A sphere of linear magnetic material is placed in an otherwise uniform magnetic field B0. Find the new field inside the sphere.

Short Answer

Expert verified

The value of new magnetic field inside the sphere is B=3μrμr+2B0.

Step by step solution

01

Write the given data from the question.

Consider asphere of linear magnetic material is placed in an otherwise uniform magnetic field B0.

02

Determine the formula of new magnetic field inside the sphere.

Write the formula of new magnetic field inside the sphere.

B=B0n=023χmχm+1n …… (1)

Here,B0 is uniform magnetic field andχm is magnetic susceptibility.

03

Determine the value of new magnetic field inside the sphere.

I'll apply the solution to issue 4.23. A magnetization is caused by the original magnetic field.

M0=χmH0=χmμB0

Determine the modifies magnetic field within the sphere:

B1=B0+23μ0M0=B01+23χmχm+1

Determine the magnetization induced by this field is:

M1=χmH1=χmμB1=χmμB01+23χmχm+1

Now, field is further modified:

B1=B0+23μ0M1=B0+23μ0χmμB01+23χmχm+1=B01+23χmχm+1+23χmχm+12

Where this song and dance will take us is fairly evident. The final magnetic field is obviously B:

Determine the new magnetic field inside the sphere.

B=B01123χmχm+1=B03(χm+1)3(χm+1)2χm=3μrμr+2B0

Therefore, the value of new magnetic field inside the sphere is B=3μrμr+2B0.

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Most popular questions from this chapter

A magnetic dipole m is imbedded at the center of a sphere (radius R) of linear magnetic material (permeability μ). Show that the magnetic field inside the sphere 0<rR is

μ4π{1r3[3(m.r^r^-m)]+2(μ0-μ)m(2μ0+μ)R3}

What is the field outside the sphere?

Suppose the field inside a large piece of magnetic material is B0, so that H0=(1/μ0)B0-M, where M is a "frozen-in" magnetization.

(a) Now a small spherical cavity is hollowed out of the material (Fig. 6.21). Find the field at the center of the cavity, in terms of B0 and M. Also find H at the center of the cavity, in terms of H0 and M.

(b) Do the same for a long needle-shaped cavity running parallel to M.

(c) Do the same for a thin wafer-shaped cavity perpendicular to M.

Figure 6.21

Assume the cavities are small enough so M, B0, and H0 are essentially constant. Compare Prob. 4.16. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite magnetization.]

Question: Of the following materials, which would you expect to be paramagnetic and which diamagnetic: aluminum, copper, copper chloride (Cucl2), carbon, lead, nitrogen (N2), salt (Nacl ), sodium, sulfur, water? (Actually, copper is slightly diamagnetic; otherwise, they're all what you'd expect.)


In Prob. 6.4, you calculated the force on a dipole by "brute force." Here's a more elegant approach. First writeB(r)as a Taylor expansion about the center of the loop:

,B(r)B(r0)+[(rr0)0]B(r0)

Wherer0the position of the dipole and 0is denotes differentiation with respect tor0. Put this into the Lorentz force law (Eq. 5.16) to obtain

.F=IdI×[(r0)B(r0)]

Or, numbering the Cartesian coordinates from 1 to 3:

Fi=Ij,k,l=13εijk{rldlj}[0lBk(r0)],

Where εijk is the Levi-Civita symbol (+1ifijk=123,231, or312; 1ifijk=132, 213, or 321;0otherwise), in terms of which the cross-product can be written (A×B)i=j,k=13εijkAjBk. Use Eq. 1.108 to evaluate the integral. Note that

j=13εijkεljm=δilδkmδimδkl

Whereoil is the Kronecker delta (Prob. 3.52).




Imagine two charged magnetic dipoles (charge q, dipole moment m), constrained to move on the z axis (same as Problem 6.23(a), but without gravity). Electrically they repel, but magnetically (if both m's point in the z direction) they attract.

(a) Find the equilibrium separation distance.

(b) What is the equilibrium separation for two electrons in this orientation. [Answer: 4.72x10-13m.]

(c) Does there exist, then, a stable bound state of two electrons?

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