Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Theory

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Electrical Engineering (EE) : Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The document Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
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Magnetostatics

Sources of magnetic field are steady currents. In such a field a moving charge experiences a sideways force. Recall that an electric field exerts a force on a charge, irrespective of whether the charge is moving or static. Magnetic, field, on the other hand, exerts a force only on charges that are moving. Under the combined action of electric and magnetic fields, a charge experiences, what is known as Lorentz force,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

where the field Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev known by various names, such as, “magnetic field of induction”, “magnetic flux density”, or simply, as we will be referring to it in this course as the “magnetic field” . Note that the force due to the magnetic field is expressed as a cross product of two vectors, the force is zero when the charges are moving perpendicular to the direction of the magnetic field as well.

Current and Current Density

Let us look at the definition of current. Current is a scalar quantity which is the amount of charge that crosses the boundary of a surface of a volume per unit time, the surface being oriented normal to the direction of flow. In the steady state there is no accumulation of charge inside a volume through whose surface the charges flow in. This results in the “equation of continuity”.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

If p is the charge density, the current is given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

where Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the current density. Recalling that Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev we get,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Here, the minus sign is taken because, we define the current to be positive when it flows from outside the volume to the inside and the surface normal, as in the previous lectures, is taken to be outward. Using the divergence theorem, we can rewrite this as

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Since the relationship is true for arbitrary volume, we can equate the integrands from both sides, which results in the equation of continuity :

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The principle is a statement of conservation of electric charge.

As we stated earlier in the lecture, magnetostatics deals with steady currents which implies Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev so that for magnetostatic phenomena, we have Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Though we will not be talking much about it, we would like to mention that this equation is also relativistically invariant.

Biot- Savart’s Law

Recall that the electric field at a point due to a charge distribution was calculated by the principle of superposition of fields due to charges inside infinitesimal volume elements which make this charge distribution. The field due to the infinitesimal charge element is given by Coulomb’s law, which is an inverse square law. We take a similar approach to calculate the magnetic field due to a charge distribution.

We consider a current distribution to comprise of infinitesimal current elements whose direction is taken along the direction of the current flow. If Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev is such a current element located at the position Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the field due to it at the position Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev given by the law of Biot and Savart,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Note that, like the Coulomb’s law, this is also an inverse square law as Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev is the distance from the current element to the position where the field is to be calculated. The constant μ0 which appears here is known as the permeability of free space, which has a value  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

We will now do some algebraic manipulation of this equation.

Writing  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and summing over the current distribution, the field at the position Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

We use the relation

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

and absorb the minus sign by interchanging the order of the cross product,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

In arriving at the last step, we have used the fact that the gradient being with respect to r it has no effect on  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and we have used the identity that for a scalar field f and a vector field Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Since the curl operator Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev does not depend on the integration variable, we can take it outside the integral sign and write,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

As divergence of a curl is zero, we have,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

This is the magnetostatic Gauss’s law. Comparing with the corresponding electrostatic formula Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev we see that this equation implies non-existence of magnetic monopoles.

We are aware that a vector field is uniquely given by specifying its divergence and curl. We will now try to find the curl of the magnetic field.

Using the form of Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev in eqn. (1) we get on taking curl,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Use the identity,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Let us examine each of the terms.

The first term can be simplified as follows. We first take the divergence operator inside the integral as the integration is with respect to the primed variable,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

where we have written the gradient with respect to the primed variable by incorporating a minus sign because it acts on the difference Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Using the identity

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The first term can be converted to a surface integral, using the divergence theorem and the surface can be taken to infinite distances making the integral vanish. The second term also vanishes because of continuity equation. Thus the first tem of eqn. (2) vanishes, leaving us with the second term

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev operator can be taken inside the integral and as it is derivative with respect to r, it acts only on Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

This is known as Ampere’s Law. Thus the magnetic field is specified by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Note that for the case of electric field, the curl is zero , which is characteristic of a conservative field.

We will now provide an integral formulation of these two relations.

Consider a closed volume V defined by a surface S. The normal to the surface is defined in the usual way. We can take the volume integral of the first relation and convert to a surface integral using the divergence theorem.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

This is the integral form of magnetostatic Gauss’s law

Take the second relation. Let S be an arbitrary surface through which the current passes. The surface integral is then given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The first relation can be converted to a line integral using the Stoke’s thorem,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

In the following, we illustrate the use of Biot Savart’s law and Ampere’s law. Just as direct use of Coulomb’s law for calculation of electric field for a charge distribution was not always easy and we had to restrict our calculation to a few cases where symmetry of the problem allowed us to use Gauss’s law to our advantage, direct use of Biot Savart’s law can be made only in a few cases, though the law is always applicable. Ampere’s law provides an attractive alternative for cases where symmetry allows us to express the magnetic field in a form which allows us to use the line integral of the magnetic field in terms of B itself.

Magnetic Field of a long straight wire carrying current :

Let us take the wire along the z direction. We will first compute the magnetic field using the integral form of Ampere’s law. By symmetry, the field cannot depend on z coordinate and its magnitude can only depend on the distance from the wire. Thus, if we take a circle of radius r, the magnetic field strength should be the same everywhere. Further, the direction of the field should be circumferential, i.e. given by the right hand rule. If we were to clasp the wire with our right palm, finger pointing in the direction of the current, the direction in which the finger curls is the direction of the magnetic field. (This is dictated by symmetry and we will see more explicitly when we relook at the problem from Biot Savart’s law).

Taking an Amperian loop in the form of a circle of radius R,

The line integral is Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Thus

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev
where Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the unit vector along the azimuthal direction.

Let us redo the problem using Biot Savart’s law.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Let us take  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev along the x-axis which is also the direction of the current. We will calculate the magnetic field at a distance r from the wire along the y-axis. Note that this is quite general as from whichever point we want to calculate the magnetic field, we can drop a perpendicular on to the x-axis and call this to be the y axis. Thus the vector Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev is in the xy plane. We have, Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

It is directed along Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev because the vectors Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev are in the xy plane. We also have,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

which gives Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Thus the magnetic field is given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev
Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Field along the axis of a circular loop

Let us take the loop in the xy plane. We wish to calculate the field along the z axis. A length element along the circle is given by Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and has the position vector Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

We take the point P along the z axis so that Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRevWith these relations, we have, Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRevMagnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Thus, the magnetic field along the axis is given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

where we have used the magnetic moment vector Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev  given by the product of the current with the area of the loop directed along the normal to the loop according to the right hand rule. It is observed that at large distances from the loop the magnetic field varies as the inverse cube of distance from the loop,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Tutorial Assignment

1. Two infinite conducting planes at z=0 and z=d carry currents in opposite directions with surface current density in opposite directions Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Calculate the magnetic field everywhere in space. 2. Part of a long current carrying wire is bent in the form of a semicircle of radius R. Calculate the magnetic field at the centre of the semi circle.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

3. A long cylindrical wire has a current density flowing in the direction of its length whose density is J = J0r, where r is the distance from the cylinder’s axis. Find the magnetic field both inside and outside the cylinder.

4. A thin plastic disk of radius R has a uniform charge density σ. The disk is rotating about its axis with an angular speed ω. Find the magnetic field along the axis of the disk at a distance z from the centre.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Solutions to Tutorial assignment

1. Let us take the yz plane to be the plane of the paper so that the direction of current flow is perpendicular to this plane. Consider a single conducting plane carrying current in the x direction. Let the plate be in the xy plane. The surface current is shown by a circles with a dot.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Consider an Amperian loop of length L and width 2z located as shown. The magnetic field direction above the plane is in –y direction while that below is in +y direction. Taking the loop to be anticlockwise, the line integral of magnetic field is

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Thus, above the plane the field is  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and below the plane it is  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Not that the field strength is independent of the distance from the plane. If we have two planes, at z=0 and z=d, the former having a linear current density  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the fields would add in the region between the plates and would cancel outside. Between the plates,  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and outside both the plates the field is zero.

2. Since the centre is along the line carrying the current, Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev for the straight line section and the contribution to magnetic field is only due to the semicircular arc.

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The field is into the page at B and is given by Biot Savart’s law to be

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

3. The total current in the wire can be obtained by integrating the current density over its cross section The current enclosed within a radius r from the axis is

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Using Ampere’s law, the field at a distance is r < R is 

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The total current carried by the wire being  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the field outside is given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

4. The current on the disk can be calculated by assuming the rotating disc to be equivalent to a collection of concentric current loops. Consider a ring of radius r and width dr. As the disc rotates, the rotating charge on this annular section behaves like a current loop carrying current Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev The field at a distance z due to this ring is

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

The total field is obtained by integrating this expression from 0 to R,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

which can be easily performed by a substitution x = r2 + z2. The result is

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Self Assessment Quiz

1. Two infinite conducting planes at z=0 and z=d carry currents in opposite directions with surface current density in the same directions Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Calculate the magnetic field everywhere in space

2. Two infinite conducting sheets lying in x-z and y-z planes intersect at right angles along the z axis. On each plane a surface current Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Find the magnetic field in each of the four quadrants into which the space is divided by the planes.

3. Consider the loop formed by two semicircular wires of radii R1 and R2 (>R1) and two short straight sections, as shown. A current I flows through the wire. Find the field at the common centre of the semicircles.

 Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

4. The current density along a long cylindrical wire of radius a is given by  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev where r is the distance from the axis of the cylinder. Use Ampere’s law to find the magnetic field both inside and outside the cylinder .

Solutions to Self Assessment Quiz

 1. Refer to the solution of Tutorial assignment 1. The field due to a single plane carrying a linear current density  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev direction above the plane while below the plane it is in +y direction. Since the magnitude of the field is independent of distance, the field cancels between the planes and add up above the planes. For z > d, we have  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev while for z<0 it is  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Its value between the planes is zero.

2. Referring to the solution of tutorial problem 1, the field due to the current in xz plane is along  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev  the plane (i.e. for y > 0) and along Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the plane (i.e. for y < 0)

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Likewise for the y-z plane, the field is along  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev the plane (i.e. for x > 0) and along Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev below (x < 0). The magnitude of the field in each case is the same and is Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev Thus the field along the first quadrant is  Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev and along the fourth is Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev One can similarly find the fields in the other two quadrants.

 3. From Tutorial problem 2 it follows that the contribution to the field due to two straight sections is zero. The smaller semicircle section gives a field into the page while the bigger semicircle gives a field out of the page. The net field is (into the page)

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

4. First find the current enclosed within a distance r from the axis,

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Thus the field inside is given by

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Outside the cylinder the total current Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 1) Electrical Engineering (EE) Notes | EduRev

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