Institutt for fysikk, NTNU TFY4155/FY1003: Elektrisitet og magnetisme Vår 2007

Øving 14

Guidance: Week 16 To be delivered by: Monday April 23

Exercise 1

In the lectures, we showed that atoms may be viewed as small current loops, i.e., as small magnetic dipoles with magnetic dipole moment m = IA, where the current I goes in a loop which encloses a (planar) area A. (”The vector area” is then A = A n̂, where n̂ is a unit vector perpendicular to the enclosed surface, with the positive direction determined by the right hand rule.) Here, we will use a quadratic current loop as a model for such an atomic magnetic dipole and look closer at how it will behave in a magnetic field B. (We could have used a circular loop, but the quadratic one is a little simpler when it comes to the calculations...) The current loop has edges with length a and transports a current I. It is placed in a homogeneous magnetic field B = B ẑ and is allowed to rotate freely around the y-axis, which in our case passes through the centre of the current loop, as shown in the figure:

The orientation of the current loop is defined through the angle θ between the z-axis and the surface normal n̂. (Positiveθ counterclockwise, as shown in the figure.)

a) What is the magnetic dipole moment m of this current loop? What is the total force due to B on the current loop?

b) Find the torque τ on the loop with respect to the y-axis and show that it can be written in the form τ = m × B.

[Hint: Find the force on each of the four straight elements of the loop and use the face that torque equals ”arm times force”.]

c) Determine the potential energy U(θ) of such a magnetic dipole in the field B. Draw a sketch of U(θ). Which orientation of the dipole with respect to B represents a stable and an unstable equilibrium, respectively?

d) In iron, each atom has a magnetic dipole moment m_{Fe} which is made up of two parallel electron spins, so that m_{Fe} = 2μ_{B}. Here, μ_{B} = eh̄/2m_{e} is the magnetic dipole moment of a single electron spin, the so-called Bohr magneton, which has the value 9.27 · 10^{−24} Am^{2}. What is then the maximum density of magnetic dipole moment, i.e., the maximum magnetic dipole moment pr unit volume, in iron? [Comment: Magnetic dipole moment pr unit volume is, by definition, the quantity magnetization. In electrostatics, we introduced polarization, which by definition is electric dipole moment pr unit volume. More about magnetism and magnetization in the lectures!]

Given information: Molar mass of iron: 55.9 g/mol. Mass density of iron: 7.9 g/cm^{3}. 1 mol = 6.02 · 10^{23}.

Exercise 2

Show, by using Ampere’s law, that the magnetic field B from a uniform ”surface current” i = i ŷ flowing in the (complete) xy-plane in the positive y direction is

B = \{ \begin{matrix}−(μ_{0}i/2) x̂ & for z < 0 \\ +(μ_{0}i/2) x̂ & for z > 0 \end{matrix}

(I.e., independent of the distance from the xy plane, just like we found for the electric field from an infinitely large uniformly charged plane.) Here, i is the current pr unit length of the x direction. In other words, in a ”stripe” of width Δx runs a current ΔI = i · Δx.

Hint:

• You have already been informed that the y and the z component of B are both zero. However, spend some time to convince yourself that it has to be like that! Such an ”investigation” of the symmetry of the problem is completely essential if you want to take advantage of Ampere’s law in order to determine the magnetic field. Often, you then need to go back to the Biot-Savart law and look at the consequences of ”current elements” I dl that give contributions dB ∼ I dl × r̂ to the total magnetic field.

• In this particular problem, you will perhaps convince yourself that a sensible choice of ”amperian loop” is a rectangular curve with surface normal in the current direction. If so, you are on the right track!

Exercise 3

An infinitely long, thin solenoid is located with its central axis on the x axis.

In the lectures, we argued that if there is a magnetic field on the outside of the solenoid, it must be directed along the direction of the solenoid, i.e., B = B x̂. Using Ampere’s law, we then showed that B = 0. This can be explained by the fact that some of the turns on the solenoid contribute with negative x component to B while others contribute with positive x component. In a distance y = 50 cm from the axis of a solenoid with 1000 turns pr meter, how many turns will contribute with a negative x component to B? (The rest of the turns, an infinite number, will contribute with positive x component to B.) Witout telling you the exact answer, I can say that if you obtain somewhat more than 700, you have probably done this one correctly. (Or: Done the same mistakes as I have...)

Hint: Each turn on the solenoid may be regarded as an ideal magnetic dipole m = m x̂, i.e., we assume that the radius of the solenoid is small compared to the distance y. In that case, the magnetic field in a distance r from a specific turn is given by

B_{dipol} = \frac{μ_{0}}{4πr^{3}} [3 (m · r̂) r̂ − m]

With a current I in the solenoid wire and a cross section of area A, we have the relation m = IA. However, you don’t need I and A to solve this problem.