new value of I. In general, the potential at P will be multivalued, and will acquire a new value after each full rotation in the xy plane: B.Hemalatha APECE 95Overcoming the Ambiguity To remove the ambiguity, we construct a mathematical barrier at any value of phi. The angle domain cannot cross this barrier in either direction, and so the potential function is restricted to angles on either side. In the present case we choose the barrier to lie at so that The potential at point P is now singlevalued: Barrier at B.Hemalatha APECE 96Vector Magnetic Potential We make use of the Maxwell equation: .. and the fact that the divergence of the curl of any vector field is identically zero (show this) This leads to the definition of the magnetic vector potential, A: Thus: and Ampere’s Law becomes B.Hemalatha APECE 97Equation for the Vector Potential We start with: Then, introduce a vector identity that defines the vector Laplacian: Using a (lengthy) procedure (see Sec. 7.7) it can be proven that We are therefore left with B.Hemalatha APECE 98The Direction of A We now have In rectangular coordinates: (not so simple in the other coordinate systems) The equation separates to give: This indicates that the direction of A will be the same as that of the current to which it is associated. The vector field, A, existing in all space, is sometimes described as being a “fuzzy image” of its generating current. B.Hemalatha APECE 99Expressions for Potential Consider a differential elements, shown here. On the left is a point charge represented by a differential length of line charge. On the right is a differential current element. The setups for obtaining potential are identical between the two cases. Line Current Line Charge Scalar Electrostatic Potential Vector Magnetic Potential B.Hemalatha APECE 100General Expressions for Vector Potential For large scale charge or current distributions, we would sum the differential contributions by integrating over the charge or current, thus: and The closed path integral is taken because the current must close on itself to form a complete circuit. For surface or volume current distributions, we would have, respectively: or in the same manner that we used for scalar electric potential. B.Hemalatha APECE 101Example We continue with the differential current element as shown here: In this case becomes at point P: Now, the curl is taken in cylindrical coordinates: This is the same result as found using the BiotSavart Law (as it should be) B.Hemalatha APECE 102UNIT III UNIT III Static Magnetic Fields Static Magnetic Fields B.Hemalatha APECE 1Motivating the Magnetic Field Concept: Motivating the Magnetic Field Concept: Forces Between Currents Forces Between Currents Magnetic forces arise whenever we have charges in motion. Forces between current carrying wires present familiar examples that we can use to determine what a magnetic force field should look like: Here are the easilyobserved facts: B.Hemalatha APECE 2BiotSavart Law BiotSavart Law • The current flowing in any path, can be considered as many infinitesimal current elements, each of length dl, flowing in a magnetic field dB, at any point P ˆ Idlr 0 dB 2 4r B.Hemalatha APECE 3Idl sin 0 • The magnitude of dB is: dB 2 4r where θ is the angle between dl and r ˆ Idlr 0 BdB • The total magnetic field at point P is: 2  2r • This is equivalent to Coulomb’s Law written in differential form: 1dq dE 2 4r 0 B.Hemalatha APECE 4Magnetic Field due to a Current Carrying Wire BiotSavart Law Hans Christian Oersted, 1820 B.Hemalatha APECE 5• Magnetic fields are caused by currents. • Hans Christian Oersted in 1820’s showed that a current carrying wire deflects a compass. No Current in the Wire Current in the Wire B.Hemalatha APECE 6Right Hand Curl Rule Right Hand Curl Rule B.Hemalatha APECE 7Magnetic Fields of Long CurrentCarrying Wires B =  I o 2r I I = current through the wire (Amps) r = distance from the wire (m)  = permeability of free space o 7 = 4 x 10 T m / A B = magnetic field strength (Tesla) B.Hemalatha APECE 8Magnetic Field of a Current Carrying Wire B.Hemalatha APECE 9What if the currentcarrying wire is not straight Use What if the currentcarrying wire is not straight Use the BiotSavart Law: the BiotSavart Law: Assume a small segment of wire ds causing a field dB: ˆ  dsr  0 dBI 2  4  r Note: dB is perpendicular to ds and r B.Hemalatha APECE 10BiotSavart Law allows us to calculate the BiotSavart Law allows us to calculate the Magnetic Field Vector Magnetic Field Vector • To find the total field, sum up the contributions from all the current elements I ds ˆ  dsr  0ii BI  2  4  r i • The integral is over the entire current distribution ˆ μIdsr o B 2  4πr B.Hemalatha APECE 11Note on BiotSavart Law Note on BiotSavart Law • The law is also valid for a current consisting of charges flowing through space • ds represents the length of a small segment of space in which the charges flow. • Example: electron beam in a TV set B.Hemalatha APECE 12Comparison of Magnetic to Electric Field Magnetic Field Electric Field 2 2 • B proportional to r • E proportional to r • Vector • Vector • Perpendicular to F , ds, r • Same direction as F B E • Magnetic field lines have no • Electric field lines begin on beginning and no end; they positive charges and end on form continuous circles negative charges • BiotSavart Law • Coulomb’s Law • Ampere’s Law (where there • Gauss’s Law (where there is is symmetry symmetry) B.Hemalatha APECE 13Derivation of B for a Long, Straight CurrentCarrying Derivation of B for a Long, Straight CurrentCarrying Wire Wire Integrating over all the current elements gives ˆ ˆ dsrdx sin θ k  θ μI 2 o Bsinθ dθ  θ 1 4πa μI o cosθcosθ  1 2 4πa B.Hemalatha APECE 14If the conductor is an infinitely long, straight If the conductor is an infinitely long, straight wire, = 0 and =  wire, = 0 and =    • The field becomes: a μI o B 2πa B.Hemalatha APECE 15B.Hemalatha APECE 16B for a Curved Wire Segment B for a Curved Wire Segment • Find the field at point O due to the wire segment A’ACC’: B=0 due to AA’ and CC’ Due to the circular arc: ˆ μIdsr o B 2  4πr μI o Bθ 4πR q=s/R, will be in radians B.Hemalatha APECE 17B at the Center of a Circular Loop of Wire B at the Center of a Circular Loop of Wire μIμI oo Bθ 2π 4πR 4πR Consider the previous result, with  = 2 μI o B 2R B.Hemalatha APECE 18Note The overall shape of the magnetic field of the circular loop Is similar to the magnetic field of a bar magnet. B.Hemalatha APECE 19B along the axis of a Circular Current Loop B along the axis of a Circular Current Loop • Find B at point P ˆ μIdsr o B 2  4πr 2 μIR o If x=0, B same as at center of a loop B x 3 2 2 2 2xR  B.Hemalatha APECE 20If x is at a very large distance away from the loop. xR: 2 2 μIRμIR oo B x 3 3 2 2 2 2x 2xR  B.Hemalatha APECE 21Magnetic Force Between Two Parallel Conductors The field B due to the current 2 in wire 2 exerts a force on wire 1 of F = I ℓ B 1 1 2 μI o 2 B 2 2πa μII o 1 2 F 1 2πa B.Hemalatha APECE 22Magnetic Field at Center of a Solenoid B =  NI o L N: Number of turns L : Length n=N/L L B.Hemalatha APECE 23Direction of Force Between Two Parallel Conductors If the currents are in the: –same direction the wires attract each other. –opposite directions the wires repel each other. B.Hemalatha APECE 24Magnetic Force Between Two Parallel Conductors, F B FμII Bo 1 2 Force per unit length:   2πa B.Hemalatha APECE 25Definition of the Ampere • When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 7 m is 2 x 10 N/m, the current in each wire is defined to be 1 A B.Hemalatha APECE 26Definition of the Coulomb • The SI unit of charge, the coulomb, is defined in terms of the ampere • When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C B.Hemalatha APECE 27BiotSavart Law: ˆ  dsr  0 Field produced by current carrying wires dBI 2  4  r I 0 B – Distance a from long straight wire 2a I 0 B – Centre of a wire loop radius R 2R NI 0 B – Centre of a tight Wire Coil with N turns 2R FII 0 1 2 • Force between two wires  l 2a B.Hemalatha APECE 28Differences between Ampere’s law the Biot Savart Law • The difference between Ampere’s Law and the Biot BdlI Savart Law is that in Ampere’s Law ( 0 e n c l), the  magnetic field is not necessarily due only to the current enclosed by the path of integration, as Ampere suggests • In the BiotSavant Law, dB is due entirely to the current element I·dl. To find the total B, it is necessary to include all currents B.Hemalatha APECE 29Magnetic Field Due to Current in a Straight Wire • To find the magnetic field near an infinitely long, straight wire, carrying a current I, the Biot Savant Law gives us: y Idysin 0 B 2  4r y • The solution of this integral yields: I 0 B 2R • This is the same as Ampere’s Law B.Hemalatha APECE 30The Magnetic Field OnAxis of a Current Loop • To find the magnetic field OnAxis, of a Current Loop, applying the BiotSavant Law yields: Idl sin  0 ˆ dB , 90becausedlr 2 4R Idl 0 dB 2 4R II 0 0 Bdl2R 2 2  4R 4R I 0 B 2R B.Hemalatha APECE 31Magnetic Field of a Loop at a Point OnAxis • B = 0 is from the symmetry of the y problem • B is calculated using the Biot Savant Law: z Savart Law 2  2RI 0 B z 3 2 2 2 4 zR • For zR then the above relationship simplifies to: 2  2RI 0 B z 3 4 z B.Hemalatha APECE 32ˆ r Magnetic Field of a Wire Segment  • Again: , the BiotSavant Law ˆ  90becausedlr gives: Idl 0 dB 2 4R ˆ r • Solving the Integral yields: dB I 1 0 B 8 R B.Hemalatha APECE 33Magnetic fields of long wires • Example for two wires. Find B at P , P , and P . 1 2 3 B.Hemalatha APECE 34Magnetic fields of long wires • P :  I/2d) +  I/2(4d) =  I/8d 1 0 0 0 • P : + I/2d) +  I/2(d) = + I/d 0 0 0 2 • P : +  I/2d)  I/2(d) =  I/3d 0 0 0 3 B.Hemalatha APECE 35Ampere’s law (general statement) • General statement of Ampere’s law B.Hemalatha APECE 36Ampere’s law (special case) • Ampere’s law for a circular path around a long straight conductor. B.Hemalatha APECE 37Applications of Ampere’s Law   Ampère’s law in differential form  B(r) J(r) 0   ( B) da B d J da I Ampère’s law in integral form 0 0 enc  Example 1  Find the magnetic field of an infinite uniform surface current ˆ , K Kx flowing over the xy plane B.Hemalatha APECE 38Example 2 Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R and carrying a steady current I.     Bds B NI Bds Bds B ds B   0 side1 side1 Where N is the number of turns in the length N B I nI 0 0  B.Hemalatha APECE 39Example 3 Find the magnetic field of a toroidal coil, consisting of a circular ring around which a long wire is wrapped.   Bds B ds B(2r) NI  0  NI 0 B 2r B.Hemalatha APECE 40Applications of Ampere’s Law Problem 1 A steady current I flows down a long cylindrical wire of radius a. Find the magnetic field, both inside and outside the wire, if (a). The current is uniformly distributed over the outside surface of the wire. (b). The current is distributed in such a way that J is proportional to s, the distance from the axis. a I (a)  B d 2sB I 0 enc   0 for s a   B  I  0 ˆ  for s a  2s B.Hemalatha APECE 41Applications of Ampere’s Law a 3 2ka 3I (b) J ks I Jda ks(2s)ds k  3 3 2a 0 3Is J ks B d 2sB I 3 0 enc  2a For s a s 3  3Is 3 I 1 Is 3 0 0 B d 2sB I 2sds s 0 enc 0 3 3 3  2a a 3 a 0 2  Is 0 B 3 2a For s a a  3Is 3 I 1 3 0 B d 2sB I 2sds a I 0 enc 0 0  3 3 2a a 3 0  I 0 B 2s B.Hemalatha APECE 42Comparison of Magnetostatics and Electrostatics The divergence and curl of the electrostatic field are  1  Gauss’s law  E    0    E 0  The divergence and curl of the magnetostatic field are    B 0   Ampère’s law  B J  0 “The electric force is stronger than the magnetic force. Only when both the source charge and the test charge are moving at velocities comparable to the speed of light, the magnetic force approaches the electric force.” B.Hemalatha APECE 43Magnetic Vector Potential   V: electric scalar potential  E 0 EV You can add to V any function whose gradient is zero (V f )VfV   A : magnetic vector potential  B 0 B A  A You can add to any function whose curl is zero   (A) A A   2  B ( A)( A) A J B A 0  We will prove that   A  0 : B.Hemalatha APECE 44Magnetic Vector Potential  Suppose that our original vector potential is not divergenceless A 0   A 0 0  Because we can add to any function whose curl is zero A 0   2 A A  A A A 0 0 0 If a function  can be found that satisfies  2  A A 0 0  2  V Mathematically identical to Poisson’s equation  0 In particular, if  goes to zero at infinity, then the solution is 1 V d'  4 0 B.Hemalatha APECE 45 By the same token, if goes to zero at infinity, then  A 0  1 A 0  d'  4 It is always possible to make the vector potential divergenceless   A 0   2 2 so  B( A) A A J 0 This again is a Poisson’s equation  Assuming goes to zero at infinity, then J     J(r') 0 A(r) d'  4 For line and surface currents        K(r')  I(r') 0 0 A(r) da' A(r) d'   4 4 B.Hemalatha APECE 46Magnetic Vector Potential Example 1 A spherical shell, of radius R, carrying a uniform surface charge , is set spinning at angular velocity . Find the vector potential it  produces at point . r  r  The integration is easier  if we let lie on the z r  r' axis, so that  is titled at an angle . B.Hemalatha APECE 47Magnetic Vector Potential       K(r') 2 2 0 where A(r) da' Kv R r 2Rr cos'  4 ˆ ˆ ˆ x y z  v r'sin 0cos R sin'cos' R sin'sin' R cos'  R(cos sin'sin')x ˆ (cos sin'cos' sin cos')y ˆ (sin sin'sin')z ˆ 2 2  ˆ Because We just consider vRsin cos' y sin'd' cos'd' 0  0 0  (Rsin cos') 2 0 ˆ A(r) R sin'd'd'y  2 2 4 R r 2Rr cos'  3  Rsin sin'cos' 0 ˆ  ( d')y  2 2 2 R r 2Rr cos' 0 B.Hemalatha APECE 48Magnetic Vector Potential Letting u  c o s  ' , the integral becomes  1 sin'cos' u I d' du  2 2 2 2 R r 2Rr cos' R r 2Rru 01 1 u 1 2 2 2 2 3/ 2   R r 2Rru (R r 2Rru) 2 2 Rr 3R r 1 1 2 2 (R r Rru) 2 2   R r 2Rru 2 2 3R r 1 1 2 2 2 2  (R r Rr) R r (R r Rr)(R r) 2 2 3R r  2r I If the point lies inside the sphere, Then R r.  r 2 3R  2R If the point lies outside the sphere, Then R r.  I r 2 3r B.Hemalatha APECE 49Magnetic Vector Potential  3 3   Rsin sin'cos' Rsin 0 0 ˆ ˆ A(r) ( d')y Iy  2 2 2 2 R r 2Rr cos' 0  2r I If the point lies inside the sphere, Then R r.  r 2 3R  2R If the point lies outside the sphere, Then R r.  I r 2 3r  ˆ ( r)r siny Noting that For the point inside the sphere 3 3   Rsin Rsin 2r R 0 0 0 ˆ ˆ A(r) Iy y ( r) 2 2 2 3R 3 For the point outside the sphere 3 3 4   Rsin Rsin 2R R 0 0 0 ˆ ˆ A(r) Iy y ( r) 2 3 2 2 3r 3r B.Hemalatha APECE 50Magnetic Vector Potential   We revert to the original coordinates, in which coincides  with the z axis and the point is at (r,,) r For the point inside the sphere   R R 0 0 ˆ A(r) ( r) r sin 3 3   2 R 2 R 0 0 ˆ ˆ ˆ B A(r) (cosr sin) z 3 3 The magnetic field inside this spherical shell is uniform For the point outside the sphere 4 4   R R 0 0 ˆ A(r) ( r) sin 3 2 3r 3r B.Hemalatha APECE 51Magnetic Vector Potential Example 2 Find the vector potential of an infinite solenoid with n turns per unit length, radius R, and current I     I(r') 0 We cannot use because the current itself extends A(r) d'  4 to infinity.   A d ( A) da B da Notice that   ˆ B nIz Since the magnetic field is uniform inside the solenoid : 0      nIs 2 0 ˆ A d A(2s) B da nI(s ) A For s R 0  2 For an amperian outside the solenoid 2    nIR 2 0 ˆ For s R A d A(2s) B da nI(R ) A 0  2s B.Hemalatha APECE 52Motivating the Magnetic Field Concept: Forces Between Currents Magnetic forces arise whenever we have charges in motion. Forces between currentcarrying wires present familiar examples that we can use to determine what a magnetic force field should look like: Here are the easilyobserved facts: B.Hemalatha APECE 53Magnetic Field The geometry of the magnetic field is set up to correctly model forces between currents that allow for any relative orientation. The magnetic field intensity, H, circulates around its source, I , 1 in a direction most easily determined by the righthand rule: Right thumb in the direction of the current, fingers curl in the direction of H Note that in the third case (perpendicular currents), I is in the same direction as H, so that their 2 cross product (and the resulting force) is zero. The actual force computation involves a different field quantity, B, which is related to H through B =  H in free space. This will be taken up in  a later lecture. Our immediate concern is how to find H from any given current distribution. B.Hemalatha APECE 54Magnetic Field Arising From a Circulating Current At point P, the magnetic field associated with the differential current element IdL is To determine the total field arising from the closed circuit path, we sum the contributions from the current elements that make up the entire loop, or The contribution to the field at P from any portion of the current will be just the above integral evaluated over just that portion. B.Hemalatha APECE 55Two and ThreeDimensional Currents On a surface that carries uniform surface current density K A/m, the current within width b is ..and so the differential current quantity that appears in the BiotSavart law becomes: The magnetic field arising from a current sheet is thus found from the twodimensional form of the BiotSavart law: In a similar way, a volume current will be made up of threedimensional current elements, and so the BiotSavart law for this case becomes: B.Hemalatha APECE 56Example of the BiotSavart Law In this example, we evaluate the magnetic field intensity on the y axis (equivalently in the xy plane) arising from a filament current of infinite length in on the z axis. Using the drawing, we identify: and so.. so that: B.Hemalatha APECE 57Example: continued We now have: Integrate this over the entire wire: ..after carrying out the cross product B.Hemalatha APECE 58Example: concluded Evaluating the integral: Current is into the page. Magnetic field streamlines are concentric circles, whose magnitudes finally: decrease as the inverse distance from the z axis B.Hemalatha APECE 59Field Arising from a Finite Current Segment In this case, the field is to be found in the xy plane at Point 2. The BiotSavart integral is taken over the wire length: ..after a few additional steps (see Problem 7.8), we find: B.Hemalatha APECE 60Another Example: Magnetic Field from a Current Loop Consider a circular current loop of radius a in the xy plane, which carries steady current I. We wish to find the magnetic field strength anywhere on the z axis. We will use the BiotSavart Law: where: B.Hemalatha APECE 61Example: Continued Substituting the previous expressions, the BiotSavart Law becomes: carry out the cross products to find: but we must include the angle dependence in the radial unit vector: with this substitution, the radial component will integrate to zero, meaning that all radial components will cancel on the z axis. B.Hemalatha APECE 62Example: Continued Now, only the z component remains, and the integral evaluates easily: Note the form of the numerator: the product of the current and the loop area. We define this as the magnetic moment: B.Hemalatha APECE 63Ampere’s Circuital Law Ampere’s Circuital Law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. In the figure at right, the integral of H about closed paths a and b gives the total current I, while the integral over path c gives only that portion of the current that lies within c B.Hemalatha APECE 64Ampere’s Law Applied to a Long Wire Symmetry suggests that H will be circular, constant valued at constant radius, and centered on the current (z) axis. Choosing path a, and integrating H around the circle of radius  gives the enclosed current, I:  so that: as before. B.Hemalatha APECE 65Coaxial Transmission Line In the coax line, we have two concentric solid conductors that carry equal and opposite currents, I. The line is assumed to be infinitely long, and the circular symmetry suggests that H will be entirely  directed, and will vary only with radius . Our objective is to find the magnetic field for all values of  B.Hemalatha APECE 66Field Between Conductors The inner conductor can be thought of as made up of a bundle of filament currents, each of which produces the field of a long wire. Consider two such filaments, located at the same radius from the z axis,  , but which lie at symmetric   coordinates,  and Their field contributions  superpose to give a net H component as shown.  The same happens for every pair of symmetrically located filaments, which taken as a whole, make up the entire center conductor. a  b B.Hemalatha APECE 67Field Within the Inner Conductor With current uniformly distributed inside the conductors, the H can be assumed circular everywhere. Inside the inner conductor, and at radius we again have: But now, the current enclosed is so that or finally: B.Hemalatha APECE 68Field Outside Both Conductors Outside the transmission line, where  c, no current is enclosed by the integration path, and so 0 As the current is uniformly distributed, and since we have circular symmetry, the field would have to be constant over the circular integration path, and so it must be true that: B.Hemalatha APECE 69Field Inside the Outer Conductor Inside the outer conductor, the enclosed current consists of that within the inner conductor plus that portion of the outer conductor current existing at radii less than  Ampere’s Circuital Law becomes ..and so finally: B.Hemalatha APECE 70Magnetic Field Strength as a Function of Radius in the Coax Line Combining the previous results, and assigning dimensions as shown in the inset below, we find: B.Hemalatha APECE 71Magnetic Field Arising from a Current Sheet For a uniform plane current in the y direction, we expect an xdirected H field from symmetry. Applying Ampere’s circuital law to the path we find: or In other words, the magnetic field is discontinuous across the current sheet by the magnitude of the surface current density. B.Hemalatha APECE 72Magnetic Field Arising from a Current Sheet If instead, the upper path is elevated to the line between 3 and 3’ , the same current is enclosed and we would have from which we conclude that So the field is constant in each region (above and below the current plane) By symmetry, the field above the sheet must be the same in magnitude as the field below the sheet. Therefore, we may state that B.Hemalatha APECE 73Magnetic Field Arising from a Current Sheet The actual field configuration is shown below, in which magnetic field above the current sheet is equal in magnitude, but in the direction opposite to the field below the sheet. The field in either region is found by the cross product: where a is the unit vector that is normal N to the current sheet, and that points into the region in which the magnetic field is to be evaluated. B.Hemalatha APECE 74Magnetic Field Arising from Two Current Sheets Here are two parallel currents, equal and opposite, as you would find in a parallelplate transmission line. If the sheets are much wider than their spacing, then the magnetic field will be contained in the region between plates, and will be nearly zero outside. These fields cancel for current H (z d/2 ) x2 sheets of infinite width. H (z d/2 ) x1 These fields are equal and K = K a 1 y y add to give H (d /2 z d/2 ) x1 H (d /2 z d/2 ) x2 H = K x a (d/2 z d/2 ) N K = K a 2 y y where K is either K or K 1 2 H (z d/2 ) x1 These fields cancel for current H (z d/2 ) sheets of infinite width. x2 B.Hemalatha APECE 75Current Loop Field Using the BiotSavart Law, we previously found the magnetic field on the z axis from a circular current loop: We will now use this result as a building block to construct the magnetic field on the axis of a solenoid formed by a stack of identical current loops, centered on the z axis. B.Hemalatha APECE 76OnAxis Field Within a Solenoid We consider the single current loop field as a differential contribution to the total field from a stack of N closelyspaced loops, each of which carries current I. The length of the stack (solenoid) is d, so therefore the density of turns will be N/d. Now the current in the turns within a differential length, dz, will be We consider this as our differential “loop current” z d/ 2 so that the previous result for H from a single loop: now becomes: d/2 in which z is measured from the center of the coil, where we wish to evaluate the field. B.Hemalatha APECE 77Solenoid Field, Continued The total field on the z axis at z = 0 will be the sum of the field contributions from all turns in the coil or the integral of dH over the length of the solenoid. z d/2 d/2 B.Hemalatha APECE 78Approximation for Long Solenoids We now have the onaxis field at the solenoid midpoint (z = 0): z d/2 Note that for long solenoids, for which , the result simplifies to: ( ) d/2 This result is valid at all onaxis positions deep within long coils at distances from each end of several radii. B.Hemalatha APECE 79Another Interpretation: Continuous Surface Current The solenoid of our previous example was assumed to have many tightlywound turns, with Several existing within a differential length, dz. We could model such a current configuration as a continuous surface current of density K = K a A/m. a  d/2 Therefore: d/2 In other words, the onaxis field magnitude near the center of a cylindrical current sheet, where current circulates around the z axis, and whose length is much greater than its radius, is just the surface current density. B.Hemalatha APECE 80Solenoid Field OffAxis To find the field within a solenoid, but off the z axis, we apply Ampere’s Circuital Law in the following way: The illustration below shows the solenoid crosssection, from a lengthwise cut through the z axis. Current in the windings flows in and out of the screen in the circular current path. Each turn carries current I. The magnetic field along the z axis is NI/d as we found earlier. B.Hemalatha APECE 81Toroid Magnetic Field A toroid is a doughnutshaped set of windings around a core material. The crosssection could be circular (as shown here, with radius a) or any other shape. Below, a slice of the toroid is shown, with current emerging from the screen around the inner periphery (in the positive z direction). The windings are modeled as N individual current loops, each of which carries current I. B.Hemalatha APECE 82Ampere’s Law as Applied to a Toroid Ampere’s Circuital Law can be applied to a toroid by taking a closed loop integral around the circular contour C at radius Magnetic field H is presumed to be circular, and a function of radius only at locations within the toroid that are not too close to the individual windings. Under this condition, we would assume: This approximation improves as the density of turns gets higher (using more turns with finer wire). Ampere’s Law now takes the form: so that…. Performing the same integrals over contours drawn in the regions or will lead to zero magnetic field there, because no current is enclosed in either case. B.Hemalatha APECE 83Surface Current Model of a Toroid Consider a sheet current molded into a doughnut shape, as shown. The current density at radius crosses the xy plane in the z direction and is given in magnitude by K a Ampere’s Law applied to a circular contour C inside the toroid (as in the previous example) will take the form: leading to… inside the toroid…. and the field is zero outside as before. B.Hemalatha APECE 84Approximation of H Along One Segment Along path 12, we may write: where: And therefore: B.Hemalatha APECE 85Stokes’ Theorem We now take our previous result, and take the limit as In the limit, this side In the limit, this side becomes the path integral becomes the integral of H over the outer perimeter of the curl of H over because all interior paths surface S cancel . The result is Stokes’ Theorem This is a valuable tool to have at our disposal, because it gives us two ways to evaluate the same thing B.Hemalatha APECE 86Obtaining Ampere’s Circuital Law in Integral Form, using Stokes’ Theorem Begin with the point form of Ampere’s Law for static fields: Integrate both sides over surface S: ..in which the far right hand side is found from the left hand side using Stokes’ Theorem. The closed path integral is taken around The perimeter of S. Again, note that we use the righthand convention in choosing the direction of the path integral. The center expression is just the net current through surface S, so we are left with the integral form of Ampere’s Law: B.Hemalatha APECE 87Magnetic Flux and Flux Density We are already familiar with the concept of electric flux: Coulombs in which the electric flux density in free space is: and where the free space permittivity is In a similar way, we can define the magnetic flux in units of Webers (Wb): Webers in which the magnetic flux density (or magnetic induction) in free space is: and where the free space permeability is This is a defined quantity, having to do with the definition of the ampere B.Hemalatha APECE 88A Key Property of B If the flux is evaluated through a closed surface, we have in the case of electric flux, Gauss’ Law: If the same were to be done with magnetic flux density, we would find: The implication is that (for our purposes) there are no magnetic charges specifically, no point sources of magnetic field exist. A hint of this has already been observed, in that magnetic field lines always close on themselves. B.Hemalatha APECE 89Another Maxwell Equation We may rewrite the closed surface integral of B using the divergence theorem, in which the right hand integral is taken over the volume surrounded by the closed surface: Because the result is zero, it follows that This result is known as Gauss’ Law for the magnetic field in point form. B.Hemalatha APECE 90Example: Magnetic Flux Within a Coaxial Line Consider a length d of coax, as shown here. The magnetic field strength between conductors is: and so: The magnetic flux is now the integral of B over the flat surface between radii a and b, and of length d along z: B d The result is: The coax line thus “stores” this amount of magnetic flux in the region between conductors. This will have importance when we discuss inductance in a later lecture. B.Hemalatha APECE 91Scalar Magnetic Potential We are already familiar with the relation between the scalar electric potential and electric field: So it is tempting to define a scalar magnetic potential such that: This rule must be consistent with Maxwell’s equations, so therefore: But the curl of the gradient of any function is identically zero Therefore, the scalar magnetic potential is valid only in regions where the current density is zero (such as in free space). So we define scalar magnetic potential with a condition: B.Hemalatha APECE 92Further Requirements on the Scalar Magnetic Potential The other Maxwell equation involving magnetic field must also be satisfied. This is: in free space Therefore: ..and so the scalar magnetic potential satisfies Laplace’s equation (again with the restriction that current density must be zero: B.Hemalatha APECE 93Example: Coaxial Transmission Line With the center conductor current flowing out of the screen, we have Thus: So we solve: .. and obtain: where the integration constant has been set to zero B.Hemalatha APECE 94Ambiguities in the Scalar Potential The scalar potential is now: where the potential is zero at ) the potential is At point P ( But wait As increases to we have returned to the same physical location, and the potential has a new value of I. In general, the potential at P will be multivalued, and will acquire a new value after each full rotation in the xy plane: B.Hemalatha APECE 95Overcoming the Ambiguity To remove the ambiguity, we construct a mathematical barrier at any value of phi. The angle domain cannot cross this barrier in either direction, and so the potential function is restricted to angles on either side. In the present case we choose the barrier to lie at so that The potential at point P is now singlevalued: Barrier at B.Hemalatha APECE 96Vector Magnetic Potential We make use of the Maxwell equation: .. and the fact that the divergence of the curl of any vector field is identically zero (show this) This leads to the definition of the magnetic vector potential, A: Thus: and Ampere’s Law becomes B.Hemalatha APECE 97Equation for the Vector Potential We start with: Then, introduce a vector identity that defines the vector Laplacian: Using a (lengthy) procedure (see Sec. 7.7) it can be proven that We are therefore left with B.Hemalatha APECE 98The Direction of A We now have In rectangular coordinates: (not so simple in the other coordinate systems) The equation separates to give: This indicates that the direction of A will be the same as that of the current to which it is associated. The vector field, A, existing in all space, is sometimes described as being a “fuzzy image” of its generating current. B.Hemalatha APECE 99Expressions for Potential Consider a differential elements, shown here. On the left is a point charge represented by a differential length of line charge. On the right is a differential current element. The setups for obtaining potential are identical between the two cases. Line Current Line Charge Scalar Electrostatic Potential Vector Magnetic Potential B.Hemalatha APECE 100General Expressions for Vector Potential For large scale charge or current distributions, we would sum the differential contributions by integrating over the charge or current, thus: and The closed path integral is taken because the current must close on itself to form a complete circuit. For surface or volume current distributions, we would have, respectively: or in the same manner that we used for scalar electric potential. B.Hemalatha APECE 101Example We continue with the differential current element as shown here: In this case becomes at point P: Now, the curl is taken in cylindrical coordinates: This is the same result as found using the BiotSavart Law (as it should be) B.Hemalatha APECE 102
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