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Chapter 9 MAXWELL'S EQUATIONS

Part Two iugaza2010.blogspot.com [email protected] conducting bar is connected via flexible leads to a pair of rails in a magnetic field B = 6 cos 10t ax mWb/m2 as in Figure 9.20. If the z-axis is the equilibrium position of the bar and its velocity is 2 cos 10t ay m/s, find the voltage induced in it.

motional transformer emf 2 Ba 6cos(10t )x mWb/m

ut 2cos(10 )ay m/s d  V  emf dt 5 y 10  B. dS  6cos(10 t )aa . dydz  6cos(10 t ) (5)( y  10)     xx   00 30(yt 10)cos(10 ) 3 dy u  dt sin(10t ) y  udt 2cos(10 t ) dt   5 sin(10t )  30(y  10)cos(10 t )  30  10 cos(10 t ) 5 6sin(10tt ) 300 cos(10 ) dB V E. dl   . dSB  ( u  ). dl emf  dt  dB B6cos(10tt ) a mWb/m2 ,   60sin(10 ) a xxdt ut 2cos(10 )ay m/s dS dydz aaxz , dl=dz 4 dB V . dSB  ( u  ). dl emf dt 5510 y  60sin(10t ) dydz    12cos2 (10 t ) dz 0 0 0 60sin(10tt ) (5)(10+y) 12cos2 (10 ) (5) 2 Vemf =300(10+y)sin(10 t ) 60cos (10 t ) dy u  dt sin(10t ) y  udt 2cos(10 t ) dt   5

5 A car travels at 120 km/hr. If the earth's magnetic field is 4.3 X 10"5 Wb/m2, find the induced voltage in the car bumper of length 1.6 m. Assume that the angle between the earth magnetic field and the normal to the car is 65°.

V( u B ). dl  | u || B | sin  emf  120,000 = ( )(4.3 105 )sin(25) (1.6) 0.96 mV 3600

6 As portrayed in Figure 9.21, a bar magnet is thrust toward the center of a coil of 10 turns and resistance 15 fl. If the magnetic flux through the coil changes from 0.45 Wb to 0.64 Wb in 0.02 s, what is the magnitude and direction (as viewed from the side near the magnet) of the induced current? d 0.64 0.45 VN   (10)   95 V emf dt 0.02 0 V 95 I emf   6.33 A R 15

7 - It is not an electric current of moving charges , but a time-varying electric field .

-This provides a mechanism whereby a time-varying electric field can create a magnetic field, complementary to Faraday's law, in which a time-varying magnetic field can produce an electric field.

- "mutual embrace" of electric and magnetic fields can produce propagating electromagnetic waves. This would not be possible without the displacement current.

8 Types of current: 1) Conduction current: through conductor J=σE

2) Convection current: through isolator J=ρvu 3) Displacement current: is due to time varying field.

Simple example(RC circuit)

- For resistor the conduction current model is valid(JR= σRER) - But the conduction current doesn’t characterized the

capacitor current for ideal capacitor (perfect insulator σc=0) - The capacitor current depends on charging and discharging of the capacitor plates.

9 dQ A it( ) , Q=CV(t)=C E (t) d= E (t) d= AE (t) c dtc d c c dE (t) dD (t) i() t Acc A c dt dt i( t ) dD (t) c  c A dt dD (t) J  c (displacement current density) D dt . The displacement current doesn't exist under static conditions. dD (t) J Jc , J= E+ u tot dt v 10 Ampere's law// Ien . H dl (   ). H dS  J. dS    tot dD (t) HJJ    c tot dt

- Without the term Jd, electromagnetic wave propagation (radio or TV waves, for example) would be impossible.

11 . For non ideal capacitor: JE  dD JE D dt J   J D 

- At low frequencies, Jd is usually neglected compared with J. However, at radio frequencies, the two terms are comparable

  good conductor (JD negligible)

  good insulator (Jconduction negligible) 12 In free space E=20cos(wt-50x) ay V/m , find: (a) Jd (b) H dD dE (a) J    20   sin(  t  50 x ) a A/m ddt o dt o y

(b)  H=J+ Jdd  J dH dH dHzyy  dH x dH z  dH x  H   ax   a y     a z dy dz  dz dx  dx dy 

dHxz dH    20o  sin( tx  50 ) dz dx dH z  20  sin( tx  50 ) dx o cos(t  50)x H  20 0.4  cos( tx 50 ) zo 50 o 13 H0.4o  cos(  t 50 x ) az A/m A 50-V voltage generator at 20 MHz is connected to the plates of an air dielectric parallel plate capacitor with plate area 2.8 cm2 and separation distance 0.2 mm. Find the maximum value of displacement current density and displacement current.

V50cos( t )  50cos(2  20  106  t ) A 2.8 104 dV A d(50cos(  t ))  A I C   50 sin( t ) d dt d dt d  (2.8 104 ) I o 50(2  20  106 )  0.0778 A d ,max 0.2 103 I 0.0778 J d ,max   277.7 A/m2 d ,max A 2.8 104

14 The ratio JIJd (conduction current density to displacement current density) is very important at high frequencies. Calculate the ratio at 1 GHz for: - sea water (µ=µo, ε = 8εo, σ = 25 S/m)

J  25  9  5.55 ,good conductor J Do(2  10 )(81  )

15 A conductor with cross-sectional area of 10 cm2 carries a conduction current 9 6 0.2 sinl0 t mA. Given that σ = 2.5 X 10 S/m and εr = 6, calculate the magnitude of the displacement current density.

0.2 103 |JEE | 0.2     8  1011 c 2.5 106 dD J j E D dt 9 11 2 |JEDo |  (2   10 )(6  )(8  10 )  26.7 pA/m

16 Maxwell’s equations in final forms   Gauss’s Law :  D d S Qenc v dv   Gauss’s Law for magnetic :  B d S 0 (Non existence of isolated magneticcharge)

  d B.dS Faraday's Law:  E d L  dt   dDc (t) Ampere's Law:  H d L Ienc   J  .dS dt Maxwell’s Equations in final forms

 D v  B0

 dB E   dt  dD HJ   dt 18