Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Theory

Created by: Machine Experts

Electrical Engineering (EE) : Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The document Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
All you need of Electrical Engineering (EE) at this link: Electrical Engineering (EE)

Vector Potential 

In this lecture we will calculate the vector potential in a few cases. We have seen that the vector potential is not unique and we have a choice of gauge in the matter. The most common gauge in which we work is the Coulomb gauge in which the divergence of the vector potential is chosen to be zero, i.e.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We obtained an expression for the vector potential starting with BiotSavart’s law and saw that there exists a much stronger relationship between the vector potential than which exists for the magnetic field itself. In many cases where the direction of the current is constant, the vector potential simply points in the direction of the current. We have the following expression for the vector potential,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Vector Potential for a long straight wire carrying current

Let the current be in the z direction. The vector potential also points the same way.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The current being linear  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev the vector potential becomes a simple one dimensional integral

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The expression diverges when the limits are evaluated. This is not a very serious issue because we have seen that the vector potential is arbitrary up to a constant which in this case is infinite. For instance, if instead of integrating from -∞to+∞ we realized that the integrand is even, we could integrate it from zero to infinity and double the result. In that case the integral diverges only in the upper limit leaving us with a finite expression in the lower limit. Discarding the infinite constant, we would then have,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

In this simple case, we can start from our knowledge of the magnetic field and calculate back. We know that the magnetic field has cylindrical symmetry and is directed along the circumferential direction,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the curl of the vector potential only has Ø component 

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

By symmetry, since the wire is infinite, the derivative with respect to z must be zero and we have

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

which gives

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

where we have explicitly added gradient of an arbitrary scalar field.

There is another trick which is often used to calculate the vector potential which is to relate the line integral of vector potential to the flux. If we take the line integral of the vector potential along any closed loop, we get, using Stoke’s theorem,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We can then use the symmetry of the problem to find the vector potential.

Vector Potential of a solenoid

We know that the field inside a solenoid is along its axis (z direction) and is zero outside. If we take a circular path of radius s centered along its axis, the flux through the circular area is  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev The line integral of the vector potential is  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev This is because A is along the direction of current which is circumferential.

The vector potential is thus given by for s<R

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Though the field outside is zero, the vector potential does not vanish outside the solenoid. This is because, if we take a circle of radius s>R , the flux through the circular area is  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev flux being contributed only from inside the solenoid. Thus for  s>R

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

which falls off as inverse of distance from the axis.

Does vector potential have any physical significance?

AharanovBohm Effect

It may be seen from the above example that the vector potential remains non-zero outside the solenoid even though the magnetic field itself has become zero. It turns out that the vector potential is not just a mathematical artifact but has physical reality. The experiment described below illustrates this though the effect that we are talking about is of quantum mechanical origin. (The source of the picture is William O. Straub (2010))

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

You are familiar with the Young’s double slit experiment done with a beam of coherent light. The experiment is not really restricted to light wave but can be performed with matter wave such as a beam of electrons. We learn from quantum theory that like light exhibits dual behavior, that of wave as well as particles known as photon, a wavelength is associated with material particles as well. This is known as de Broglie wavelength and is given by the ratio of the Planck’s constant to the momentum of the particle.

The experiment is performed with an electron beam in place of light. What is done is to put a small solenoid just beyond the slits between the slits and the screen. Initially the solenoid does not carry any current and its dimensions are small enough so that it does not disturb the interference pattern produced on the screen because of the phase difference between electron waves arriving at the screen from the slits.

The current in the solenoid is switched on. The solenoid is of very small dimensions so that most of the electron beam passes outside the solenoid. Since the magnetic field outside is zero, the electron beams does not experience any force due to the magnetic field and should reach the screen undeflected. This should not then affect the interference pattern. However, what is found is that the pattern on the screen shifts suggesting a change in the phase relationship. It can be shown in quantum mechanics that the agent responsible for this phase change is the vector potential which is non-zero outside the solenoid though the magnetic field is zero.

Vector potential for a Uniform Magnetic Field 

Uniform magnetic field is of practical importance. Let us take the field to be in an arbitrary direction having a magnitude B. One of the possibilities for the vector potential is 

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

It can be seen that

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The first term gives  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev since the  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev The second term is zero because  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev The thirdand the fourth terms are calculated as follows,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

because the field is uniform.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Adding all the four terms, result follows. It can be seen that the Coulomb gauge condition is satisfied.The divergence of the vector potential is given by

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

because the first term vanishes as the field is uniform while the second term vanishes because the curl of position vector is zero. Thus the expression satisfies Coulomb gauge condition. If the magnetic field is along the z direction, we can take either of the following expressions for the vector potential components,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Both these expressions are valid expressions for the vector potential and they differ by gradient of a scalar field. It can be checked that if we add a term   Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev to the first tem where  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev we get the second expression.

Vector Potential of a current sheet

Let the current sheet be in the xy plane with the current flowing along the x direction with a linear current density K,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We have seen that the magnitude of the magnetic field is constant both above the pland and below the plane with a discontinuity at the boundary.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the field is constant, we can use expression for the vector potential from the last section, For z>0

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The vecor potential below the plane is obtained by changing the minus sign in the beginning of the expression to a plus. Note that the components of the vector potential (both tangential and normal) are continuous across the boundary, though the magnetic field itself is discontinuous.

Vector Potential of a circular current carrying loop

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The current being in the azimuthal direction, the current density vector in this case can be written as

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

(Remember  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev not being a constant vector, its direction at the point of observation is not the same as its direction on the current element. As a result this expression cannot be directly substituted in the general expression for vector potential.)

Contribution of a current element  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev to the vector potential is given by

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The vector potential due to the loop is given by 

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

In order to evaluate it, we will expand  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev in Legendre polynomials, as we did in electrostatic problems.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

For large values of r, we need to keep only leading terms. The lowest order term l=0 for which  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev vanishes because  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev Let us keep the next order term with l=1which gives,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The scalar product inside the integral can be simplified and expressed as a sum of two quantities ,one of which is a perfect integral ,which on integration would give zero.

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

 Remember that in this expression, Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is a fixed vector (being the observation point) and  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is change in the value of the vector  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We can write, using chain rule,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Substituting this in the preceding relation, we get,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The loop integral of the first term vanishes as it is a perfect integral. The second term gives,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Note that the loop integral, along with the factor of 1/2 is just the area vector of the loop, which, when multiplied by the current , gives the magnetic moment I,gives the magnetic moment  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev of the loop. We thus have, interchanging the order of the cross product to take care of the minus sign, 

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

This form has the advantage that it can be used to calculate the magnetic field at large distances even away from the axis. If we write this in spherical coordinates, the vector potential clearly has only the azimuthal component, and we can write,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

This gives for the magnetic field components,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We can also obtain a coordinate independent form for the magnetic field from the expression for the vector potential,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Proof of this is left as an exericise.

 

Tutorial Assignment

1. Show that the coordinate free form of the magnetic field derived from the vector potential for the circular loop is given by

    Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

2. A vector potential is given by  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev where  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is a unit vector along an arbitrary but given direction. Calculate the corresponding magnetic field.

3. A spherical shell of radius R carries a uniform surface charge density σ The sphere is rotated with a uniform angular velocity ω about an axis (z direction).). Find the vector potential and the magnetic field of induction at an arbitrary point both outside and inside the shell.

Solutions to Tutorial Assignment

1. Magnetic field can be written as

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

(The full expansion of  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev has four terms of which two vanish because  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is constant, leaving us with two terms above. The first term gives zero because  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev the remaining term is easily expanded to get the desired result).

2. The magnetic field is given by

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The vector potential corresponds to a constant magnetic field in the given direction.

3. Since the charges are on the surface, we can write the current density as follows:

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The vector potential can be written as

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The integral above can only depend on the direction  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev Let us write,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Taking the dot product of both sides withMagnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev , and performing the delta function integration, we get

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

with  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev where θ' is the angle between the variable vector  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev and the directionMagnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRevWe now, expand  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev in terms of spherical harmonics and recognize that  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using the orthogonality of spherical harmonics, the term that survives is  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev and we are left with

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the vector potential is given by

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

[ The integral can be done in more direct method. Let us take the point of observation along the z axis and let the sphere be rotating about an axis which lies in the xz plane and which makes an angle  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev with the z axis. If the  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev have the coordinates  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev with respect to a spherical system of coordinates with its origin at the centre of the sphere. We can expand  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev in Cartesian coordinates ( expending a cross product in spherical coordinates being not a straightforward exercise),

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

when these are inserted into the integral, the integration over  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev would give zero. We are left with only the contribution coming from the last term of the second term of the cross product. The term which survives is

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

the last relation follows because Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is in xz plane. The rest of the derivation is same as above]

Self Assessment Quiz

  1. For the case of vector potential derived for the solenoid show that the expression satisfies Coulomb gauge condition.
  2. Find the vector potential inside a cylindrical wire of radius R.
  3. A charge q is moving slowly with a uniform velocity  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev Obtain an expression for the vector potential and the magnetic field at a distance r from it.

Solutions to Self Assessment Quiz

1. In cylindrical coordinates  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

      Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev we only have to consider the second term. As  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is independent of Ø, the divergence is zero.

2. The field inside a cylindrical wire carrying a current in the z direction is given by

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the magnetic field is along the azimuthal direction, we have on equating it with the azimuthal component of the vector potential,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the wire is long there cannot be a variation with respect to z and the only non-zerocomponent of  Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev is

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

which gives, Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

3. The moving charge can be considered to be equivalent to a current qv with the current density being directed along the direction of the velocity. The vector potential can be written as

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The magnetic field is obtained by taking the curl of this equation. Using,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

and using the fact that the velocity vector being constant, the first term of above is zero, we get,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

 

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
29 videos|49 docs|70 tests

Up next >

Dynamic Test

Content Category

Related Searches

Sample Paper

,

Exam

,

mock tests for examination

,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

,

Summary

,

study material

,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

,

Viva Questions

,

Free

,

shortcuts and tricks

,

ppt

,

video lectures

,

practice quizzes

,

Important questions

,

Semester Notes

,

Extra Questions

,

past year papers

,

MCQs

,

Objective type Questions

,

pdf

,

Previous Year Questions with Solutions

,

Magnetostatics (Part - 3) Electrical Engineering (EE) Notes | EduRev

;