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Question 6.20: How would you go about demagnetizing a permanent magnet (suc...

How would you go about demagnetizing a permanent magnet (such as the wrench we have been discussing, at point c in the hysteresis loop)? That is, how could you restore it to its original state, with M = 0 at I = 0?

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Place the object in a region of zero magnetic field, and heat it above the Curie point—or simply drop it on a hard surface. If it’s delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it back and forth many times; each time you reverse the direction, reduce the field slightly.

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Notice the following parallel: { ∇.D=0,∇×E=0, ε0E=D-P,(on free charge); {∇.B=0,∇×H=0, μ0H=B-μ0M,(no free current 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 knowl)

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(a) The electric field inside a uniformly polarize...

At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that tan θ2/ tan θ1 = μ2/μ1, assuming there is no free current at the boundary. Compare Eq. 4.68.

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At the interface, the perpendicular component of B...

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 < r ≤ R) is μ/4π {1/r^3 [3(m · rˆ)rˆ − m] +2(μ0-μ)m/(2μ0+μ)R^3} . What is the field outside the sphere?

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In view of Eq. 6.33, there is a bound dipole at th...

You are asked to referee a grant application, which proposes to determine whether the magnetization of iron is due to “Ampère” dipoles (current loops) or “Gilbert” dipoles (separated magnetic monopoles). The experiment will involve a cylinder of iron (radius R and length L = 10R), uniformly

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The problem is that the field inside a cavity is n...

Place the object in a region of zero magnetic field, and heat it above the Curie point—or simply drop it on a hard surface. If it’s delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it back and forth many times; each time you reverse the direction, reduce th

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(a) The magnetic force on the dipole is given by E...

In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: B(r) ≅ B(r0) + [(r − r0) · ∇0]B(r0),where r0 is the position of the dipole and ∇0 denotes differentiation with respect to

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F =I \oint d l \times B =I(\oint d l ) \tim...

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-back magnets on the rod, the upper one will “float”

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(a)  B _{1}=\frac{\mu_{0}}{4 \pi} \frac{2 m...

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.

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(a) Forces on the upper charge: F _{q}=\fra...

Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ, P, and M are uniform, the same integral is involved in all three: ∫ᴫˆ/ᴫ^2 dτ. Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized ob

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Eq. 2.15 :  E =\rho\left\{\frac{1}{4 \pi \e...

A uniform current density J = J0 zˆ fills a slab straddling the yz plane, from x = −a to x = +a. A magnetic dipole m = m0 xˆ is situated at the origin. (a) Find the force on the dipole, using Eq. 6.3. (b) Do the same for a dipole pointing in the y direction: m = m0yˆ. (c) In the electrostatic case

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(a) B =\mu_{0} J_{0} x \hat{ y } (P...