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

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Most popular questions from this chapter

Derive Eq. 8.39. [Hint: Treat the lower loop as a magnetic dipole.]

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

Figure 8.6

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

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