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Prove the following uniqueness theorem: If the current density J isspecified throughout a volume V ,and eitherthe potential A orthe magnetic field B is specified on the surface Sbounding V,then the magnetic field itself is uniquely determined throughout V.[Hint:First use the divergence theorem to show that

[(×U).(×V)-U.(××)]dr=(U××V)da

for arbitrary vector functions Uand V ]

Short Answer

Expert verified

It is proved If the current density is specified throughout a volume and eitherthe magnetic vector potential orthe magnetic field is specified on the surface bounding the volumethen the magnetic field itself is uniquely determined throughout throughout the volume.

Step by step solution

01

Given data

The specified current density is J.

To prove,

×U.×V-U.××Vdτ=U××V.da

02

Vector product, divergence theorem and curl of magnetic field

Vector product rule of two arbitrary vector functions Uand Vis

localid="1658559820207" .(U××V)=(×V).(×U).U.(××V)

According to divergence theorem, the volume integral of the divergence of a vector function is

localid="1658559830383" .Udτ=U.da........(3)

The curl of the magnetic field is

localid="1658559839318" ×B=μ0J.......(4)

Here, μ0is the permeability of free space.

03

Proof of continuity equation

Take volume integral of both sides of equation (2) and use equation (3) on the left hand side to get,

U××V.da=×V.×U-U××Vdτ

Assume that there are two values of magnetic fields B1 and role="math" localid="1657773504274" B2 and two corresponding magnetic vector potentials A1 and A2. The difference between the values is defined as

B3=B1-B2A3=A1-A2B3=×A3

Since the current density is uniquely specified, from equation (4),

×B3=×B1-×B2=μ0J-μ0J=0

Set U,V=A3in equation (1), to get,

×A3.×A3-A3××A3dτ=A3××A3.daB3.B3-A3.×B3dτ=A3×B3.da

Use equation (5) to get,

B32dτ=A3×B3.da

If either magnetic field or the potential is uniquely specified on the surface then either A3=0 or B3=0. In both cases the right hand side of the previous equation becomes zero. Hence B3=0 in the volume.

Thus, the magnetic field is uniquely determined in the volume.

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Most popular questions from this chapter

Use the Biot-Savart law (most conveniently in the form of Eq. 5.42 appropriate to surface currents) to find the field inside and outside an infinitely long solenoid of radiusR, with n turns per unit length, carrying a steady current I.

Prove Eq. 5.78, using Eqs. 5.63, 5.76, and 5.77. [Suggestion: I'd set up Cartesian coordinates at the surface, with Z perpendicular to the surface and X parallel to the current.]

Question: Using Eq. 5.88, calculate the average magnetic field of a dipole over

a sphere of radius Rcentered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the ambiguity at . (If you get stuck, refer to Prob. 3.48.) Evidently the truefield of a magnetic dipole is

Bdip(r)=μ04πr3[3(m·r^)r^-m]+2μ03mδ3(r)Bdip(r)=μ04πr3[3m·r^r^-m]+2μ03mδ3(r)

Compare the electrostatic analog, Eq. 3.106.

(a) By whatever means you can think of (short of looking it up), find the vector potential a distance from an infinite straight wire carrying a current . Check that .A=0and ×A=B.

(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.

A thin uniform donut, carrying charge Qand mass M, rotates about its axis as shown in Fig. 5.64.

(a) Find the ratio of its magnetic dipole moment to its angular momentum. This is called the gyromagnetic ratio (or magnetomechanical ratio).

(b) What is the gyromagnetic ratio for a uniform spinning sphere? [This requires no new calculation; simply decompose the sphere into infinitesimal rings, and apply the result of part (a).]

(c) According to quantum mechanics, the angular momentum of a spinning electron is role="math" localid="1658120028604" 12, where is Planck's constant. What, then, is the electron's magnetic dipole moment, in role="math" localid="1658120037359" A×M2 ? [This semi classical value is actually off by a factor of almost exactly 2. Dirac's relativistic electron theory got the 2right, and Feynman, Schwinger, and Tomonaga later calculated tiny further corrections. The determination of the electron's magnetic dipole moment remains the finest achievement of quantum electrodynamics, and exhibits perhaps the most stunningly precise agreement between theory and experiment in all of physics. Incidentally, the quantity (e2m ), where e is the charge of the electron and m is its mass, is called the Bohr magneton.]

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