Chpter 4 Syfruddin Hsn
Chpter Objectives Mgnetic force The totl electromgnetic force, known s Lorentz force Biot Svrt lw Guss s lw for mgnetism Ampere s lw Vector mgnetic potentil 3diff different ttypes of mteril Boundry between two different medi Mgnetic energy density
Chpter Outline 4-1) Mgnetic Forces nd Torques 4-) The Biot Svrt Lw 4-3) Mgnetic Force between Two Prllel Conductors 4-4) 4) Mxwell s Mgnetosttic t ti Equtions 4-5) Vector Mgnetic Potentil 4-6) Mgnetic Properties of Mterils 4-7) Mgnetic Boundry Conditions 4-8) Inductnce 4-9) Mgnetic Energy
Mgnetism nd electricity were considered distinct phenomen until 180 when Hns Christin Oersted introduced d n experiment thtt showed compss needle deflecting when in proximity to current crrying wire. He used compss to show tht current produces mgnetic fields tht loop round the conductor. The field grows weker s it moves wy from the source of current.
INTRODUCTION (Cont d) A represents current coming out of pper. A represents current heding into the pper. Mgnetic fields cn be esily visulized by sprinkling iron filings on piece of pper suspended over br mgnet.
INTRODUCTION (Cont d) The principle of mgnetism is widely used in mny pplictions: Motors nd genertors Mgnetic memory Microphones nd spekers Mgneticlly levitted high-speed vehicle. The field lines re in terms of the mgnetic field intensity, H in units of mps per meter. This is nlogous to the volts per meter units for electric field intensity, E. Mgnetic field will be introduced in mnner prlleling our tretment to electric fields.
Mxwell s equtions for mgnetosttic: where J is the current density The mgnetic flux density B nd the mgnetic field intensity H re relted by where μ is medium mgnetic permebility μ o free spce permebility 4π 10-7 H/m μ r reltive permebility of medium
4-1 Mgnetic Forces nd Torques When chrged prticle moving with velocity u, mgnetic force F m is produced. F m qu B ( N) where B mgnetic flux density (N/C-m/s or Tesl T) When chrged prticle hs E nd B, totl electromgnetic force is ( E + u B) F Fe + Fm qe + qu B q This force is known s Lorentz force
D 9.1 The point chrge Q 18 nc hs velocity of 5x10-6 m/s in the direction v 0.60 x + 0.75 y + 0.30 z. Clculte the mgnitude of force exerted on the chrge by the field : ) B - 3 x + 4 y + 6 z mt; 660 μn b) E - 3 x + 4 y + 6 z kv/m 140 μn c) B nd E cting together. 670 μn
4 1.1 Mgnetic Force on Current Crrying Conductor To find force on current element, consider line conducting current in the presence of mgnetic field with differentil segment dq of chrge moving with velocity u: But, u dl dt df dqu so B dq df m dl dt Since dq/dt corresponds to the current I in the line, df m IdL B We cn find the force from collection of current elements F m I dl B B
For surfce current For volume current One simple result is obtined to stright conductor in uniform mgnetic field, D 9. F I L x B
4-1.1 Mgnetic Force on Current-Crrying Crrying Conductor For closed circuit of contour C crrying I, totl mgnetic force F m is F m I dl B ( N) F m is zero for closed circuit. C ( F m I( dl ) B 0 C Curved wire in uniform B field
Exmple 4.1 Force on Semicirculr Conductor The semicirculr conductor shown lies in the x y plne nd crries current I. The closed circuit is exposed to uniform mgnetic field B ŷb yb 0. Determine () the mgnetic force F 1 on the stright section of the wire nd (b) the force F on the curved section. Solution ) b) ( Ir) yb ˆ zˆ ( N) ˆ 0 0 F 1 x IrB π π F I dl B zi ˆ rb0 sin φ d φ zˆ IrB0 φ 0 φ 0 ( N ) We note tht F 1 - F net Force on the closed loop is zero 1 p
EXAMPLE 1 The mgnetic flux density in region of free spce is given by B 3x x + 5y y z z T. Find the totl force on the rectngulr loop shown which lies in the plne z 0ndisboundedbyx 1,x 3, y, nd y 5, ll dimensionsi in cm. Try to sketch this!
SOLUTION TO EXAMPLE 1 The figure is s shown.
SOLUTION TO EXAMPLE 1 (Cont d) First, note tht in the plne z 0, the z component of the given field is zero, so will not contribute to the force. We use: F loop IdL Which in our cse becomes with, I 30A nd x y z x B B 3x + 5y z
SOLUTION TO EXAMPLE 1 (Cont d) So, F 0.03 30dx x 3x x + 5y y 0.0 y 0.01 ( ) + 0.0505 ( 3x + 5y ) 30dy y + x 0.03 x y 0.0 0.01 ( ) 3x + 5y 30dx x x + y 0.05 y 0.03 0.0 ( 3x 5y ) 30dy y + x 0.01 x y 0.05
SOLUTION TO EXAMPLE 1 (Cont d) Simplifying these becomes: F 0.03 0.05 30(5)(0.0) ) dx + 30(3)(0.03) ( )dy) z ) z y 0.01 0.01 0.03 0.0 0.0 30(5)(0.05) 05) dx 01) ( )d) z + 30(3)(0.01) dy + 0.05 ( 0.06 + 0.081081 0.150 0.0707 ) N z z F 36 z mn
4-1. Mgnetic Torque on Current-Crrying Loop Applied force vector F nd distnce vector d re used to generte torque T T d F (N m) Rottion direction is governed by right-hnd hnd rule. Mgnetic Field in the Plne of the Loop F 1 nd F 3 genertes torque in clockwise direction. T d 1 x F 1 + d x F
4-1. Mgnetic Torque on Current-CrryingCrrying Loop Mgnetic Field Perpendiculr to the Axis of Loop When loop consists of N turns, the totl torque is T N I AB sinθ 0 where N I A mgnetic moment, m The vector m with norml vector is expressed s m Δ nn ˆ I A ( A m ) T cn be written s T m B ( N m)
Biot-Svrt s Lw Stte: At ny point P, the mgnitude of the mgnetic field intensity produced by the differentil element is proportionl to the product of the current, the mgnitude of the differentil length, nd the sine of the ngle lying between the filment nd line connecting the filment to the point P t which h the field is desired; d lso the mgnitude of the mgnetic field intensity is inversely proportionl to the squre of the distnce from the differentil element to the point P.
3.1 BIOT-SAVART S LAW Jen Bptiste Biot nd Felix Svrt rrived mthemticl reltion between the field nd current. dh 1dL L1 I 4π R 1 1 Anlogous to Coulomb s lw de dq 1 4πε R 1 1 1 dl Rˆ d H 4 π R ( A/m ) where dh differentil mgnetic field dl differentil length
To determine the totl H we hve H I 4π l dl Rˆ R ( A/m ) (Line current) 4-.1 Mgnetic Field due to Surfce nd Volume Current Distributions 1 J Rˆ 4π R s H ds ( for surfce current ) H 1 4π S v J Rˆ dv R ( for volumecurrent )
DERIVATION Let s pply H Id L R 4πR to determine the mgnetic field, H everywhere due to stright current crrying filmentry conductor of finite length AB.
DERIVATION (Cont d)
DERIVATION (Cont d) We ssume tht the conductor is long the z-xis with its upper nd lower ends α1 α respectively subtending ngles nd t point P where H is to be determined. The field will be independent of z nd φ nd only depend on.
DERIVATION (Cont d) The term dl is simply dz z nd the vector from the source to the test point P is: R R z + R z Where the mgnitude is: R z + And the unit vector: R z z z z + +
DERIVATION (Cont d) Combining these terms to hve: H Id L Id L R 3 4πR 4πR Idz z + R B A z ( ) z ( z + ) 4 π 3
DERIVATION (Cont d) ( z ) Cross product of dz z z z + : d L R 0 0 dz This yields to: H B A 4π φ z 0 z dz ( z + ) 3 φ dz φ
DERIVATION (Cont d) Trigonometry from figure, tnα So, z cotα z Differentite to get: dz sec αdα H α I sec αdα 4 π α + cot α 1 ( ) 3 φ
DERIVATION (Cont d) Simplify the eqution to become: α I 4 π sec H 3 α α I sin 4π I 4π 1 α 1 α d α 3 φ ( sec α ) α d α ( cos α cos α 1 ) φ φ
DERIVATION 1 Therefore, I ( cos α α ) cos φ H 1 4 π This expression generlly pplicble for ny stright filmentry conductor of finite length.
DERIVATION As specil cse, when the conductor is semifinite with respect to P, A t ( 0,0,0) B t The ngle become: So tht, H 4π φ I ( 0,0, ) or ( 0,0, ) α 0 0 1 90, α 0
DERIVATION 3 Another specil cse, when the conductor is infinite with respect to P, A t ( 0,0, ) B t The ngle become: So tht, H π φ I ( 0,0, ) α 0 0 1 180, α 0
Exmple 4.3 Mgnetic Field of Pie-Shped Loop Determine the mgnetic field t the pex O of the pieshped loop s shown. Ignore the contributions to the field due to the current in the smll rcs ner O. Solution For segment AC, dl Rˆ zdl ˆ ˆ zdφ Consequently, 1 zd ˆ φ 1 H zˆ φ whereφ is in rdins 4 π 4 π
4-. Mgnetic Field of Mgnetic Dipole To find H in sphericl coordinte system, we hve where R >> m H + θ' 3 4 π R ' ( Rˆcosθ' θˆsin ) ( A/m)
Mgnetic Force between Differentil Current Elements The mgnetic field t point due to current element t point 1 The differentil force on differentil current element is df I dl x B The differentil force on element s d(df ): d(df ) I dl xdb since db μ o dh D9.4 Hyt
4-3 Mgnetic Force between Two Prllel Conductors Force per unit length on prllel current-crrying conductors is μ I I F' 0 1 yˆ 1 πd where F 1 -F (ttrct ech other with equl force)
4-44 Mxwell s Mgnetosttic Eqution There re importnt properties: Guss s nd Ampere s Lw. 4-4.14.1 Guss s Lw for Mgnetism Guss s lw for mgnetism sttes tht B 0 (differentil form) Bds Net electric mgnetic flux through closed surfce is zero. S 0 (integrl form)
4-4. AMPERE S LAW In mgnetosttic problems with sufficient symmetry, we cn employ Ampere s Circuitl Lw more esily tht the lw of Biot-Svrt. The lw sys tht the integrtion of H round ny closed pth is equl to the net current enclosed by tht pth. i.e. H d L I enc
for nd b pths: H.dL I for c pth: H.dL < I
AMPERE S LAW (Cont d) The line integrl of H round the pth is termed the circultion of H. To solve for H in given symmetricl current distribution, it is importnt to mke creful selection of n Amperin Pth (nlogous to gussin surfce) tht is everywhere either tngentil or norml to H. The direction of the circultion is chosen such tht the right hnd rule is stisfied.
DERIVATION 3 Find the mgnetic field intensity everywhere resulting from n infinite length line of current situted on the z-xis using Ampere s Lw. Solution Select the best Amperin pth, where here re two possible Amperin pths round n infinite length line of current. Choose pth b which hs constnt vlue of H φ round the circle specified by the rdius
Using Ampere s circuitl lw: We could find: So, H Solving for H φ : H d L H φ φ d φ π φ H d L I enc dl I enc H φ φ dφφ φ 0 H φ I π Where we find tht the field resulting from n infinite length line of current is the expected result: H I π φ Sme s pplying Biot-Svrt s Lw! I
DERIVATION 4 Use Ampere s Circuitl Lw to find the mgnetic field intensity resulting from n infinite extent sheet of current with current sheet K K x x in the x-y plne. solution Rectngulr mperin pth of height ΔhndwidthΔw. According to right hnd rule, perform the circultion in order of b c d
We hve: + + + d c b enc d d d d I d L H L H L H L H L H d c b From symmetry rgument, there s only H y component exists. So, H z will be zero nd thus the expression reduces e sts. So, z be e o d t us t e e p ess o educes to: + d c b enc d d I d L H L H L H c So, we hve: d d d d b + L H L H L H ( ) dy H dy H w y y y y y y c + Δ 0 w H y y y y w y y y Δ Δ 0
The current enclosed by the pth, I This will give: Δw w KdS 0 K x dy K H d L I enc H Δw K Δw y K x H y x x Δw or generlly, 1 H K N
EXAMPLE 3 An infinite sheet of current with K 6 z the x-z plne t y 0. Find H t P (3,,5). solution A/m exist on Use previous expression, tht is: 1 H K N N is norml vector from the sheet to the test point P (3,4,5), where: N y nd K 6 z so, 1 H 6z y 3 x A m
EXAMPLE 4 Consider the infinite length cylindricl conductor crrying rdilly dependent current Find H everywhere. J J 0 z
SOLUTION TO EXAMPLE 4 Wht components of H will be present? Finding the field t some point P, the field hs both nd components. φ The field from the second line current results in cncelltion of the components
To clculte H everywhere, two mperin pths re required: Pth #1 is for Pth # is for > Evluting the left side of Ampere s lw: H π d L H φ φ d φ φ π H 0 This is true for both mperin pth. φ
The current enclosed for the pth #1: 0 φ d d J d I w z z Δ S J 3 3 0 0 0 0 π φ π J d d J z z 3 0 0 φ Solving to get H φ : 3 0 φ J H or φ H 3 J 0 for The current enclosed for the pth #: 3 0 0 J d d J d I π φ π S J 3 0 0 0 d d J d I φ φ S J Solving to get H φ : 3 0 J > H 3 φ J 0 for >
SOLUTION TO EXAMPLE 4 (Cont d) Solving to get H φ : J 0 J H φ Or H 0 φ for 3 3 The current enclosed for the pth #: I π J ds J0 0φ 0 ddφ πj 0 3 3 Solving to get H φ : 3 H J 0 φ for > 3
EXAMPLE 5 Find H everywhere for coxil cble s shown.
SOLUTION TO EXAMPLE 5 Even current distributions re ssumed in the inner nd outer conductor. Consider four mperin pths.
It will be four mperin pths: < b b < c c > Therefore, the mgnetic field intensity, H will be determined for ech mperin pths. As previous exmple, only H φ component is present, nd we hve the left side of mpere s circuitl lw: H π d L H φ φ d φ φ π H 0 φ
For the pth #1: I enc J ds We need to find current density, J for inner conductor becuse the problem ssumes n event current distribution (< is solid volume where current distributed uniformly). J π I ds Where, So, ds J dφd, S dφd π I ds z I ππ φ 0 0 z z
We therefore hve: I enc π J ds φ 0 0 I π Equting both sides to get: H I φ z dφd I for π π z I For the pth #: The current enclosed is just I, I enc I, therefore: H dl πh Ienc φ I H φ for < b π I
NOTES The rdius of Amperin Pth#1 is smller thn the rdius of conductor (wire) The current I enc flowing through the re enclosed by < is equl to the totl current multiplied by the rtio of the re enclosed by the to the totl cross section re on the wire ( ) I enc π π I I
For the pth #3: For totl current enclosed by pth 3, we need to find the y p, current density, J in the outer conductor becuse the problem ssumes n event current distribution (<<b is solid volume where current distributed uniformly) given by: ( ) ( ) z z I I J ( ) ( ) ( ) z z b c ds π We therefore hve: ( ) ( ) 0 S J b c b I d d b c I d b z z φ π π φ But, the totl current enclosed is: S J b c c I b c b I I d I I enc + + b c b c
So we cn solve for pth #3: c H d L π H φ Ienc I c b I c H φ π c b for b < c For the pth #4: the totl current is zero. So, H 0 for > c φ This shows the shielding bility by coxil cble!!
SOLUTION TO EXAMPLE 5 (Cont d) Summrize the results to hve: Summrize the results to hve: I φ < b I π φ φ < c b c I π φ H > < c c b b c π φ 0
Fundmentl of Electromgnetics with Engineering Applictions : Wentorth Suppose in the coxil exmple just exmined.0 cm, b 4.0 cm, c 5.0 cm, nd I 1.0 A. Plot H Φ versus from 0 to 6 cm using MATLAB
AMPERE S CIRCUITAL LAW (Cont d) Expression for curl by pplying Ampere s Circuitl Lw might be too lengthy to derive, but it cn be described s: H J The expression is lso clled the point form of Ampere s Circuitl Lw, since it occurs t some prticulr point.
Find J t (m, 1m, 3m) if H xy z A/m. z / Find J t (3m, 90 o, 0) if H r sin Φ d J t (3, 90,0) s Φ θ A/m.
AMPERE S CIRCUITAL LAW (Cont d) The Ampere s Circuitl Lw cn be rewritten in terms of current density, s: H d L J d S Use the point form of Ampere s Circuitl Lw to replce J, yielding: H d L ( H ) d S This is known s Stoke s Theorem.
Infinitely long idel solenoid of rdius nd uniform current density K Φ H K z ( < ) H 0 ( > ) If the solenoid hs finite length d nd consists of N closely wound turns of filment tht crries current I, An idel toroid crrying surfce current K For N-turn toroid: (inside toroid) H 0 (outside) (inside toroid) H 0 (outside)
The point or differentil form of Ampere s circuitl Lw is: H J A closed line integrl is relted to surfce integrl by Stoke s Theorem: H dl ( H) ds
Exmple 4.6 Mgnetic Field inside Toroidl Coil A toroidl coil (lso clled torus or toroid) is doughnut-shped structure (clled its core) with closely spced dturns of wire wrpped round dit s shown. For toroid with N turns crrying current I, determine the mgnetic field H in ech of the following three regions: r <, < r < b, ndr > b, ll in the zimuthl plne of the toroid.
Solution 4.6 Mgnetic Field inside Toroidl Coil H 0 for r < s no current is flowing through the surfce of the contour H 0 for r > b, s equl number of current coils cross the surfce in both directions. Appliction of Ampere s lw then gives C H. dl π Hence, H NI/(πr) nd 0 ( ˆ φ H ) ˆ φ rd φ π rh NI NI H ˆ φh ˆ φ for < πrπ r ( < r b)
Solution 4.6 Mgnetic Field inside Toroidl Coil H 0 for r < s no current is flowing through the surfce of the contour H 0 for r > b, s equl number of current coils cross the surfce in both directions. Appliction of Ampere s lw then gives C H. dl π ( ˆ ) ˆ φh φ rdφ πrh NI 0 Hence, H NI/(πr) nd NI H ˆ φh ˆ φ for < πr ( < r b)
SCALAR AND VECTOR MAGNETIC POTENTIAL Sclr mgnetic potentil For ny vector of vector mgnetic potentil A (Wb/m), ( A ) 0 We re ble to derive B A ( Wb/m ) where nd nd Vector Poisson ss equtionisis given s
4-6 MAGNETIC PROPERTIES OF MATERIALS Mgnetic behvior is due to the interction of dipole nd field. 6 types of mgnetic mterils: Dimgnetic, Prmgnetic, Ferromgnetic. Antiferromgnetic Ferrimgnetic Superprmgnetic p
4-6.1 Orbitl nd Spin Mgnetic Moments Electron genertes round the nucleus nd spins bout its own xis. Orbiting electron Spinning electron
4-6. Mgnetic Permebility Mgnetiztion vector M is defined s where χ m mgnetic susceptibility (dimensionless) Mgnetic permebility is defined s μ μ + ( 1 χ ) ( H/m) 0 m nd reltive permebility is μ defined s μ r 1+ χ m μ μ 0
4-6. Mgnetic Permebility Mgnetiztion vector M is defined s where χ m mgnetic susceptibility (dimensionless) Mgnetic permebility is defined s μ μ + ( 1 χ ) ( H/m ) 0 m nd reltive permebility is defined s μ μ 1+ r χ m μ0
4-6.3 Mgnetic Hysteresis of Ferromgnetic Mterils Ferromgnetic mterils is described by mgnetized domins. Properties of mgnetic mterils re shown below.
4-6.3 Mgnetic Hysteresis of Ferromgnetic Mterils Comprison of hysteresis curves for () hrd nd (b) soft ferromgnetic mteril is shown.
3.3 MAGNETIC FLUX DENSITY In electrosttics, it is convenient to think in terms of electric flux intensity nd electric flux density. So too in mgnetosttics, where mgnetic flux density, B is relted to mgnetic field intensity by: nd μ μ o μ r B μ H Where: B mgnetic flux density (Wb/m ) or tesl T H mgnetic field intensity (A/m) μ permebility of medium with: μ0 4π 10 7 H m μ r reltive permebility of medium
MAGNETIC FLUX DENSITY (Cont d) The mount of mgnetic flux, Φ in Webers from mgnetic field pssing through surfce is found in mnner nlogous to finding electric flux: Φ B ds Guss Lw for electrosttics: The net electric flux through ny closed surfce is equl to the totl chrge enclosed by tht surfce D.dS Q enc
()Find B for n infinite length line of 3.0 A current going in the + z direction long the z-xis in free spce. (b)find the mgnetic flux through surfce defined by 1.0 m <40m 4.0 m, 0<z<30m < 3.0 m, φ 90 o
EXAMPLE 6 Find the flux crossing the portion of the plne φπ/4 / defined by 0.01m < r < 0.05m nd 0 < z < m in free spce. A current filment of.5a is long the zxis in the z direction. Try to sketch this!
SOLUTION TO EXAMPLE 6 The reltion between B nd H is: B I μ0 H μ0 φ π To find flux crossing the portion, we need to use: Φ B d S where ds is in the φ direction.
SOLUTION TO EXAMPLE 6 (Cont d) So, d S d dz Therefore, Φ φ B ds 0.0505 z 0 0.01 μ0 I π φ d dz μ0 ln 1.61 10 π 0.01 I 0.05 6 φ Wb
4-7 Mgnetic Boundry Conditions For different medi when pplying Guss s lw, we hve D1 n Dn B1 n Bn s μ 1H μ H Boundry condition for H is 1n n. Vector defined by the right-hnd rule is ( H H ) J s n ˆ 1 At interfce between medi with finite conductivities, Js 0 nd H1 1tt H tt.
EXAMPLE 9 The mgnetic field intensity is given s: H 1 6 x + y + 3 A m In medium with µ r1 6000 tht exist for z < 0. Find H in medium with µ r 3000 for z>0. x
SOLUTION TO EXAMPLE 9
4-8 Inductnce An inductor is mgnetic cpcitor. An exmple is solenoid s shown below.
4-8.1 Mgnetic Field in Solenoid For cross section of solenoid, B μniμ zˆ ( sinθ θ ) sin When l >, θ 1 90 nd θ 90,, B zˆ μnl zˆ μni >> l ( long solenoid with l / 1) 1
4-8. Self-Inductnce Mgnetic flux is given by Φ Φ Bds S ( Wb ) To compute the inductnce we need re S.
4-8. Self-Inductnce The self-inductnce L of conducting structure is defined s L Λ ( H) I Λ where totl mgnetic flux (mgnetic flux linkge) For solenoid, L N μ l S ( solenoid) For two-conductor configurtions, L Λ I Φ I 1 Bds I S
4-8.3 Mutul Inductnce Mgnetic field lines re generted by I 1 nd S of loop. The mutul inductnce is L Λ I N I 1 1 1 1 s B ds 1 ( H) Trnsformer uses torodil coil with windings.
4-9 Mgnetic Energy Totl energy (Joules) expended in building up the current in inductor is W l m pdt ivdt L idi 0 1 LI Some of the energy is stored in the inductor, clled mgnetic energy, W m. m The mgnetic energy density w m is defined s w m W 1 H ( 3 J/m ) m μ v
Exmple 4.9 Mgnetic Energy in Coxil Cble Derive n expression for the mgnetic energy stored in coxil cble of length l nd inner nd outer rdii nd b. The insultion mteril hs permebility µ. Solution The mgnitude of the mgnetic field is H B 1 μ πr Mgnetic energy stored in the coxil cble is given by W m 1 μiμ I μh dv 8π V 1 r μ I 1 μ I b ln 8 πrldr π r 8π V V dv ( ) ( J)