/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q6.18P A sphere of linear magnetic mate... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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+2B→0.

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→∞=B→0∑n=0∞23χmχm+1n …… (1)

Here,B→0 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.

M→0=χmH→0=χmμB→0

Determine the modifies magnetic field within the sphere:

B→1=B→0+23μ0M→0=B→01+23χmχm+1

Determine the magnetization induced by this field is:

M→1=χmH→1=χmμB→1=χmμB→01+23χmχm+1

Now, field is further modified:

B→1=B→0+23μ0M→1=B→0+23μ0χmμB→01+23χmχm+1=B→01+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→∞=B→011−23χmχm+1=B→03(χm+1)3(χm+1)−2χm=3μrμr+2B→0

Therefore, the value of new magnetic field inside the sphere is B→∞=3μrμr+2B→0.

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 91Ó°ÊÓ!

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 familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass md and dipole moment m.

(a) If you put two back-to-hack magnets on the rod, the upper one will "float"-the magnetic force upward balancing the gravitational force downward. At what height (z) does it float?

(b) If you now add a third magnet (parallel to the bottom one), what is the ratio of the two heights? (Determine the actual number, to three significant digits.) [Answer:(a)3μ0m2(b)0.8501]

(a)Show that the energy of a magnetic dipole in a magnetic field B is

U=−m⋅B.

[Assume that the magnitude of the dipole moment is fixed, and all you have to do is move it into place and rotate it into its final orientation. The energy required to keep the current flowing is a different problem, which we will confront in Chapter 7.] Compare Eq. 4.6.

Figure 6.30

(b) Show that the interaction energy of two magnetic dipoles separated by a displacement r is given by

U=μ04π1r3[m1⋅m2−3(m1⋅r^)(m2⋅r^)]

Compare Eq. 4.7.

(c) Express your answer to (b) in terms of the angles θ1 and θ2 in Fig. 6.30, and use the result to find the stable configuration two dipoles would adopt if held a fixed distance apart, but left free to rotate.

(d) Suppose you had a large collection of compass needles, mounted on pins at regular intervals along a straight line. How would they point (assuming the earth's magnetic field can be neglected)? [A rectangular array of compass needles aligns itself spontaneously, and this is sometimes used as a demonstration of "ferromagnetic" behaviour on a large scale. It's a bit of a fraud, however, since the mechanism here is purely classical, and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism. 13]

Notice the following parallel:

{∇·D=0∇×E=0,ε0E=D-P(Nofreecharge)∇·B=0∇×H=0,μ0H=B-μ0M(Nofreecharge)

Thus, the transcription D→B,E→H,P→μ0M,ε0→μ0,, turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive.

(a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16);

(b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18);

(c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93).

A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility χm. A currentI flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field.

Figure 6.24

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.)

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.