EE468G NOTES (5C) Reading assignment: Contents: Magnetic material and inductanceMagnetic material and boundary condition Rotation of electron around nucleus form magnetic dipole kM. The total magnetic dipole per unit volume is defined as the magnetization vector: Magnetization vector: 0,limkkvMMv [A/m] The magnetic potential produced by the magnetization vector is SnVmdSRaMdVRMA 'ˆ4''400 Meanwhile, the magnetic potential produced by the imposed volume current and surface current is given by SsVidSRJdVRJA '4'400 We conclude that the effect of magnetization vector is equivalent to an induced volume and surface current: nSMMaMJMJˆ, naˆ is the outward normal direction of S.Using the induced currents in the Ampere’s law, we have  MJJJJBM000total0 This means:  JMB0/ Comparing this equation with Maxwell equation: JH, Hence we established that ,/0MBH or MHB00μ Simple magnetic material (isotropic), M is parallel to H,  mmmHHHBHM10000 Example: A material has 2000r and atom density of 27 -32 10 [m ]. If it is put in a uniform magnetizing field with 0.15mB  [T], (1) calculate Mand the magnetic moments per atom. Solution: (1) From 0BHM, we have 0 0 0 0111rrB B BM H B           M is in the same direction of B, hence 67111 0.15 0.12 10 [A/m]4 10 2000M     (2) Moment per atom is: 621270.12 100.06 10 [A-m]2 10m  Magnetic field boundary condition is derived from IldHC and 0SSdB The boundary conditions are: SJHHn )(ˆ21 0)(ˆ21- BBn or 0)(ˆ2211- HHnExample (8.3): At the material interface as shown, derive the relationship between the magnitudes and angles of B on the two sides of the interface. Assume there is no surface current on the interface. Solution: The normal component of 1B is 11cosB The normal component of 2B is 22cosB Boundary condition for B requires that: 2211coscosBB  (1) The tangent component of 1H is  111sin/B The tangent component of 2H is  222sin/B Since there are no imposed electric surface current on the interface, the boundary condition for H requires that:    222111sin/sin/BB  (2) Solving (1) and (2), we have: 21 2 2222 1 2 1 1 111tan tan , sin cosBB               1 2 2B 1 2 1BNumerical example: let 01 1 1 0 2 00.1[T], 10 , 2 , 1000B        we find from the above expression that  1 1 0022110122222 1 1 112 2 21000tan tan tan tan102tan 88.1689.35 ,sin cos0.1 500 sin (10 ) cos (10 )0.1 7539.48.68 [T]oooBB  Faraday’s Law of inductance: --SCSdBdtdldE The differential form of the Farady’s law is: BEt   Example (9.1): A high-permeability core that carries a uniform, time-varying flux density:  0ˆcoszB B t a, calculate the electric field generated inside the core. Solution: From Farady Law (differential equation form):  0ˆsinzBE B t at    By symmetry, E-field has -component only, and depends ononly. Hence the curve operation give rise to:  1ˆzEa. This means:     0011sin , sin2EB t E B t       B EExample (9.2): Calculate the gap voltage in the rectangular loop formed by a conducting wire if the flux density B passing through the loop is uniform and given by:  0ˆsinzB B t a. Solution: For a counterclockwise path around the loop, dxdyzSdˆ dxdytBdxdyzztBSdB )sin(ˆˆ)sin(00-- Faraday’s law yields  00sin( )( ) cosC S SddE dl B dS B t dxdydt dtab B t        On the other hand, the line integral of the electric field around the path evaluates to the voltage on the gap. Hence  0cosgCV E dl ab B t   . a b gV - + x yExample (9.3): A magnetic field  2ˆsin [Wb/m ]zB t a links the circuit as shown. Calculate the voltage recorded by the voltage meter. (a) (b) Solution: Use Faraday’s law   26(100 200) 2 10 cos 0.01CSdE dl i B dS tdt        Solve for the current, we have: 0.666 cos [A]it Hence the voltages on the two resistances are:   100200100 66.6 cos [ ]200 133.3 cos [ ]V i t VV i t V    (1) For this case, we evaluate Farady’s law on the path: abcd, since there is no magnetic flux passing this loop, we have 100100066.6cos( ) [ ]mabcdmE dl V VV V t V      (2) For this case, we evaluate the integral on the path of adcb,   200200200cos200cos 66.6cos( ) [ ]mabcd SmdE dl V V B dS tdtV V t t V           Meter 100V 200V mV a b c d 200V 100V Meter a d c i b iFlux linkage: total magnetic flux passing through a surface. It is defined as the following integral: Flux linkage (single turn): SB dS   [Web] For N-turn wire wind on a cylinder, they form N-faces. The total flux linkage passing through this N-turn wire is then N   Using the flux linkage, the Faraday’s law can be written as: CdE dldt   S BSelf-inductance: 1 1 111111Flux linking C due to current in C [H]Current in CLI Circuit model for self inductance: divLdt To calculate self inductance: 1. Set current I on a closed loop C 2. Calculate the magnetic flux linkage 3. Take the ration of the linkage