Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Theory

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Electrical Engineering (EE) : Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The document Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
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Magnetic field due to a solenoid and a toroid

A solenoid is essentially a long current loop with closely packed circular turns. The length of the solenoid is very large compared to the diameter of the turns. In diagram shown below, current enters the page from the top and goes behing the page at the bottom. It leads to addition of the magnetic field due to the turns inside the solenoid and cancellation outside.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The magnetic field can be calculated by Ampere’s law and it gives,

Field inside = μ0nl, where n is the number of turns per unit length of the solenoid. The field outside thec solenoid is zero. The same result can be obtained by use of Biot Savart’s law, which is left as an exercise.

A toroid looks like a doughnut with a central hole and current carrying wires wound over its core. The core is usually made of iron or some such magnetic metal.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Take a circular loop of radius r concentric with the toroid. The amount of current enclosed by the loop is NI where N is the total number of turns and I is the current in each turn. By Ampere’s law, the field is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

where in the last step we have explicitly shown that the field is cuircumferential. The field outside the toroid is zero because the Amperian loop would enclose zero current as current both goes in and comes out through the loop. Likewise, inside the hole the field would also be zero as it would not thread any current. Note that unlike in the case of a solenoid, the field is not uniform.

Force Between Current Loops

We have seen that a magnetic field exerts a force on a moving charge. Since a current loop contains moving charge, it follows that such a loop would experience a force in a magnetic field. Further, since the magnetic field in which a loop experiences a force must have its origin in another current source, two current carrying circuits would exert force on each other.

Consider two such circuits, numbered 1 and 2.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Consider the field due to the current element  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev in circuit 1. At a position  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev the field is given by Biot Savart’s law,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The net magnetic field at this point is given by summing over the contribution due to all elemental current comprising the circuit. Since the circuit 2 contains moving charges, the element   Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev experiences a force due to this field which is given by  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev By summing over all the current elements in circuit 2, we get the force exerted by circuit 1 on circuit 2 is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

This is a very clumsy expression and can be evaluated only in cases of simple geometry.

By symmetry it follows that the force on circuit 1 due to current in circuit 2 is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

It is not obvious from these two expressions that  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev as it ought to be from Newton’s third law. We will show that the third law remains valid. This requires a bit of algebraic manipulation. Let us start with the expression for Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev  The integrand can be written as follows:

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Where we have used the vector identity,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

As  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is negative gradient of Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev with respect to Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev we can write the integral over the first term as 

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

If we now perform the second line integral it can be rewritten as a surface integral of the curl of a gradient which is identically zero,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

We are thus left only with the integral of the second term of the simplified form of integral, which clearly shows the symmetry which we are looking for.

Force between two parallel wires carrying current

Consider two parallel wires carrying current in the same direction. We take both wires in the plane (xy) of the paper. The first wire (to the left) creates a magnetic field which is into the plane Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev at the location of the second wire and is given by  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The force exerted on a unit length of the second wire is thus given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus the two wires attract each other.

Electric and Magnetic Fields are different manifestations of the same physical phenomenon

We have seen that electric field is produced by static charges whereas the source of a magnetic field is steady current. However, static charge in one reference frame can be seen as moving charges in another reference frame moving with a relative velocity with respect to the first frame. This requires some comments. In the following, we assume familiarity with basic concepts of the special theory of relativity. (For those who are not familiar, this section may be dropped without affecting understanding of subsequent lectures. Such readers should directly go to the section on vector potential in this lecture itself).

If we have a static charge q, it would experience a force on another charge q’, whether or not that charge is moving. The force on the second charge is electrostatic and is given by Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is the field at the location of the second charge. If the second charge happens to be moving, the field is the instantaneous field at the , location of the second charge. If we have a steady current, it will create a magnetic field around it which will not exert a force on a static charge though it will exert a sidewise force on a moving charge. This force is given by  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev and is very much real. Suppose now there is an observer who is moving with the same velocity as that of this charge, i.e. if the observer is in the frame of this moving charge, in the observer’s frame the velocity of the charge being zero, he would not expect any magnetic force . However, both the observers will notice the sidewise deflection of the charge and hence since the force exerted on the charge is not fictitious. How does one resolve this apparent contradiction?

Let us consider the following situation. Suppose we have two, infinite, charged parallel wires carrying linear charge densities λ1 and λ2 separated by a distance d. Let the wires be in the xy plane with their lengths parallel to the x direction.

In a frame S, which is at rest with respect to these wires, these charged wires exert a force on each other (repulsive, if λ1 and λ2 are of the same sign.) We can calculate this force. The electric field due to λ1 at the location of  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev where the unit vector is directed from the first wire to the second.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

If we consider a length L of the second wire (with charge density λ2) the force on the wire (as seen by S) will be  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Lets us now look at the situation with respect to an observer moving with a speed v along x direction. We know that according this observer, all lengths along x direction are contracted by a factor Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus in this frame a length L parallel to the x direction is shortened to L/γ.

Note that the charge density increases because the charge itself does not depend on observer and the same charge is now distributed over a shorter length.  ThusMagnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The electric force on a section which had a length L in S is seen by S’ to be 

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Note that the length d being in a transverse direction is not contracted.

This result is not consistent with what we learnt regarding the way a transverse force transforms under Lorentz transformation. Which is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

where u is the velocity of the object with respect to the moving observer. In this case u=0, so that  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev calculated by S’.

The anomaly can be removed if S’ invokes the existence of another force in his frame of reference, the magnetic force. Let that force be  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev so that the total force in S’ frame is Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev which should be equated to Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus S’ invokes a magnetic force given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Using the expression for F γ,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

the minus sign shows that the force is attractive. This makes perfect sense when viewed from S’ because in his frame the charge densities produce two currents Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev which are parallel and hence attract. Thus what was a purely electric force in S becomes partly electric and partly magnetic forces in S’. The observers S and S’ do not agree on origin of the forces but measure the same force.

Conclusion : Electric and magnetic fields are interchangeable, depending on the state of motion of the observer.

Vector Potential 

We had seen that the electrostatic field is irrotational (conservative) i.e., Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev This allowed us to introduce a scalar potential φ corresponding to the electric field in terms of which the field is expressed as  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev This simplified our efforts a lot because a scalar field subject to superposition principle can be easier to deal with.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev i.e. the field is not irrotational. Instead we have the magnetostatic Gauss’s law which gives the field to be solenoidal, Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev  This allows us to express the magnetic field as a curl of a vector field because divergence of a curl is identically zero. Thus we define the vector potential corresponding to a magnetic field by the relation

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

We are aware that as per Helmholtz Theorem, a vector potential is uniquely given by specifying its curl and divergence. The curl of the vector potential, being the magnetic field has definite physical meaning. We have the freedom to choose the divergence of the vector potential at our will and usually this is done to make calculations easier. For instance, if we consider the Ampere’s law and express it in terms of the vector potential, we would get

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

We now use our freedom in the choice of the divergence of the vector potential. Since the divergence has no physical significance, let us choose this to be zero. The Ampere’s law then takes the form

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev
We find that each component of the vector potential satisfies a Poisson’s (or Laplace’s) equation. This freedom in choosing the divergence of the vector potential is known as “gauge choice” and the particular choice of  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is known as the “Coulomb Gauge”.

A question naturally arises does one always have the gauge choice ? Suppose we have calculated a vector potential such that its divergence is non-zero. Since to any such vector potential, we can always add gradient of an arbitrary scalar without affecting the value of the magnetic field (because curl of a gradient is identically zero), let us define a new vector potential

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

where ψ is some scalar field. We then have,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Clearly, the new vector potential will satisfy Coulomb gauge condition Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev if we choose ψ such that

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

This being a Poisson’s equation, a solution can always be found. Thus even if we start with an arbitrary vector potential, we can by suitable transformation, arrive at a new vector potential which satisfies the Coulomb gauge condition.

Let us express the BiotSavart’s law in terms of the vector potential.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

In arriving at the last expression we have used

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

In this case the vector field Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is constant in so far as the derivatives with respect to r is concerned so that the first term gives zero. The relative minus sign comes because the order of the cross product is reverse of what we have in the vector identity above. Thus we have,

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Note the similarity of this expression with that of the expression for the potential for the electrostatic field

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The role of charge density is taken by the current density here.

Note that unlike the magnetic field, no useful purpose is served by calculating the vector potential at a single point, one needs to calculate it in a region of space to get an expression for the magnetic field. We will do so in the next lecture.

 

Tutorial Assignment

1. Two long parallel wires carry charge density λ (in the rest frame of the wires) each, are moving with a velocity v each parallel to their length. Calculate the force per unit length on either of the wires in a reference frame that is (a) at rest in the laboratory and (b) moving along with the wires with a velocity v (i.e. in the rest frame of the moving wires).

2. Two parallel current carrying wires of length L andmass m per unit length each are suspended from the ceiling by means of massless strings of length l each. The wires carry equal currents in opposite directions. If the angle between the wires suspended from the same point is 2θ, obtain an expression for the current flowing in each wire in terms of the given quantities.

Solutions to Tutorial Assignment

1. In a frame that is at rest with the wire (i.e. the observer is moving with the same velocity as the wires) , the only force is electrostatic and the force on either of the wires per unit length is given by  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The force is repulsive.

For the observer who is at rest in the laboratory, in addition to the electric force, there is also a magnetic force because the moving charge constitutes a current. Since, in this frame (which is moving with a velocity –v with respect to the wires), the lengths are contracted, the charge density is Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev the electric force is  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The current being Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev the magnetic force between the wires is attractive (parallel currents)

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The net force is then given by Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

2. Let the tension in the strings be T. In equilibrium, we have, T cos θ = mLg, T sin θ = Fm which gives F = mg tan θ.

  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The magnetic force is  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev is the separation between the wires in equilibrium. Substituting these in the equation F = mg tan θ, the result follows.

Self Assessment Quiz

1. Obtain the expression for the magnetic field on the axis of a finite solenoid carrying n turns per unit length.

2. A rectangular loop of dimension  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev carries a current l1. A long wire carrying a current l2 is  is at a distance d  from one of the sides of length a it, the rectangle and the wire are in the same plane. The direction of current in the wire and that flowing in the nearer side of the rectangle are parallel. Calculate the force on the loop

3. A single current loop is in the shape of a right triangle of sides a, b, c. A uniform magnetic field of strength B0 is directed parallel to the hypotenuse. Calculate the total force on the loop as well as the torque acting on the loop.

4. A semicircular loop of radius R lies in the xy plane carrying a current I in the counterclockwise direction. Magnetic field B points perpendicular to the diameter (along the y direction). Calculate the force on the straight-line segment as well as on the semi circular arc.

Solutions to Self Assessment Quiz

1. The turns of the coil are shown below with a dot indicating that the current is coming out of the page and a cross indicating that it is going into the page. Consider a point P on the axis of the solenoid.

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let a be the radius of the solenoid. Consider a length dx at a distance x from P. The amount of current that this section contains is I n (dx). This section can be considered as a circular loop of radius a, the field due to which at P is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Let tanMagnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The denominator of the above expression is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

The field due to the solenoid at P is given by

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

As the length of the solenoid approaches infinity both the angles θ1 = θ2 → 0 and we get back B = μ0nl as expected.

2. The force on the two shorter sides cancel because they carry current in the opposite direction and the force on each current element on one of the sides is canceled by the force on the corresponding element on the other side. The force of attraction on the nearer side is  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev and the repulsive force on the far  side is Thus the net force is attractive and is given by Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

3. Let the triangle be in the xy plane. The unit vector along c is Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Force on side c is zero. Force on side  Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

It can be shown that the force on the side b is equal and opposite so that the net force is zero. This is also seen by adding up the force vectorially since Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The torque is not zero however. As the torque can be written as

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

4. The force on the straight-line segment is Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRevTake a length element Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev The force on the length element is Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

Integrating, the force on the arc is

Magnetostatics (Part - 2) Electrical Engineering (EE) Notes | EduRev

(Of course, the net force on the loop has to be zero!

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