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Compare to principle of superposition applied to determine the electric field inside and outside an infinite parallel plate capacitor ELEC 3105 BASIC EM AND POWER ENGINEERING Force and torque on Magnetic Dipole Magnetic dipole = product of current in loop with surface area of loop DEFINITION OF MAGNETIC DIPOLE I = current in loop A “S” = surface area of the loop 𝑛 = unit vector normal to loop surface - RHR Magnetic moment 𝑚 vector in the direction of 𝑛 𝑚 = 𝐼𝐴 = 𝐼"𝑆" Units of Am2 ẑ FORCE ON A MAGNETIC DIPOLE Consider a circular ring of current I placed at the end of a solenoid as shown in the figure. The current in the solenoid produces a magnetic field in which the current loop is placed into. By postulates 1 and 2 of magnetic fields, the current ring will be subjected to a magnetic force. out of page into page B I ẑ FORCE ON A MAGNETIC DIPOLE F Circular ring F out out F down F out F B F down I Cancel in pairs around the ring out F down F down Will add in same direction on ring giving a net force. ẑ FORCE ON A MAGNETIC DIPOLE F Circular ring F out out F 2r down F down F B F down I Will add in same direction on ring giving a net force. down Using postulate 1: Gives: We need for find Br dF I Bd Fdown 2rIBr B B z B r F down FORCE ON A MAGNETIC DIPOLE z Gaussian cylinder B z We will relate Br to z z Total magnetic flux through Gaussian cylindrical surface must be zero. As many magnetic field lines that enter the surface, leave the surface. No magnetic charges or monopoles. B 0 Another important property of B Recall B 0 everywhere No net magnetic flux through any closed surface. Closed surface S B da 0 S Using divergence theorem B dv vol 0 vol 0 11 3-D view FORCE ON A MAGNETIC DIPOLE z Flux through top: r B z z 2 z top z Flux through side: 2rzB r side Flux through bottom: r B z 2 z bottom side top bottom 0 3-D view FORCE ON A MAGNETIC DIPOLE side top z 0 bottom z 2rzB r B z z r B z 0 2 r 2 z z r B z z B z B 2 z z z r r B B 2 z z r We can now use this in our force on current ring expression 3-D view ẑ FORCE ON A MAGNETIC DIPOLE F Circular ring B F out out F down 2r I F down We have found Br r B F 2rIB B z down r 2 z r B B z F down r B 2rI 2 z B r z F down ẑ FORCE ON A MAGNETIC DIPOLE Circular ring F F out out 2r F I F down down F r B 2rI 2 z z down F down F down B B A I z z F down B r I z 2 z B m z B B z z B F m z Force pulls dipole into region z z of stronger magnetic field B r F down FORCE ON A MAGNETIC DIPOLE z In general Fx m Fy m Fz m Bx x B y y Bz z Solenoid with axis along x Solenoid with axis along y Solenoid with axis along z Form suggests some sort of dot product with the del operator 3-D view FORCE ON A MAGNETIC DIPOLE z In general F m B F m B F m B x x y y z z F m B 3-D view TORQUE ON A MAGNETIC DIPOLE m B Side view We will consider a dipole in a uniform magnetic field. We can use any shape we want for the dipole. Here we will select a square loop of wire. I out of page a I into page a I Wire loop Top view TORQUE ON A MAGNETIC DIPOLE a a 2 a a I Wire loop Top view TORQUE ON A MAGNETIC DIPOLE m F Side view Top view a B F a Pivot point I Pivot line Torque attempts to align dipole Wire loop moment m with B. TORQUE ON A MAGNETIC DIPOLE m F B a 2 Side view F r F Total torque 2F Pivot point a sin 2 F => Magnetic force on wire of length a Torque attempts to align dipole moment m with B. TORQUE ON A MAGNETIC DIPOLE m F a 2 Side view 2F F a sin 2 Pivot point F => Magnetic force on wire of length a F IBa Then Through postulate 1 for magnetic fields a 2 IB sin B TORQUE ON A MAGNETIC DIPOLE m F B a 2 Side view a 2 IB sin m B sin m B F Pivot point a a m a2 I I Wire loop ELEC 3105 BASIC EM AND POWER ENGINEERING Boundary conditions Inductance Magnetic energy Principle of virtual work 21 NORMAL COMPONENT OF B FIELD AT BOUNDARY B dA 0 S Gaussian surface B 2 Area A B normal 2 Interface Very thin T 0 2 1 B normal 1 B Area A 1 Net flux through a closed surface is zero. B2 normal B1 (normal ) The normal components of B are continuous across the interface 22 TANGENTIAL COMPONENT OF H FIELD AT BOUNDARY H H d I 0 B H 2 H tangential S 2 Square closed path Length L Interface Very thin 2 1 Length L T 0 H H tangential 1 1 Integral of H around closed path is equal to the current enclosed (I = 0) H 2 tangential H1 ( tangential) The tangential components of H are continuous across the interface THIS BOUNDARY CONDITION ASSUMES NO SURFACE CURRENT AT THE INTERFACE. 23 TANGENTIAL COMPONENT OF H FIELD AT BOUNDARY H d I H 2 Square closed path Very thin 2 H tangential S Interface B H Length L X X X X X X H H tangential H (tangential) X X X X X X X X X 1 Length L T 0 Surface Current K2 1 X 2 H tangential 1 1 H 2 tangential H1 ( tangential) K The tangential components of H are discontinuous across the interface THIS BOUNDARY CONDITION ASSUMES A SURFACE CURRENT AT THE INTERFACE. 24 SUMMARY OF BOUNDARY CONDITIONS (GENERAL) 25 SUMMARY OF BOUNDARY CONDITIONS (CONDUCTORS) 26 SUMMARY OF BOUNDARY CONDITIONS (CONDUCTORS) 27 SELF INDUCTANCE Introduction A transformer is a device in which the current in one circuit induces an EMF in a second circuit through the changing magnetic field. B, H, and M relationship 17 To understand how current in one circuit induced EMF in another, we will first examine how a current in a circuit can induce an EMF in the same circuit. 28 SELF INDUCTANCE INDUCTANCE SELF Consider a single wire loop Enclosed surface S Current in loop produces a magnetic field B , giving a flux through the loop. B Bi Biot-Savard Law This expression is The Biot-Savard Law Consider a small segment of wire of overall length d Thus: i i The Biot-Savard law applied to the small segment gives an element of magnetic field dB at the point P. I d rˆ21 dBr1 o 4 r21 2 i dB P Magnetostatics Same result as postulate 2 for the magnetic field POSTULATE 2 FOR THE MAGNETIC FIELD A current element I d produces a magnetic field B which at a distance R is given by: I Rˆ dB o d 4 R 2 dB r 21 v I Units of {T,G,Wb/m2} 26 Lecture 15 slid 26 d 13 From Biot-Savard Law WRITE: Li 29 SELF INDUCTANCE Consider a single wire loop Enclosed surface S Current in loop produces a magnetic field B , giving a flux through the loop. B Li i i L is the self inductance of the loop d di v L dt dt v emf t 30 SELF INDUCTANCE Consider a single wire loop Enclosed surface S Current in loop produces a magnetic field B , giving a flux through the loop. B Li i It is difficult to compute L for a simple wire loop since the magnetic field produced by the loop is not constant across the surface of the loop. A possible solution is to find B at center of loop and then approximate: B S center i v v d di L dt dt 31 SELF INDUCTANCE A simple example for the calculation of a self inductance is the long solenoid. Magnetic field of a long solenoid Current out of page Axis of solenoid Magnetic field of a long solenoid Magnetic field of a long solenoid In the vicinity of the point P Bb 0 Current out of page N : number of turns enclosed by length L Current out of page P 1 3 2 4 11 Current into page 3 5 Axis of solenoid 41 B 0 P 2 P Infinite coil of wire carrying a current I Evaluate B field here 5 4 resultant B Current into page Expect B to lie along axis of the solenoid Implies that B field has no radial component. I.e. no component pointing towards42 or away from the solenoid axis. Magnetic field of a FINITE solenoid Magnetic field of a FINITE solenoid Current out of page NIr dI d L sin Axis of solenoid a sin r B dB P Current into page dB L z a NIr 2 2r finite solenoid start 2 3 La Magnetic field of a FINITE solenoid d 2 L 2 1 1 2 3 4 5 2 z r a NI o 2r L r d d L dB We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid. NI cos cos B NI 2 L B sin d 2 L NI cos cos zˆ B 2 L z a o z 2 o 50 3 d NI •B is independent of distance from the axis of the long solenoid as we are inside the solenoid! •B is uniform inside the long solenoid. d z 28 2 o dB Evaluate B field here Cross-section cut through solenoid axis o z finite coil of wire carrying a current I Radius of solenoid is a. dI a L infinite solenoid (36) sub in a o NI 1 2 o sin d z 1 dB o 1 z 34 2 35 32 SELF INDUCTANCE Current out of page Long solenoid of length B N turns of wire carrying current I AREA A NI B o B is uniform over the cross-section of the solenoid 33 SELF INDUCTANCE Long solenoid of length NI B o B AREA A Flux through one loop of area A 1 NIA o 34 SELF INDUCTANCE Long solenoid of length B NI B AREA A Flux through all N loops of solenoid o From LI N N Then L 1 N IA 2 o N A 2 o 35 SELF INDUCTANCE Long solenoid of length NI B o LI L N A 2 o AREA A Self inductance of a long solenoid of N turns with a current I in the windings. The solenoid has cross-sectional area A. 36 EXAMPLE: SELF INDUCTANCE 𝐼 Calculate the “self inductance” per unit length for a segment of a coax cable. Inner radius (a), outer radius (b). Example completed in class 37 Energy in Magnetic Field Consider a long solenoid in order to develop a general expression for the energy stored in a magnetic field. Magnetic field of a long solenoid Current out of page Axis of solenoid Magnetic field of a long solenoid Magnetic field of a long solenoid In the vicinity of the point P Bb 0 Current out of page N : number of turns enclosed by length L Current out of page P 1 3 2 4 11 Current into page 3 5 Axis of solenoid 41 B 0 P 2 P Infinite coil of wire carrying a current I Evaluate B field here 5 4 resultant B Current into page Expect B to lie along axis of the solenoid Implies that B field has no radial component. I.e. no component pointing towards42 or away from the solenoid axis. Magnetic field of a FINITE solenoid Magnetic field of a FINITE solenoid Current out of page NIr dI d L sin Axis of solenoid a sin r B dB P Current into page dB L z a NIr 2 2r finite solenoid start 2 3 La Magnetic field of a FINITE solenoid d 2 L 2 1 1 2 3 4 5 2 z r a NI o 2r L r d d L dB We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid. NI cos cos B NI 2 L B sin d 2 L NI cos cos zˆ B 2 L z a o z 2 o 50 3 d NI •B is independent of distance from the axis of the long solenoid as we are inside the solenoid! •B is uniform inside the long solenoid. d z 28 2 o dB Evaluate B field here Cross-section cut through solenoid axis o z finite coil of wire carrying a current I Radius of solenoid is a. dI a L infinite solenoid (36) sub in a o NI 1 2 o sin d z 1 dB o 1 z 34 2 35 38 Energy in Magnetic Field Current out of page May have core with constant permeability Long solenoid of length NI B AREA A N turns of wire carrying current I Find work done by current source in building up magnetic field: Power V I V d dI L dt dt 39 ENERGY IN MAGNETIC FIELD Energy in Magnetic d dI V L dt dt dW Power V I dt THEN I W LI dI Field d dW I dt dt THEN 0 THEN LI W 2 2 d dW I dt dt Energy stored 40 ENERGY IN MAGNETIC FIELD Energy in Magnetic LI W 2 W 2 Field N A 2 L Energy stored N AI 2 B 2 2 NI For core solenoid 1 N I W A 2 2 2 2 2 enclosed volume 1 W B A 2 2 41 ENERGY IN MAGNETIC FIELD Energy in Magnetic Field Total magnetic energy stored in solenoid W 1 B 2 A 2 W Energy density 1 2 B dv 2 vol W VOLUME 2 W B VOLUME 2 EXPRESSION VALID FOR ALL 42 ENERGY IN MAGNETIC FIELD Energy in Magnetic Energy in Magnetic Field Energy in Magnetic Field Current out of page May have core with constant permeability Long solenoid of length NI B dW Power V I dt AREA A THEN dW L I W LI dI N turns of wire carrying current I d dI L dt dt d I dt dt THEN dW L THEN Find work done by current source in building up magnetic field: d dI V L Power V I dt dt Energy in Magnetic Field V 0 W LI 2 2 Field Total magnetic energy stored in solenoid Energy density 1 B A 2 2 W VOLUME 2 d I dt dt W W B VOLUME 2 Expression valid for all Energy stored Energy in Electric Field 43 Energy in Electric Field For electric fields, we argued that the energy was really stored in the potential energy of the charged particle’s positions, since it would require that much energy to take separate charges and form that distribution from a universe with equally distributed charges. Energy in Magnetic Field This is harder to do for magnetic fields since there are no magnetic charges. But one possible approach is to take current loops enclosing zero area, and consider the forces on the wires as we expand the loops so as to form the current distributions which generate the magnetic field. 44 PRINCIPLE OF VIRTUAL WORK (MAGNETIC) We can use the principle of virtual work to determine forces as we did for electric forces. Gives correct magnitude F mag U s mag Energy stored in magnetic field Position variable Forces in Electrostatics Conductor caries a surface charge of density Find force on plates of a parallel plate capacitor. Plate area A x U d L 2 xLS yd y F F E d s oE 2 U o E 2 xD y 2 L Force pulling metal insert into capacitor Be very careful using the principle of virtual work F o E 2 xD 2 45 PRINCIPLE OF VIRTUAL WORK (MAGNETIC) Magnetic Relays 46 PRINCIPLE OF VIRTUAL WORK (MAGNETIC) Use principle of virtual work to obtain expression for the magnetic force on the movable contact. I Magnetic Relays GAP Movable contact V Metal spring provides restoring force when current is zero Example completed in class 47 MUTUAL INDUCTANCE Enclosed surface S1 B Enclosed surface S2 2 1 i i 2 1 i v v Loop 1 Loop 2 1 2 We shall consider two current loops close together. 48 MUTUAL INDUCTANCE B Suppose current i1 flows in loop 1, creating a flux in the loop and a flux in loop 2. We will set the source current i2 zero for now. 1 12 1 S S 1 2 2 B da i 1 12 1 2 S2 v 1 Loop 1 Loop 2 Integral over loop 2 surface Magnetic field of loop 1 in the region of loop 2 Flux of loop 2 produced by current in loop 1 49 Now some math!!!! MUTUAL INDUCTANCE B da Using magnetic vector potential A da Using Stoke’s theorem A d Using definition of magnetic vector potential i d d 4 r d d Rearrange terms i 4 r 12 1 2 S2 1 12 2 S2 12 2 1 2 o 1 12 2 1 1 2 21 o 12 1 1 2 1 21 50 2 MUTUAL INDUCTANCE d d i 4 r o 12 1 1 2 1 i M 12 1 2 21 12 Constant that depends on loop geometry FLUX IN LOOP 2 DUE TO CURRENT IN LOOP 1 51 MUTUAL INDUCTANCE B Suppose current i2 flows in loop 2, creating a flux in the loop and a flux in loop 1. We will set the source current i1 zero for now. 2 21 1 S S 1 2 2 i 2 B da 21 2 1 S1 v 1 Loop 1 Loop 2 Integral over loop 1 surface Magnetic field of loop 2 in the region of loop 1 Flux of loop 1 produced by current in loop 2 Now some math!!!! MUTUAL INDUCTANCE B da Using magnetic vector potential A da Using Stoke’s theorem A d Using definition of magnetic vector potential i d d 4 r d d i Rearrange terms 4 r 21 2 1 S1 2 21 1 S1 21 1 2 1 o 2 21 1 2 2 1 12 o 21 2 2 1 2 1 12 53 MUTUAL INDUCTANCE d d i 4 r o 21 2 2 1 2 i M 21 2 1 12 21 Constant that depends on loop geometry FLUX IN LOOP 1 DUE TO CURRENT IN LOOP 2 54 MUTUAL INDUCTANCE d d i 4 r o 12 1 1 2 1 i M 12 1 d d i 4 r o 2 21 21 12 2 Conclusion M’s are geometrical factors 2 1 1 2 12 i M 21 2 21 M M M 12 21 MUTUAL INDUCTANCE BETWEEN LOOPS 55 MUTUAL INDUCTANCE General result Mutual Inductance Enclosed surface S1 B Enclosed surface S2 v 1 d d di di L M dt dt dt dt 1 21 1 2 1 2 1 i i v 2 1 2 i v v2 Loop 1 Loop 2 d d di di M L dt dt dt dt 2 12 1 2 2 Sign convention 1 We shall consider two current loops close together. Indicates v2 positive when v1 is positive v 1 i i 1 2 primary v 2 56 ELEC 3105 BASIC EM AND POWER ENGINEERING START Extra 57 EXTRA ON INDUCTANCE 58 EXTRA ON INDUCTANCE 59 EXTRA ON INDUCTANCE 60 EXTRA ON INDUCTANCE 61 EXTRA ON INDUCTANCE 62 EXTRA ON INDUCTANCE 63 EXTRA ON INDUCTANCE 64