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Consider an ideal stationary magnetic dipole m in a static electric field E. Show that the fields carry momentum

p=-ε0μ0(m×E) (8.45)

[Hint: There are several ways to do this. The simplest method is to start with p=ε0(E×B)dτ, write E=-V, and use integration by parts to show that

p=ε0μ0VJdτ.

So far, this is valid for any localized static configuration. For a current confined to an infinitesimal neighbourhood of the origin, we can approximate V(r)V(0)-E(0)·r. Treat the dipole as a current loop, and use Eqs. 5.82 and 1.108.]21

Short Answer

Expert verified

The field carry momentum isp=-ε0μ0m×E .

Step by step solution

01

Expression for electromagnetic momentum per unit volume:

Write the expression for electromagnetic momentum per unit volume.

p=ε0(E×B) …… (1)

Here, E is the electric field, and B is the magnetic field.

02

Determine the equation for electromagnetic momentum:

Substitute the known values in equation (1).

p=ε0-V×Bdτ

Use integration parts in the above equation.

p=-ε0×VB-V×Bdτp=ε0VB×da+ε0μ0VJdτp=0+ε0μ0VJdτp=ε0μ0VJdτ

Use the given equation VrV0+V.r=V0-E0.r.

localid="1657364211619" p=ε0μ0V0Jdτ-ε0μ0E0.rJdτp=0-ε0μE0.rJdτ ……. (2)

For a current loop,

Jdτd=Idl=0

Solve the integral as,

E0.rJdτ=E0.rJdτE0.rJdτ=E0.rldlE0.rJdτ=E0.rdlE0.rJdτ=la×E0

……. (3)

Write the equation for the magnetic dipole.

m = la

Substitute the known value in equation (3).

E0.rJdτ=m×E

From equation (2),

p=-ε0μ0m×E

Therefore, the field carry momentum isp=-ε0μ0m×E.

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Most popular questions from this chapter

In Ex. 8.4, suppose that instead of turning off the magnetic field (by reducing I) we turn off the electric field, by connecting a weakly conducting radial spoke between the cylinders. (We’ll have to cut a slot in the solenoid, so the cylinders can still rotate freely.) From the magnetic force on the current in the spoke, determine the total angular momentum delivered to the cylinders, as they discharge (they are now rigidly connected, so they rotate together). Compare the initial angular momentum stored in the fields (Eq. 8.34). (Notice that the mechanism by which angular momentum is transferred from the fields to the cylinders is entirely different in the two cases: in Ex. 8.4 it was Faraday’s law, but here it is the Lorentz force law.)

A sphere of radius R carries a uniform polarization P and a uniform magnetization M (not necessarily in the same direction). Find the electromagnetic momentum of this configuration. [Answer:49ττμ0R3(M×P)]

(a) Consider two equal point charges q, separated by a distance 2a. Construct the plane equidistant from the two charges. By integrating Maxwell’s stress tensor over this plane, determine the force of one charge on the other.

(b) Do the same for charges that are opposite in sign.


Imagine two parallel infinite sheets, carrying uniform surface charge +σ(on the sheet at z=d) and -σ(at z=0). They are moving in they direction at constant speed v (as in Problem 5.17).

(a) What is the electromagnetic momentum in a region of area A?

(b) Now suppose the top sheet moves slowly down (speed u) until it reaches the bottom sheet, so the fields disappear. By calculating the total force on the charge q=σA, show that the impulse delivered to the sheet is equal to the momentum originally stored in the fields. [Hint: As the upper plate passes by, the magnetic field drops to zero, inducing an electric field that delivers an impulse to the lower plate.]

Calculate the force of magnetic attraction between the northern and southern hemispheres of a uniformly charged spinning spherical shell, with radius R, angular velocity ω, and surface charge density σ. [This is the same as Prob.5.44, but this time use the Maxwell stress tensor and Eq.8.21.]

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