Problem 5.13 Application of Ampere’s Law.

A cylindrical wire is carrying current I. Two situations arise. Case A: uniform surface current on the outside surface. Case B: a volume current density that depends upon radial distance from axis of wire. Whats the magnetic field of such a configuration?

You can get pdf and ppt version of this solution if you scroll to the bottom.

Yesterday we saw an interesting application of the Ampere’s Law (– in magnetostatics and sometimes called Ampere’s circuital law also) for the infinite uniform surface current.

( Infinite uniform surface current ) See that application of Ampere’s law first?

Today we will see yet another display of the elegance and efficacy of this law in the following problem. This problem is inherited from Griffith’s text on Electrodynamics (3rd edition)

I have tried to be a bit more explanatory than the basic solution available (in instruction manual, if you have a copy). Thats the whole idea of this labor I have taken up. I also strongly suggest anyone who want to sharpen his saber to try the problem on his/her own effort before looking into the solution. That way one can prepare oneself for the pitfalls of one’s own understanding before taking up help and damaging the opportunity of developing of a better sense of solving such problems.

Griffith Problems from 3rd edition Electrodynamics Text. Problem 5.13 Application of Ampere's Law

Griffith Problems from 3rd edition Electrodynamics Text. Problem 5.13 Application of Ampere’s Law.

A steady current I flows down a cylindrical wire of radius a. What would be the magnetic field outside the wire and inside of it? We need to find the same in two different scenarios given.

Griffith Problems from 3rd edition Electrodynamics Text. Problem 5.13 Application of Ampere's Law

Griffith Problems from 3rd edition Electrodynamics Text. Problem 5.13 Application of Ampere’s Law

Here are the two different scenarios.

A. Its a surface current density on the outside surface and its uniform across the surface.

B. Its a volume current density and its distributed in the volume of the  wire, but this time its not uniform. In-fact the volume current density J is directly proportional to s; the distance from the axis of the wire to where we are referring the value of J.

Application of Ampere's Law: Surface current density on the outside surface is uniform. Volume current density J is directly proportional to s; the distance from the axis of the wire where we are concerned about the value of J.

Application of Ampere’s Law: Surface current density on the outside surface is uniform. Volume current density J is directly proportional to s; the distance from the axis of the wire where we are concerned about the value of J.

Now we are ready to solve the problem for required answers. We can recognize that we can apply Ampere’s law instead of the Biot-Savart law due to the symmetry which is present in the situation. Biot-Savart law would entail a more tedious path for achieving the solution. For achieving the solution we will therefore resort to the Ampere’s Law.

We need to first off all find the direction of the resulting magnetic field from the given source of current. For part a) where there is an uniform surface current density on the outside of the wire, its clear that magnetic field would be directed along the circumference of a circle whose center lies on the axis of the wire and which includes the point on its circumference where we would like to evaluate the magnetic field.

In other words we determined the direction by applying the right hand thumb rule where our thumb points along the axis of the cylinder, this incidentally we can name as the z-axis of our r-φ-z cylindrical coordinates (according to Griffith this would be s-φ-z and in some texts ρ-φ-z). So curl of palm would give the magnetic field direction which is what we have specified by the circle around the cylindrical wire.

One can also say z-component and radial (r/s) component have to be zero by symmetry argument as reversing the direction of current should mean these components are reversed.  (I in Biot-Savart law takes opposite sign) But physically this means inverting the wire which does not change the symmetry of the situation.

Direction of magnetic field: there are 3 ways to know this, right hand thumb rule, Biot Savart rule expression or symmetry arguments. This gives circumferential directed magnetic field. Application of Ampere's Law.

Direction of magnetic field: there are 3 ways to know this, right hand thumb rule, Biot Savart rule expression or symmetry arguments. This gives circumferential directed magnetic field. Application of Ampere’s Law.

We could also see the same explicitly from the Biot-Savart law where the magnetic field is a cross product along z-axis unit vector and radial direction unit vector. Thus magnetic field B is along azimuthal unit vector (φ-cap).

Direction of magnetic field: Biot Savart rule expression gives circumferential directed magnetic field. Due to cross product of surface current density which is along z-axis and separation vector (difference of reference point vector r and source point r') we get magnetic field along phi-cap. Application of Ampere's Law.

Direction of magnetic field: Biot Savart rule expression gives circumferential directed magnetic field. Due to cross product of surface current density which is along z-axis and separation vector (difference of reference point vector r and source point r’) we get magnetic field along phi-cap. Application of Ampere’s Law.

We also see from symmetry that magnetic field magnitude can’t vary along any circle around the cylinder whose center lies on the axis of the wire’s cylindrical body. That means we can evaluate LHS and RHS of Ampere’s law to determine the magnitude of the magnetic field.

From symmetry that magnetic field magnitude can't vary along any circle around the cylinder whose center lies on the axis of the wire's cylindrical body. Application of Ampere's Law.

From symmetry that magnetic field magnitude can’t vary along any circle around the cylinder whose center lies on the axis of the wire’s cylindrical body. Application of Ampere’s Law.

LHS gives a constant unknown value of magnetic field = B multiplied by circumference of the circle (the Amperian loop). RHS is in terms of total value of current enclosed by this circle. This differs depending on we are out side the cylinder or inside of it.

Since this is a surface current inside of cylinder no current flows. While if we are outside of the cylinder all of current I is enclosed by the Amperian circle.

We can evaluate LHS and RHS of Ampere's law to determine the magnitude of the magnetic field. Inside the cylinder there is no surface current. But outside the cylinder all the surface current I is enclosed by Amperian loop. Application of Ampere's Law.

We can evaluate LHS and RHS of Ampere’s law to determine the magnitude of the magnetic field. Inside the cylinder there is no surface current. But outside the cylinder all the surface current I is enclosed by Amperian loop. Application of Ampere’s Law.

We need to find the volume current density J in order to find enclosed current. Volume current density J is defined as current per unit area perpendicular to the current. The diagram shows in the chosen coordinate system the elemental surface da = rdrdρ.

Both J and I are vectors which are directed along z-axis. J is given to be proportional to the distance from axis (take r or s) we can take this as some constant times r (or s). We then integrate J over perpendicular area (the circle with radius a, which is cross section of the wire) to get total current I. This therefore gives the value of j and therefore J.

Both J and I are vectors directed along z-axis. We integrate J over perpendicular area (the circle with radius a, which is cross section of the wire) to get total current I. This way we determine value of j and therefore J. Application of Ampere's Law.

Both J and I are vectors directed along z-axis. We integrate J over perpendicular area (the circle with radius a, which is cross section of the wire) to get total current I. This way we determine value of j and therefore J. Application of Ampere’s Law.

Depending on whether we are outside of the wire or inside the total enclosed current will now differ, based on the value of J. We get accordingly total current I as the enclosed current (by the Amperian loop) if we are outside. If we are inside we need to integrate over the radial variable (from zero to r or s) thus enclosed current varies with cubic power of this distance (r/s).

The enclosed current is found from the value of the volume current density J. But since J depends on distance of reference point from the axis of the wire, the enclosed current is now a function of the distance r/s. J varies as cubic power of s inside of wire. Application of Ampere's Law..

The enclosed current is found from the value of the volume current density J. But since J depends on distance of reference point from the axis of the wire, the enclosed current is now a function of the distance r/s. J varies as cubic power of s inside of wire. Application of Ampere’s Law..

We note that the direction of magnetic field for case B is no different than the same for case A, due to same symmetry. Only the magnitude differs.

But the expression for magnitude of the field differs only if we are inside of the wire (in case-a it was zero, now its quadratic power).

In case of outside the expression is same for both surface current and volume current. (Same in terms of total current I, although technically the situation is quite different in both cases, only the formula matches).

magnetic field outside and inside of the cylinder. Outside value depends on total value of current: I. Inside value varies as the quadratic power of distance from axis. Application of Ampere's Law.

magnetic field outside and inside of the cylinder. Outside value depends on total value of current: I. Inside value varies as the quadratic power of distance from axis. Application of Ampere’s Law.

Here is a pdf file of the solution. Griffith Problem 5.13

and here a power-point file: Griffith Problem 5.13



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