Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Electromagnetic Theory

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Electrical Engineering (EE) : Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The document Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
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As before, we keep in view the four Maxwell’s equations for all our discussions.

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

In the last lecture, we have seen that the electromagnetic field carries both energy and momentum and any discussion on conservation of these two quantities must keep this into account. In this lecture, we will first talk about the possibility of angular momentum being associated with the electromagnetic field and in the second half, introduce the electromagnetic waves.

Angular Momentum

Electromagnetic field, in addition to storing energy and momentum, also has angular momentum and this can be exchanged with the charged particles of the system.

We have seen that the momentum density of the electromagnetic field is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Based on this, we define the angular momentum density of the electromagnetic field about an arbitrary origin as

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

so that the angular momentum of the field is given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Let us start with a collection of charges and currents. The expression for the force density (i.e. the rate of change of momentum density) for the particles is given by the Lorentz force expression

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

We will do some aklgebraic manipulation on the right hand side by using Maxell’s equations, We take the medium to be the free space.

First, we substitute  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev for  ρ and replace for the current density,  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev With this we have,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

We next add some terms to makes this expression look symmetrical in electric and magnetic fields. To match the first term we add a term  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev which is identically zero. To make the second term symmetric, we add and subract a term  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev With these we get,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRevAngular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where we have interchanged the order of cross product in a couple of places by changing sign. 

The last two terms on right can be simplified as follows. Using  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev we can write,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Thus we have,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

We recall that the elements of the stress tensor was defined as

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

It can be shown (see Tutorial assignment 1) that the right hand side simplifies to  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev so that

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

On integrating over the volume one gets a statement of conservation of angular momentum.

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

which sates that the total rate of change of momentum (of particles and field) is equal to the flux of the torque through the surface.

Example : The Feynman Paradox – a variant of Feynman disk

We discuss a variant of the famous Feynman disk problem, which is left as an exercise. There is an infinite line charge of charge density  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is surrounded by an insulating cylindrical surface of radius a having a surface charge density  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev so that the net charge of the system is zero. The cylinder can freely rotate about the z axis, which coincides with the line charge. Because of Gauss’s law, the electric field exists only within the cylinder. The system is immersed in a uniform magnetic field  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev along the z axis. The system is initially at rest. If the magnetic field is now reduced to zero, the cylinder will be found to rotate.

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The explanation of rotation lies in conservation of angular momentum. As the system was initially at rest, the initial mechanical angular momentum is zero. The field angular momentum can be calculated as follows. The electric field is given by 

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev The magnetic field is circumferential and is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Thus, the initial field angular momentum per unit length is given by, (taking the angular momentum about the axis)

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where we have used  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

This is the net angular momentum of the field plus the mechanical system because the latter is at rest. If the magnetic field is reduced to zero, because of a changing flux through any surface parallel to the xy plane, there is an azimuthal current generated. This is caused by the rotating cylinder which rotates with an angular speed ω, giving rise to an angular momentum |ω where I is the moment of inertia per unit length of the cylinder about the z axis. If the time period of rotation is T, the current per unit length is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Since the current is circumferential, the magnetic field (like in a solenoid) is along the z direction and is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev The final angular momentum is thus given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Thus we must have,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

which allows us to determine ω

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Example : Radiation Pressure 

An experimentally verifiable example of the fact that the electromagnetic field stores momentum is provided by the pressure exerted by radiation confined inside a cavity. Consider an enclosure in the shape of a rectangular parallelepiped and let us consider the right most wall of the enclosure. The normal direction to the wall is to the let and the fieldexerts a force on this wall along the x direction.

The force exerted on an area dS of the wall is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Since the force is exerted in x direction we only need

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

If the radiation is isotropic, we can write,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where u is the energy density. This, incidentally, was the starting point for proving Stefan Boltzmann’s law.

Plane Wave solutions to Maxwell’s Equations

In the following we obtain a special solution to the Maxwell’s equations for a linear , isotropic medium which is free of sources of charges and currents.

We have then the following equations to solve:

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Taking the curl of the third equation, and substituting the first equation therein, we have,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Substituting last equation in this, we get,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

In a similar way, we get an identical looking equation for the magnetic field,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

These represent wave equations with the velocity of the wave being  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

It may be noted that, in a general curvilinear coordinate system, we do not have,  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev However, it would be true in a Cartesian coordinates where the unit vectors are fixed.

Let us look at “plane wave” solutions to these equations. A plane wave is one for which the surfaces of constant phases, viz. wavefronts, are planes. Time harmonic solutions are of the form sin  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev or cosAngular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev However, mathematically it turns out to be simple to consider an exponential form and take, at the end of calculations, the real or the imaginary part.

We take the solutions to be of the form,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Note that the surfaces of constant phase are given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

At any time t, since ωt= constant, we have, the surface given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Represent the wave-fronts. This is obviously an equation to a plane. Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is known as the “propagation vector”. As time increases these wave fronts move forward, I.e the distance from source increases. One could also look at the backward moving waves which would be of the form  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev .

As time progresses, the surfaces of constant phase satisfy the equation

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Thus the angular frequency ω ιs given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is known as the “phase velocity”.

Substituting our solutions into the two divergence equations, we get,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Thus the direction of both electric and magnetic fields are perpendicular to the propagation vector. If we restrict ourselves to non-conducting media, we would, in addition, have, from the curl equation,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

This also shows that the electric field is perpendicular to the magnetic field vector. Thus the electric field, the magnetic field and the direction of propagation form a right handed triad. If the propagation vector is along the z direction, the electric and magnetic field vectors will be in the xy plane being perpendicular each other.

Note that using the exponential form has the advantage that the action of operator  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is equivalent to replacing it by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev and the time derivative is equivalent to a multiplication by -iω.

We can take the cross product of the equation  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

expanding the scalar triple product,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

substituting  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev we get,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The ratio of the strength of the magnetic field to that of the electric field is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Where we have used the expressions for the velocity of the wave,  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev In free space this velocity is the same as the speed of light which explains why the magnetic field associate with the electromagnetic field is difficult to observe.

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

In the above figure the electric field variation is shown by the green curve and that of magnetic field by the orange curve.

We have seen that electromagnetic field stores energy and momentum. A propagating electromagnetic wave carries energy and momentum in its field. The energy density of the electromagnetic wave is given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where we have used the relation  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The Poynting vector is given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Recall that the difference between H and B is due to bound currents which cannot transport energy. Replacing the real form of the electromagnetic field, we get, taking the electric field along the x direction, the magnetic field along the y direction and the propagation vector along the z direction,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where we have assumed the electromagnetic wave to be travelling in free space. The intensity of the wave is defined to be the time average of the Poynting vector,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

where the factor of ½ comes from the average of the cos2(kz − ωt) over a period. 

In general, given the propagation direction, the electric and the magnetic fields are contained in a perpendicular to it. The fields can point in arbitrary direction in the plane as long as they are perpendicular to each other. If the direction of the electric field is random with time, the wave will be known as an “unpolarized wave”. Specifying the direction of the electric field (or equivalently of the orthogonal magnetic field) is known as a statement on the statement of polarization of the wave.

Consider the expression for the electric field at a point at a given time. Assuming that it lies in the xy plane, we can write the following expression for the electric field

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

In general, the quantities inside the bracket are complex. However, we can specify some specific relations between them.

1. Suppose E0x and E0y are in phase, i.e., suppose,

 Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

  then we can express the field as

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

 The magnitude of the electric vector changes from 0 to  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev but its direction remains constant as the wave propagates along the z direction. Such a wave is called “linearly polarized” wave. The wave is also called plane “plane polarized” as at a given time the electric vectors at various locations are contained in a plane.

2. Instead of the phase between the x and y components of the electric field being the same, suppose the two components maintain a constant phase difference as the wave moves along, we have

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

If the phase difference happens to be  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev we can write the expression for the electric field as

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Consider the time variation of electric field at a particular point in space, say at z=0. As t increases from 0 to  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Edecreases from its value |E0x| to zero while Eincreases from zero |E0y| the electric vector rotating counterclockwise describing an ellipse. This is known as “elliptic polarization”. (In the general case of an arbitrary but constant phase difference, the state of polarization is elliptic with axes making an angle with x and y axes. Alternatively, a state of arbitrary polarization can be expressed as a linear combination of circular or linear polarizations.)

3. A special case of elliptic polarization is “circular polarization” where the amplitudes of the two components are the same, viz.,|E0x| = |E0y| when the electric vector at a point describes a circle with time.

 

Tutorial Assignment

1. Prove the relation 

  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRevAngular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

2. Two concentric shells of radii a and b carry charges  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev At the centre of the shells a dipole of magnetic moment  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is located. Find the angular momentum in the electromagnetic field for this system.

Solutions to Tutorial Assignments

1. We will simplify only the electrical field term, the magnetic field term follows identically. We have,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

 Let us consider the x component of both sides,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRevAngular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

                                                 Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The y and z components follow by symmetry. The x component of the divergence of the stress tensor can be seen to be the same, as,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

2. In a spherical coordinate system, the magnetic field due to the dipole is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRevAngular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev The electric field is confined only within the shells and is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev The Poynting vector is given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The angular momentum density about the centre of the shells is then given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The total angular momentum of the electromagnetic field can be obtained by integrating this over the volume. However, since  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev is not a constant unit vector, we first convert this to Cartesian and write the total angular momentum as

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The first two terms, when integrated gives zero because the integral over Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev vanishes. We are left with

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Self Assessment Questions

  1. A metal sphere of radius R has a charge Q and is uniformly magnetized with a magnetization  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Calculate the angular momentum of the field about the centre of the sphere.
  2. In Problem 1, if the magnetization of the sphere is gradually and uniformly reduced to zero (probably by heating the sphere through the Curie temperature), calculate the torque exerted by the induced electric field on the sphere and show that the angular momentum is conserved in the process.

Solutions to Self Assessment Questions

1. The charge being uniformly distributed over its surface, the electric field exists only outside the sphere and is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev The magnetic field inside is constant and is given outside the sphere by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev One can calculate the Poynting vector and angular momentum density closely following Problem 2 of the tutorial. The total angular momentum is obtained by integrating over all space outside the sphere. Following this we get 

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Substituting  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev we get

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

2. When the sphere is slowly demagnetized from M to zero, there is a changing flux which induces an electric field. The induced electric field is circumferential and can be calculated from Faraday’s law. The magnetic field due to the magnetized sphere inside the sphere is constant and is given by  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev Consider a circumferential loop on the surface between polar angles θ and θ+ dθ The radius of the circular loop is R sin θ and the area of the loop is π (R sin θ)The changing flux through this area is  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev and this changing flux is equal to the emf induced in the loop of radius R sin θ so that we have, the electric field magnitude to be given by

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

which gives,

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

If we consider an area element  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev sin θdθ of the surface, the charge on the surface  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev experiences a force

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The torque about the z axis is obtained by multiplying this with the distance R sin θ of this element about the z axis.

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

Total torque is obtained by integrating over the angle

Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev

The change in angular momentum is  Angular Momentum, Electromagnetic Waves Electrical Engineering (EE) Notes | EduRev as was obtained in Problem 1.

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