Magnetic field and Poynting Flux in an Inductor

[Rafimah]

Hi everyone,

Lately I have been studying the Poynting Flux and I am familiar with the classic examples of how it can be used to describe the power being dissipated by a resistor and the energy flowing into a capacitor, but I have never come across a similar analog for how the Poynting flux applies to an inductor. Is there some comparable use for it in this scenario?

 

[Delta2]

Its sort of the same as with the case of a capacitor, but with the roles of Electric and Magnetic Field switched over:
Inside a (solenoidal) inductor the magnetic field lines are straight lines parallel to the axis of the solenoid, while electric field lines form closed loops around the magnetic field lines.

if you wondering what's happening in the poynting vector outside the solenoid, it still points towards the axis of the solenoid, the electric and magnetic field are not zero outside the solenoid (as long as the solenoid is not infinite) (the magnetic field outside a solenoid inductor resembles that of a bar magnet
https://en.wikipedia.org/wiki/File:VFPt_Solenoid_correct.svg https://en.wikipedia.org/wiki/File:Magnet0873.png).

However if we consider an infinitely long solenoid with time varying current , it seems we might have a mini paradox regarding the poynting vector.

 

[Rafimah]

Thats interesting, so to go about solving for the energy stored in the magnetic field, here is what I was thinking:

B = mu*n*I
E = ?

S = E cross B /mu

Integrate S over the surface area of the cylindrical space inside the solenoid (?)

So the two issues I am having are that I am not sure what the appropriate formula for electric field should be, and I am not totally sure what surface the integral should be over.

 

[Delta2]

Try $E=-\frac{1}{2\pi r}A\frac{dB}{dt}$ (this comes from properly applying Faraday's law of induction) where A the cross section area of solenoid and r its radius.

The surface of the (spatial) integration will be the solenoid view as a cylinder with radius r and cross section area A and length l. And don't forget to integrate over time as well at the end (poynting vector is energy per unit area per unit time).

The integral over time would be something like $\int_0^{t_0}{cI\frac{dI}{dt}dt}$ where c a constant that depends on A,r and length l and I(t) is the current.

Ok sorry if I edited the post too many times while maybe you reading it. I hope it is clear now.

 

[Rafimah]

So based on this, I get the following result after integrating over surface area and time:

U = -Lr ∫I*dI/dt*dt

I believe the energy stored in an inductor is given by the following formula.

U = (B^2*l*A)/(2*mu)

Are these equivalent? Should they be equivalent, as the integral of the Poynting flux over area and time in a capacitor is equal to the energy stored in the capacitor?

 

[Delta2]

They should be equal. The second formula is correct.

The first formula seems incorrect by the factor of r. The factor of r should vanish with the spatial integration over the lateral surface of the cylinder(because we have the lateral surface of the cylinder $2\pi rl$ and the electric field is proportional to $\frac{1}{2\pi r}$ ,r always the radius of the solenoid.

The integral of $\int I\frac{dI}{dt}dt$ is equal to $\frac{I^2(t)}{2}$.

So, the first formula should be (ignoring the minus sign) $U=\frac{1}{2}LI^2(t)$. You can check that this formula is equivalent to $U=\frac{1}{2\mu_0}B^2lA$ by using that $L=\mu_0\frac{N^2A}{l}$ and $B=\mu_0\frac{nI}{l}$.

EDIT: I see now your original expression for the B field in the post #3 of this thread, you forgot the l in the denominator. Don't know if that was the cause of the miscalculation .

 

[Rafimah]

Yep! That was the issue. I believe this gives a complete solution to the problem. I see how the Poynting flux in L behaves like the Poynting flux in C.