MAXWELL S EQUATONS Maxwell s four equations n the 870 s, James Clerk Maxwell showed that four equations constitute a complete description of the electric and magnetic fields, or THE ELECTROMAGNETC FELD Gauss s Law for E Gauss s Law for B We ve met these three already Faraday s Law Maxwell s modified version of Ampere s Law Why does Ampere s law need to be modified? a) Apply it to a continuous conductor: B d L around this path = 0 = current passing through an arbitrary bag-like surface whose edge is the path. B = magnetic field created by the moving charges in the wire. Surface b) Apply it to a capacitor being charged: Let the bag-like surface pass BETWEEN the plates Surface There are two surfaces that may be considered. For surface there is a conduction current ic that passes through it acting to charge the capacitor. t is therefore the case that B 0 ic is non-ero.
For surface there is no current passing through it and B 0 being in contradiction! But clearly B 0 because charges are in motion. the two results Conclusion: The result for surface needs to be modified. Maxwell s modification The surface does not intercept any current, but it DOES intercept ELECTRC FLUX How much flux is intercepted? Let Q = charge on capacitor E Q 0 d 0 d ε 0 d 0 f we want B L as for a straight wire, we must claim that B 0iC 00 dφ MAXWELL S 4th EQUATON Term Term Continuous wire : Term = 0 i C OK Term = 0
Capacitor : Term = 0 dφ Term = 0 0 OK Maxwell showed that this is a general relation which holds ALWAYS. Note: The modified Ampere s Law can be written as B 0 i C i d dφ id 0 DSPLACEMENT CURRENT Another way of thinking about this physically is to view the displacement current as the current due to the charge changing on the capacitor; For a parallel plate capacitor And i d dqc id 0 A de d d V Ed and dv C A C 0 d de d id 0A 0 de B 0 ic id 0ic 00 Maxwell s 4 th equation Where the current density is now divided into a conduction current i c and a displacement current i d What does this new version of Ampere s law imply about the relationship between E and B? dφ de E Area f 0 then 0 ie.
dφ represents a CHANGNG ELECTRC FELD. f dφ 0 then B d L 0 and therefore B 0 i.e., if there is a changing electric field, then the magnetic field cannot be ero or A CHANGNG ELECTRC FELD PRODUCES A MAGNETC FELD Recall: Faraday s Law: d E Changing B causes E This Causes this Now, the modified version of Ampere s Law implies that the reverse is also true: B 0iC 00 dφ Changing E causes B This Causes this 4
Example: A parallel plate capacitor, radius R, is connected to a source of alternating emf. Alternating emf alternating electric field E = E0sin(t) What is the magnetic field i) nside the capacitor (r < R)? ii) Outside the capacitor (r > R)? r R r E (inwards) E SDE VEW TOP VEW The current flowing across the capacitor, = 0 (plates are separated by a vacuum or insulator). i) r < R: Symmetry considerations the magnetic field must have the same magnitude and direction at all points on the dotted circle of radius r. Direction of B will be tangential because the field is associated with current flowing to the plane of the path. r B Apply modified Ampere s Law to the circular path: ntegrate B around the path. is parallel to B everywhere along the path, And B is also the same all around the path. r So r r B B B rb 0 0 5
The electric flux through the path is; = (Field)(Area) = E sinωt 0 r so dφ d r E0sin ωt 0iC 00 0 00 rb r 00E0 cos ωt B 0 0 0 t r E cosωt ii) r > R The same analysis applies except that R E 0 sin ωt (no contribution from the area outside R) B t 00R E0 cos ωt r Sketch of the amplitude of the magnetic field vs. radius: B 00RE0 B r B /r R r 6
Summary of Maxwell s Equations (in integral form) Q E da enclosed ε0 Gauss s Law for the Electric Field B da 0 Gauss s Law for the Magnetic Field d E Faraday s Law 4 de B 0iC 00 Maxwell s modification of Ampere s Law NB: Changing B generates E 4 Changing E generates B Changing B Changing E OSCLLATON OF ENERGY BETWEEN THE ELECTRC AND MAGNETC FELDS n fact, Maxwell s equations imply the existence of ELECTROMAGNETC WAVES 7
Throughout this module we have gradually developed the set of four equations that constitute Maxwell s equations from a disparate series of experiments and theory created by other scientists in the field of electricity and magnetism. All of these equations first brought together by Maxwell have been introduced in integral form. They also exist and are useful in differential form; Maxwell s Equations in Differential form. These four equations may be written in a differential form which may in many circumstances prove more useful. To do this we need to find a different way to express surface integrals and line integrals. Q E da enclosed ε0 Gauss s Law for the Electric Field We may obtain this in differential form by considering an infinitesimal cube of side. da y x da The integral refers to the flux of the E field across a closed surface and we can calculate this flux across the infinitesimal volume by pairing contributions of the opposite faces of the cube. For instance the top and bottom faces where we have; 8
top bot E E t t E da EdA b b E da E da EMF Notes ; Maxwell s Equations Where t E b E are the values of E on the top (bottom) face and as da is small E varies very little on each face. top t E E E E E da dda bot b E E da dv t b Where we approximated the increment E E at first order in which is very small. E The derivative may be calculated at any point inside the cube (again because E varies very little over this small volume. We can do the same for the other two pairs of faces to find the infinitesimal flux through the cube; S E E x x Ey y E dv where da dv is the infinitesimal volume. We can use this result in our integral form of the Maxwell equation; E E da x x Ey y E dv Q dv enclosed E dv ε0 0 f is the charge density inside the cube then Q enclosed dv Q dv enclosed E dv ε0 0 E 0 9
The above equation is the differential form of Maxwell s first Law. We can follow an identical approach for the nd law To obtain B da 0 B 0 Gauss s Law for the Magnetic Field A similar analysis may be used to deduce the differential form of Maxwell s rd Law (Faraday s Law of induction). d E Faraday s Law n this case we consider an infinitesimal square of side as our contour. y Similar to previously we may combine contributions to the line integral from opposite edges x E top E bot E E E d L x t x b x y Doing the same for the right and left hand edges gives E R E L Ey E E R y L y x Combining these two contributions to the circuit integral; dey E dx dex dy E Using the area of the square as the surface through which the magnetic field passes in order to calculate the RHS of Faraday s Law 0
d B t Combining these two results gives the differential form of Faraday s Law E B t This same procedure can be applied to the modified form of Amperes Law, Maxwell s 4 th equation 4 B 0ic 00 de Maxwell s modification of Ampere s Law Just as with the line integral of the E field we find for the B field; dby B dx dbx dy B d B 0 c 00 J A is the current density and EA E Using these as the variables on the RHS E B 0J 00 t Summary of Maxwell s Equations (in differential form) 4 E B 0 0 B E t E B 0J 00 t Gauss s Law for the Electric Field Gauss s Law for the Magnetic Field Faraday s Law Maxwell s modification of Ampere s Law