Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The current being linear  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) the vector potential becomes a simple one dimensional integral

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus the curl of the vector potential only has Ø component 

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which gives

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The line integral of the vector potential is  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) This is because A is along the direction of current which is circumferential.

The vector potential is thus given by for s<R

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) flux being contributed only from inside the solenoid. Thus for  s>R

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

It can be seen that

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

because the field is uniform.

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) to the first tem where  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

(Remember  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) to the vector potential is given by

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The vector potential due to the loop is given by 

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

In order to evaluate it, we will expand  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in Legendre polynomials, as we did in electrostatic problems.

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

We can write, using chain rule,

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substituting this in the preceding relation, we get,

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) of the loop. We thus have, interchanging the order of the cross product to take care of the minus sign, 

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This gives for the magnetic field components,

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

2. A vector potential is given by  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) where  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

2. The magnetic field is given by

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The vector potential can be written as

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The integral above can only depend on the direction  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Let us write,

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Taking the dot product of both sides withMagnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) , and performing the delta function integration, we get

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

with  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) where θ' is the angle between the variable vector  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and the directionMagnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)We now, expand  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in terms of spherical harmonics and recognize that  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Using the orthogonality of spherical harmonics, the term that survives is  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and we are left with

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus the vector potential is given by

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

[ 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) with the z axis. If the  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) have the coordinates  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) with respect to a spherical system of coordinates with its origin at the centre of the sphere. We can expand  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in Cartesian coordinates ( expending a cross product in spherical coordinates being not a straightforward exercise),

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

when these are inserted into the integral, the integration over  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

the last relation follows because Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

      Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Since  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we only have to consider the second term. As  Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which gives, Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 

The document Magnetostatics - 3 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Electromagnetic Fields Theory (EMFT).
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