Find out the self inductance of the following system :
A uniformly distributed current 'I' flows through a wire of radius 'a' and returns along the surface of the wire (assume that the surface has a very thin insulation). Also assume that the wire is long.
Method 1 : Using \Phi = LI
B.2\pi s = \frac{\mu _{o}.\pi s^{2}}{\pi a^{2}}
\Phi = \int B.da = \int_{0}^{a}{\frac{\mu _{o}s}{2\pi a^{2}}ds}\int dz
Forgot I in above expressions ... sorry :)
L_{per unit length} = \frac{\Phi}{I}=\frac{\mu _{o}}{4\pi}
Method 2 : Using U = \frac{1}{2}LI^{2} = \frac{1}{2\mu _{o}}\int B^{2}dV
This method gives the answer as
L_{per unit length} =\frac{\mu _{o}}{8\pi}
Which Is correct ?
1Quite obviously , the first is the correct method as per my judgement . Why ? Because the energy equation has a term missing !!!
66In fact, the second method is the one that's correct. The magnetic field is only inside the wire and so the entire magnetic energy is inside it as well.
In the first method, you considered the flux passing through a plane passing along one of the radii, but its not clear why you should do that. The reason is that the flux linkage will be along many such planes as you have taken. Imagining rotating the plane you have taken about the edge coinciding with the axis of the wire.
1I must have been barking mad in saying that the first one was correct . But sir , isn't there going to be some inductance due to the thickness of the wire ? I later thought that , that might be the reason .
66And what does that mean Ricky? The thickness of wire ... ? (Of course, we are assuming the relative permeability to be 1.) Look at the definition of the self inductance. In this case, its quite difficult to count the flux linkage due to the magnetic field of the wire. No surface comes easily to mind which would take into account the entire flux linkage. That's why the energy method is better.
106sir can you clarify a bit more about the planes ?
btw our sir, (harbola sir) said that first method is correct . His reason was that the second method, the energy required to form the system is not taken into account (i.e the fact that initially the system will take some time to set up which will create changing B and hence induce E whose energy we have not added) . initially, current propagates through the wire with speed of light. We have excluded this energy in our calculation so he said that if we include this also (he calculated ) then this energy also comes out to be same as B(mag). hence energy = twice of that
1That ' s what I thought too ( in the first post of mine ) !! But Anant sir is absolutely correct as well ! I am confused seriously .
66@Asish
I do not quite agree with Harbola Sir. Its true that we didn't take the energy required to create the E field. But why should we? In steady state that E field is not present. The induced field is only present when the current has not reached the steady state value. Once the steady state value of I has been achieved, the B field stabilizes and so no induced E. As long as the current has not reached steady state, even the E field is fluctuating, so basically the extra work needed to set up the current is gone as an electromagnetic wave. Once things have reached steady state, only a static magnetic field remains. And so we are justified in considering the stored energy which is present only in this static B field.
And my vote, therefore, is still with method 2. [1]