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)]

Short Answer

Expert verified

The electromagnetic momentum of the configuration isitot=49μ0πR3M×P .

Step by step solution

01

Expression for the total electromagnetic momentum:

Write the expression for the total electromagnetic momentum.

itot=iin+iout ….. (1)

Here, iinis the electromagnetic momentum from inside the sphere andiout is the electromagnetic momentum from outside the sphere.

Write the expression for the electromagnetic momentum from inside the sphere.

iin=ε0(Ein×Bin)dτ ….. (2)

Write the expression for the electromagnetic momentum from outside the sphere.

iout=ε0(Eout×Bout)dτ ….. (3)

02

Determine the electromagnetic momentum from inside the sphere:

Write the expression for the electric field inside the sphere of uniform polarization.

Ein=13ε0ρ

Here, the polarization of the sphere is p=43πR3P.

Write the expression for the magnetic field inside the sphere of uniform magnetization.

Bin=23μ0M

Here, Mis the magnetization of the sphere.

Substitute the known values in equation (2).

iin=ε0-p3ε0×23μ0Mdτiin=ε023ε0-p×Mdτiin=-2343πR3P×Miin=827πR3P×M

03

Determine the electromagnetic momentum from outside the sphere:

Write the expression for the electric field outside the sphere of uniform polarization.

Eout=14πε0133p.rr-p

Write the expression for the magnetic field outside the sphere of uniform magnetization.

Bout=μ04π1r33m.rr-m

Here, the magnetization is m=43πR3M.

Substitute the known values in equation (3).

iout=ε014πε0133p.rr-p×μ04π1r33m.rr-mdτiout=μ016π21r69p.r×r×m.rr-3p.rr×m-p×3m.rr+p×mdτiout=μ016π21r60-3p.rr×m+3m.rr×p+p×mdτiout=μ016π202π0π0a1r6-3p.rr×m+3m.rr×p+p×mr2sinθdθdrdϕ

On further solving,

role="math" localid="1657612741238" iout=μ016π20a0π02π1r43r.r.p×m-p×m+p×msinθdθdrdϕiout=μ016π20a0π02π1r4-2p×m+3rr.p×mp×msinθdθdrdϕ......4

Here,r=sinθcosϕx+sinθsinϕy+cosϕzandr.p×m=p×mcosθ.

Substitute the known values in equation (4).

iout=μ016π2Ra1r4dr-2p×m02π0πsinθdθdp+3p×mz02π0πcos2θsinθdθdϕiout=μ016π213r4ra-2p×m4π+3p×m4π3iout=μ016π23R3-8πp×m+4πp×miout=μ016π23R3-4π43πR3P×43πR3M

On further solving,

iout=μ03R319R64πM×Piout=4μ0π27R3M×P

04

Determine the electromagnetic momentum of the configuration:

Substitute the values in equation (1).

itot=827πR3P×M+4μ0π27R3M×Pitot=49μ0R3M×P

Therefore, the total electromagnetic momentum of the configuration is

itot=49μ0R3M×P

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A charged parallel-plate capacitor (with uniform electric field E=Ez^) is placed in a uniform magnetic field B=Bx^, as shown in Fig. 8.6.

Figure 8.6

(a) Find the electromagnetic momentum in the space between the plates.

(b) Now a resistive wire is connected between the plates, along the z-axis, so that the capacitor slowly discharges. The current through the wire will experience a magnetic force; what is the total impulse delivered to the system, during the discharge?

(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.

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.)

Consider the charging capacitor in Prob. 7.34.

(a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis and the timet. (Assume the charge is zero at t=0).

(b) Find the energy density uemand the Poynting vector S in the gap. Note especially the direction of S. Check that Eq.8.12is satisfied.

(c) Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap (Eq 8.9—in this case W = 0, because there is no charge in the gap). [If you’re worried about the fringing fields, do it for a volume of radius b<awell inside the gap.]

A very long solenoid of radius a, with n turns per unit length, carries a current ls. Coaxial with the solenoid, at radiusb>>a , is a circular ring of wire, with resistance R. When the current in the solenoid is (gradually) decreased, a currentir is induced in the ring.

a) Calculate role="math" localid="1657515994158" lr, in terms ofrole="math" localid="1657515947581" dlsdt .

(b) The power role="math" localid="1657515969938" (lr2R)delivered to the ring must have come from the solenoid. Confirm this by calculating the Poynting vector just outside the solenoid (the electric field is due to the changing flux in the solenoid; the magnetic field is due to the current in the ring). Integrate over the entire surface of the solenoid, and check that you recover the correct total power.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free