Buku Classical Electromagnetic Theory (Jack Vanderlinde)

The implications of the radiation reaction for causality and other limitations of the theory are discussed in the final chapter. Appendix B addresses this common gap by explaining the basics of tensor analysis.

Static Electric and Magnetic Fields in Vacuum

Static Charges

The Electrostatic Force
The Electric Field
Gauss’ Law
The Electric Potential

We therefore conclude that charges lying outside the surface do not contribute to the surface integral of the electric field. We therefore conclude that the electric field at the top surface of the pill box is Ez=σ/ε0.

Figure 1.1: When q and q are situated at r and r respectively, the vector pointing from q to q is ( r − r ).

Moving Charges

The Continuity Equation
Magnetic Forces
The Law of Biot and Savart
The Magnetic Vector Potential
The Magnetic Scalar Potential

The magnetic induction field of a circular current loop is of recurring importance both experimentally and theoretically. Calculation of the magnetic induction field along the axis of the solenoid is now simple.

Figure 1.8: The current is assumed to run along the z axis, and we pick the observer in the x-y plane.

Charge and Current Distributions

Multipole Moments

The Cartesian Multipole Expansion
The Spherical Polar Multipole Expansion

Solution: The x and y components of the dipole moment clearly vanish, and the z component is easily obtained from the definition (2–15). The dipole field is obtained if the area of the loop is shrunk to zero.

Figure 2.1: The dipole moment of a pair of equal but opposite charges points from −q to q .

Interactions with the Field

Electric Dipoles
Magnetic Dipoles

If the net force on the dipole vanishes, the torque calculated is independent of the choice of origin. It is worth pointing out that the shape of the force on the magnetic dipole is subtly different from that on the electric dipole.

Figure 2.5: A small dipole in a nonhomogeneous electric ﬁeld.

Potential Energy

2-9 Find the quadrupole moment of a rod of length L carrying charge density ρ = η(z2−L2/12) , with z measured from the center of the rod. -14 Show that the quadrupole term of the multipole expansion of the potential can be written.

Figure 2.8: The square has alternating posi- posi-tive and negaposi-tive charges along the edges.

Slowly Varying Fields in Vacuum

Magnetic Induction

Electromotive Force
Magnetically Induced Motional EMF
Time-Dependent Magnetic Fields
Faraday’s Law

Displacement Current
Maxwell’s Equations
The Potentials

Gauge Transformations

The Wave Equation in Vacuum

Plane Waves
Spherical Waves

The magnetic induction field is invariant under this change and we can use (3–30) to express the electric field in terms of the changed () potentials. The alignment of E (perpendicular to the propagation vector, k) is known as the polarization of the wave. 3-1 Show that the flux Φ passing through a loop Γ can be written in terms of the vector potential axis.

Show that the total charge carried by the coil as it is turned over is independent of the speed of the turn. 3-14Fill in the missing steps in the last step of the footnote (page 59) to demonstrate this. Find the EMF generated between the shaft and a fixed point on the edge of the disk.

Figure 3.1: A small segment of the loop increases the area at a rate v × d .

Energy and Momentum

Energy of a Charge Distribution

Stationary Charges
Coeﬃcients of Potential
Forces on Charge Distributions in Terms of Energy
Potential Energy of Currents

Let us now express the energy of a charge distribution in terms of the established field. However, if the energy of the system can be evaluated in terms of the physical parameters, it is relatively easy to obtain the forces or torques. This work must be performed at the expense of the electrostatic energy W(e) (note that we have used W to denote energy rather than work) stored in the system.

The energy of a capacitor with an inserted central plate at a distance ζ is given by W(es)=ε0. As with electric charges, we want to express energy in terms of potential and field. The constant term, µ0/4π, represents the self-inductance of the current in each wire on it.

Figure 4.1: The ﬁeld between the equipotential plates vanishes. Only where the grounded sheet is interposed is there a nonzero ﬁeld.

Poynting’s Theorem

As we did before for electric charge distributions, we now proceed to express the potential energy of the current distribution only in terms of fields. 2µ0 (A×B)·dS (4–43) B2 is positive definite, implying that the volume integral of B2 can only increase as the region of integration expands. To interpret (4-49) we note that we have already filled in the terms in the volume integral in (4-18) and (4-44) as the field energy density of the electric and magnetic induction fields respectively.

It is easily verified that the field surrounding a charged magnetic dipole has a non-destructive Poynting vector, but one would be most reluctant to associate an energy flow with this. Example 4.5: Find the energy flow (radiation) of an electromagnetic plane wave with electric field amplitude E. Therefore, it would be more correct to use the average value of the electric field in (Ex 4.5.1), namely E2= 12E2.

Momentum of the Fields

The Cartesian Maxwell Stress Tensor for Electric Fields
The Maxwell Stress Tensor and Momentum

Expressing this force as the volume integral of the force per unit volume (or force density) f,. So, by analogy with the force being found as minus the gradient of the potential energy, we can find it as minus the divergence of the stress tensor (the comments in the footnote again apply). One can now use this formalism to calculate the force on a charge by drawing a box around the charge and integrating the stress tensor over the surface of the box.

The zz component of the stress tensor Tzz is ε0(Ez2−12E2), and the surface element is dSz=−2πrdr. So it is clear that Tij also represents the impulse flow of the electromagnetic field leaving the volume. Nevertheless, as will be clear from the derivations, the impulses and energies can just as easily be related to the sources of the fields.

Figure 4.4: The force derived from the stress tensor when E is (a) perpen- perpen-dicular, (b) parallel, and (c) makes angle θ with the normal to the surface.

Magnetic Monopoles

Therefore, we are forced to think that fields do have some reality. Even if the magnetic and electric monopoles are separated by galactic distances, this intrinsic angular momentum of their fields remains. We assume that, consistent with quantum mechanics, the smallest allowed angular momentum is ¯h/2, leading to the smallest nonzero magnetic monopole charge.

Duality Transformations

On the other hand, the Dirac monopole would have a different mixing angle than ordinary charges. Find the radial force in the coils and hence the tensile force required by the coils. Note that the radial force resulting from interior B is directed outward, while the force on the end faces is inward.).

Find the force in the axial direction on the inner cylinder when it is partially withdrawn from the outer one. Capacity is varied by rotating one set of blades about an axis centered on the blade diameter relative to the other set. Given that the torque in the needle is τ = kI1I2 where I1 is the current induced in the loop by the oscillating current in the coaxial wire and I2 is the current required to capacitively charge the plate adjacent to the central conductor, write an expression for the torque rotating the needle in terms of frequency, loop area, and plate surface, curved to maintain constant distance from center conductor and distance of each from center.

Figure 4.6: The concentric cylinders of prob- prob-lem 4-15.

Static Potentials in Vacuum – Laplace’s Equation

Laplace’s Equation

Uniqueness Theorem
Cartesian Coordinates in Two Dimensions
Plane Polar Coordinates
Spherical Polar Coordinates with Axial Symmetry
Conformal Mappings
Schwarz-Christoﬀel Transformations
Capacitance
Numerical Solution

Solution: Assume the general solution (5–24) and adapt it to the appropriate boundary conditions in each of the three regions. Eliminating all divergent terms except lnr, we write the expansion of the potential as Example 5.7: Determine the magnetic scalar potential of a circular current loop of radius a at a point near the center (r < a).

Integrating ∇ · B = 0 over the thin volume of figure 5.7, we have with the help of the divergence theorem. Finally, dividing the load by the potential difference between the conductors would give us the capacitance of the system. Deviation from the infinite capacitor result will give the fringe field contribution to the capacitance.

Figure 5.1: The rectangular box has the potential Φ as indicated along each of its sides.

Cylindrical Polar Coordinates

We can make an initial guess of the solution within a region whose boundary potential is given and then improve our guess by successively replacing the potential at each point by that of the average of its four nearest neighbors. Inspection of the method by which we arrived at equation (5–79) leads to the immediate generalization in three dimensions where we find that the potential at any point must be well approximated by the average of those of the six nearest neighboring points. Z on the right-hand side of the equation we have an expression independent of z on the left-hand side, from which we deduce that any expression (on the right-hand side or on the left-hand side) must be a constant.

Explicitly inserting the sign (which still needs to be determined from the boundary conditions) of the separation constant, we have. L dϕdz (5–92) (The stacked sinmϕ and cosmϕ are intended to be read as either the top line in both sides of the equation or the bottom line in both sides of the equation, similarly±used). The orthogonality of the trigonometric functions and the weighted orthogonality of the Bessel functions (E-20) can be exploited to evaluate the coefficients of the expansion.

Figure 5.18: The cylinder has potential V L ( r, ϕ ) on its top face, 0 on the bottom, and V a ( ϕ, z ) on the curved face.

Spherical Polar Coordinates

Example 5.15: Find the potential inside a hollow sphere of radius given that the potential on the surface is Va(θ, ϕ). Example 5.17: Find the potential inside and outside a sphere of radius R that has surface charge with density σ=σ0sin2θcos 2ϕ distributed on its surface. At very large values of ρ (where ρ is indistinguishable from r) we expect the potential to disappear.

Setting the potential on the surface of the ellipsoid defined by ρ=ρ0 to V0, we find V0=−ksin−1(1/ρ0). 5-1 Show that the potential at the center of a charge-free sphere is exactly the average of the potential on the surface of the sphere. Express the potential in the pipe in closed form and compare it with the result obtained by summing the first five non-zero terms of the series of exercise 5-5 for the pointed = 0.2 and x = 0.

Figure 5.19: The rectangular pipe has three sides grounded, while the fourth has uniform potential V 0 .

Static Potentials with Sources—Poisson’s Equation

Poisson’s Equation

Image Charges

The Inﬁnite Conducting Plane
The Conducting Sphere
Conducting Cylinder and Image Line Charges

The potential on the surface of the sphere arising from the two charges must be constant (say zero, since it can always be changed to another constant by placing a charge at the center of the sphere). Find the image dipole whose field mimics the field of the induced charge on the sphere. We would like to find the image line charge λ that would make the surface of the cylinder an equipotential surface.

The expression for the position of the image line charge is identical to that for the image of the point charge and the conducting sphere. In contrast, the size of the image line charge is the same as that of the source line charge. The potential on the surface of the wire bearing +λ is then V(r=R) = − λ. Ex. 6.4.3) The other wire has a potential of the same magnitude but opposite sign.

Figure 6.2: A charge placed near the corner of a large grounded conducting plate with a 90 ◦ bend requires three image charges to mimic the ﬁeld.

Green’s Functions

Green’s Theorem
Poisson’s Equation and Green’s Functions

We now proceed to obtain the Green's function for the special case of the Dirichlet problem with a spherical (inner or outer) boundary. Except for the size of the source inhomogeneity, this is exactly the same equation and boundary condition that the Green's function must satisfy. Now using (F-47)) to rewrite P(cosγ) in terms of randr polar coordinates, we obtain the desired Green's function for an off-sphere charge distribution.

6-3 Show when the potential across a charge placed near a grounded conducting sphere due to its image charge is expressed in terms of the position of the charge. Find the dipole energy due to the induced dipole field of the sphere. 6-15 A line charge λ is placed inside a grounded cylinder, conducting parallel to the axis of the cylinder.

Figure 6.9: A conducting sphere of radius b has a uniformly charged con- con-centric ring of radius a enclosed.

Static Electromagnetic Fields in Matter

The Electric Field Due to a Polarized Dielectric

Empirical Description of Dielectrics
Electric Displacement Field

Magnetic Induction Field Due to a Magnetized Material

Polar Molecules (Langevin-Debye Formula)

Microscopic Properties of Matter

Nonpolar Molecules
Dense Media—The Clausius-Mosotti Equation
Crystalline Solids
Simple Model of Paramagnetics and Diamagnetics
Conduction

Boundary Conditions for the Static Fields
Electrostatics and Magnetostatics in Linear Media

Noting that the surface of the pillbox inside the dielectric is directed opposite to the outward normal ˆn, we find. This discontinuity in the field is exactly the effect that a surface charge, σ = P ·n, at the surface of the dielectric would have. In either case, the resulting polarization will be a function of the local electric field (including that of neighboring dipoles).

In terms of the electric field in air and the electric field set in the dielectric, the potential difference V between the plates is The stored charge is most easily found from the air value of the electric field sinceσ=ε0E⊥. In (b), the loop is placed in the they-z leftx plane to point out of page.