/*! 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} Problem 43 A familiar demonstration of supe... [FREE SOLUTION] | 91Ó°ÊÓ

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A familiar demonstration of superconductivity (Prob. \(7.42\) ) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. \({ }^{19}\) Treat the magnet as a perfect dipole \(\mathrm{m}\), a hcight \(z\) above the origin (and constrained to point in the \(z\) direction), and pretend that the superconductor occupies the entire half-space below the \(x y\) plane. Because of the Meissner cffect, \(\mathbf{B}=0\) for \(z \leq 0\), and since \(\mathbf{B}\) is divergenceless, the normal \((z)\) component is continuous, so \(B_{z}=0\) just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at \(-z\), as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region \(z>0 .\) (a) Which way should the image dipole point \((+z\) or \(-z)\) ? (b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to \(M g\) (where \(M\) is the mass of the magnet) to determine the height \(h\) at which the magnet will "float."

Short Answer

Expert verified
(a) The image dipole points in the \(-z\) direction. (b) The levitation height is \( h = \left( \frac{3\mu_0 m^2}{32\pi Mg} \right)^{1/4} \).

Step by step solution

01

Analyze the Image Dipole Direction

To determine the direction of the image dipole, recall that the Meissner effect in superconductors causes the normal component of the magnetic field at the surface to be zero. Since the field contribution from the real dipole at height \( z \) acts to increase \( B_z \), the image dipole must oppose it. Thus, the image dipole must point in the \(-z\) direction to ensure \( B_z = 0 \) at the surface \( z = 0 \).
02

Calculate the Force on the Real Dipole

The image dipole at \( -z \) exerts a force on the real dipole. The interaction energy between two dipoles, \( \mathbf{m}_1 \) and \( \mathbf{m}_2 \), is given by \( U = -\frac{\mu_0}{4\pi} \frac{\mathbf{m}_1 \cdot \mathbf{m}_2}{r^3} + 3 \frac{(\mathbf{m}_1 \cdot \mathbf{r})(\mathbf{m}_2 \cdot \mathbf{r})}{r^5} \), where \( \mathbf{r} \) is the separation vector from one dipole to the other. Given their alignment, simplifications occur as only components aligned with the separation vector matter.
03

Apply the Force Law to the Image Dipole-Dipole System

The force on a dipole due to another dipole is \( \mathbf{F} = abla (\mathbf{m}_1 \cdot \mathbf{B}_2) \). Substituting in the magnetic field due to the image dipole at \( z \), and recognizing that the real dipole is at \( 2z \) distance from the image, leads to the force formula: \( F_z = \frac{3\mu_0 m^2}{4\pi (2z)^4} \).
04

Equate the Magnetic Force to Gravitational Force for Levitation Height

For levitation, the magnetic force \( F_z \) must equal the gravitational force \( Mg \). Setting \( \frac{3\mu_0 m^2}{4\pi (2h)^4} = Mg \), solve the equation for \( h \), the height at which the dipole floats: \[ h = \left( \frac{3\mu_0 m^2}{32\pi Mg} \right)^{1/4} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Meissner Effect
The Meissner effect is a fascinating property of superconductors. It results in the expulsion of magnetic fields from the interior of a superconducting material. This occurs when a material transitions below its critical temperature and enters the superconducting state. Regardless of the magnetic field applied to it before, a superconducting material will repel these fields completely.
As a result, the region inside the superconductor remains field-free. The Meissner effect can be visualized as the material "pushing out" magnetic field lines that attempt to penetrate it. This all-encompassing expulsion is what differentiates a superconductor from a perfect conductor, the latter only preventing changes in magnetic flux. It is this property that allows for unique phenomena such as magnetic levitation.
Magnetic Levitation
Magnetic levitation, often called "maglev," is the process of lifting an object using magnetic forces alone. It's an astounding application of the Meissner effect and other electromagnetic principles. In the demonstration with a superconductor, when a magnet is placed above a superconducting material, it can float in mid-air.
This levitation happens because the superconductor expels the magnetic field lines, creating currents on its surface that generate opposing magnetic fields. The balance between magnetic repulsion and gravitational forces allows the magnet to remain suspended. This phenomenon is employed in several practical applications, such as maglev trains and futuristic transportation systems, where frictionless movement is highly desirable.
Dipole Forces
Dipole forces, in the context of electromagnetism, refer to the interaction between two magnetic or electric dipoles. A dipole consists of two charges or magnetic poles of opposite nature, separated by a distance. The simple analogy is a magnet with a north and south pole.
These forces arise due to the torque exerted on each dipole by the other's field. The strength and direction of these forces depend on the orientation and separation distance between the dipoles. In our context, the real dipole (magnet) and the image dipole (created by the superconductor) interact through their fields, leading to the levitation force experienced. Understanding dipole interactions helps us to calculate the stable position and height at which magnetic levitation occurs.
Image Method
The image method is a powerful mathematical technique used to solve boundary problems in electromagnetism. It involves replacing complex boundary conditions with simpler equivalent systems by introducing 'image' charges or dipoles outside the problem region.
This concept is particularly useful when dealing with infinite planes or surfaces like a superconductor in contact with a magnetic field. By imagining an "image dipole" below the surface of the conductor, we simulate how the field would behave as if the surface did not exist. This allows for simpler calculations of forces and fields, helping to determine critical parameters like the height of magnetic levitation in our exercise.
Electromagnetism
Electromagnetism is one of the four fundamental forces of nature, encompassing both electricity and magnetism. It governs the behavior of charged particles, electric and magnetic fields, and the interaction between them.
It is the principle behind many technological advancements today, including superconductivity and magnetic levitation. By using principles from electromagnetism, such as Ampère's Law and Faraday's Law, we can understand how changing magnetic fields induce currents, and how these interactions can lead to phenomena like the Meissner effect. Mastering the fundamentals of electromagnetism provides insight into the forces that enable a magnet to float above a superconductor, effectively bringing theoretical physics into practical, observable reality.

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Most popular questions from this chapter

A small loop of wire (radius \(a\) ) lies a distance \(z\) above the center of a large loop (radius \(b\) ), as shown in Fig. 7.36. The planes of the two loops are parallel. and perpendicular to the common axis. (a) Suppose current \(I\) flows in the big loop. Find the flux through the litle loop. (The little Ioop is so small that you may consider the field of the big loop to be essentially constant.) (b) Suppose current \(I\) flows in the liule loop. Find the flux through the big loop. (The little loop is so small that you may treat it as a magnetic dipole.) (c) Find the mutual inductances, and confirm that \(M_{12}=M_{21}\).

A square loop (side \(a\) ) is mounted on a vertical shaft and rotated at angular velocity \(\omega\) (Fig. 7.18). A uniform magnetic field \(\mathbf{B}\) points to the right. Find the \(\mathcal{E}(t)\) for this alternating current generator.

A toroidal coil has a rectangular cross section, with inner radius \(a\), outer radius \(a+w\), and height \(h\). It carries a total of \(N\) tightly wound turns, and the current is increasing at a constant rate \((d l / d t=k)\). If \(w\) and \(h\) are both much less than \(a\), find the electric field at a point \(z\) above the center of the toroid.

The current in a long solenoid is increasing linearly with time, so that the flux is proportional to \(t: \Phi=\alpha t\). Two voltmeters are connected to diametrically opposite points (A and \(B\) ), together with resistors \(\left(R_{1}\right.\) and \(\left.R_{2}\right)\), as shown in Fig. \(7.53 .\) What is the reading on cach voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter registers \(\int_{a}^{b} \mathbf{E} \cdot d^{\prime}\) between the terminals and through the meter. [Answer: \(V_{1}=\alpha R_{1} /\left(R_{1}+R_{2}\right): V_{2}=-\alpha R_{2} /\left(R_{1}+R_{2}\right) .\) Notice that \(V_{1} \neq V_{2}\), cven though they are connected to the same points! See R. H. Romer, \(A m . J\). Phys. \(50,1089(1982) .]\)

A square loop of wire (side \(a\) ) lies on a table, a distance \(s\) from a very long straight wire, which carries a current \(I\), as shown in Fig. \(7.17\). (a) Find the flux of \(\mathrm{B}\) through the loop. (b) If someone now pulls the loop directly away from the wire, at speed \(v\), what emf is generated? In what direction (clockwise or counterclockwise) does the current flow? (c) What if the loop is puiled to the right at speed \(v\), instead of away?

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