In calculating the current enclosed by an Amperian loop, one must,in general, evaluate an integral of the form

Ienc=sJda

The trouble is, there are infinitely many surfaces that share the same boundary line. Which one are we supposed to use?

Short Answer

Expert verified

As a result, any particular surface can be considered for an endless number of surfaces with the same boundary line because the integral is independent of the surface.

Step by step solution

01

Define function

Here, The sum of the enclosed currents times the permeability of free space is equal to the closed line integral of the magnetic field multiplied by the length of the curve, according to Ampere's law.

Write the expression for the ampere’s law.

Bdl=μ0Ienc …… (1)

Here,μ0 is the permeability for free space,B is the magnetic field, dlis the length of curve, Iencis the enclosed current.

Write the expression for value of current enclosed in terms of current density.

Ienc=sJda …… (2)

Here, Jis the current density andIenc is the enclosed current.

02

Determine solution

The integral is surface independent, according to the divergence less field’s theorem. Any given boundary line's integralJda value will be the same. For an enclosed surface, the integral value will be 0. In addition, the current density should have a lower divergence than the following criterion.

J=0 …… (3)

As a result, any particular surface can be considered for an endless number of surfaces with the same boundary line because the integral is independent of the surface.

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Most popular questions from this chapter

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 12, where is Planck's constant. What, then, is the electron's magnetic dipole moment, in Am2? [This semi classical value is actually off by a factor of almost exactly 2. Dirac's relativistic electron theory got the 2 right, 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 (e /2m), where eis the charge of the electron and mis its mass, is called the Bohr magneton.]

Just as V.B=0allows us to express B as the curl of a vector potential (B=×A), so .A=0permits us to write A itself as the curl of a "higher" potential:A=×W. (And this hierarchy can be extended ad infinitum.)

(a) Find the general formula for W (as an integral over B), which holds whenB0 at .

(b) Determine for the case of a uniform magnetic field B. [Hint: see Prob. 5.25.]

(c) Find inside and outside an infinite solenoid. [Hint: see Ex. 5.12.]

Question: Find the magnetic field at point Pfor each of the steady current configurations shown in Fig. 5.23.

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 ]

(a) Prove that the average magnetic field, over a sphere of radius R,due to steadycurrents inside the sphere, is

Bave=μ04π2mR3

wheremis the total dipole moment of the sphere. Contrast the electrostatic

result, Eq. 3.105. [This is tough, so I'll give you a start:

Bave=143πR3Bdτ

WriteBas×A ,and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the

surface integral first, showing that

1rda=43πr'

(b) Show that the average magnetic field due to steady currents outsidethe sphere

is the same as the field they produce at the center.

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