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Find the vector potential above and below the plane surface current in Ex. 5.8.

Short Answer

Expert verified

The vector potential above and below the plane surface current is -0K2|z|x^.

Step by step solution

01

Significance of the vector potential

The vector potential is described as a particular vector field that has a curl in order to identify the required vectors. Moreover, the vector potential involves a gradient in a particular vector field.

02

Determination of the vector potential

The equation in the example 5.8 is expressed as:

K=Kx^

Here, Kis the uniform surface current, K is the constant and x^is the unit vector in the x direction.

The equation of the magnetic field in example 5.8 is expressed as:

B=0K2y^z<0=-0K2y^z>0

Here, B is the magnetic field,0is the permeability, z is the coordinate on the z axis and y^is the unit vector in the y direction.

The vector potential is parallel to the constant K and it is dependent on the function z .

The equation of the vector potential is expressed as:

A=A(z)x^

The equation of the magnetic field is expressed as:

B=A

Here, is the curl and A is the vector potential.

Substitute the values in the above equation.

B=x^y^z^lxlylzAz00=Azy^=0K2y^

From the above equation, the vector potential can be identified.

The vector potential can be expressed as -0K2|z|x^.

Thus, the vector potential above and below the plane surface current is role="math" localid="1657522544831" -0K2|z|x^.

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Most popular questions from this chapter

Two long coaxial solenoids each carry current I , but in opposite directions, as shown in Fig. 5.42. The inner solenoid (radius a) has turns per unit length, and the outer one (radius b) has .n2Find B in each of the three regions: (i) inside the inner solenoid, (ii) between them, and (iii) outside both.

Find the magnetic vector potential of a finite segment of straight wire carrying a current I.[Put the wire on the zaxis, fromz1 to z2, and use Eq. 5.66.]

Check that your answer is consistent with Eq. 5.37.

I worked out the multipole expansion for the vector potential of a line current because that's the most common type, and in some respects the easiest to handle. For a volume current J:

(a) Write down the multipole expansion, analogous to Eq. 5.80.

(b) Write down the monopole potential, and prove that it vanishes.

(c) Using Eqs. 1.107 and 5.86, show that the dipole moment can be written

m=12(rJ)d

A thin glass rod of radius Rand length Lcarries a uniform surface charge . It is set spinning about its axis, at an angular velocity. Find the magnetic field at a distances sRfrom the axis, in the xyplane (Fig. 5.66). [Hint: treat it as a stack of magnetic dipoles.]

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.]

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