Flux linked with circular coil vs solenoid

[Aurelius120]

Homework Statement

Circular Coil: $20$ turns, $8$ cm radius
Rotated about vertical diameter $\omega=50rad\ s^{-1}$
Horizontal Magnetic Field $=3.0×10^{-2}T$
Resistance$=10\Omega$
Find average induced and maximum induced emf, maximum current and average power loss in the coil.

Relevant Equations

Flux linked with circular coil $$\phi=NBA=\frac{\mu_○N^2I\pi r^2}{2r}$$

$$E_{max}=N^2πr^2Bω=N×0.603V$$
$$<E_{avg.}>=<N^2πr^2Bω\sin⁡(ωt)>=E_{max.}<sin⁡(ωt)>=0$$
$$I_{max.}=N×0.0603A$$
Average Power loss=$$\frac{<(E_{ind}^2)>}{R}=N^2×0.018W$$

The correct solutions are : $0.603$ ,$0$ ,$0.0603$, $0.018$

Why am I gettting an extra N? Is the emf of thin circular coil $N\frac{d\phi}{dt}$OR $\frac{d\phi}{dt}$
? Why?

I know it is dependent on no of turns for solenoid and in fact I got incorrect answer on a problem because of not taking E=N$\frac{dϕ}{dt}$( for a solenoid)

 

[kuruman]

In an external field,
One coil has flux $\Phi$ through it.
Two coils have flux $2\Phi$ through them.
How much flux would $N$ coils have through them?

Now in a solenoid, $N$ coils produce flux $N\Phi$ through a single coil. Since there are $N$ single coils the flux linking all of them is $N \times (N\Phi)=N^2\Phi$

Did I answer your question?

 

[Aurelius120]

In an external field,
One coil has flux $\Phi$ through it.
Two coils have flux $2\Phi$ through them.
How much flux would $N$ coils have through them?

Now in a solenoid, $N$ coils produce flux $N\Phi$ through a single coil. Since there are $N$ single coils the flux linking all of them is $N \times (N\Phi)=N^2\Phi$

Did I answer your question?

I understand that for a solenoid. Does it also hold for a thin circular coil of negligible length(width)?
Cause the answers differ in this question

 

[hutchphd]

If the coils could all be coincident but distinct then yes it is exactly true. (Stacked coils are one long piece of wire with only two end points)
Cause what answers differ where??????

 

[Aurelius120]

@hutchphd In this question

Relevant Equations: Flux linked with circular coil $$\phi=NBA=\frac{\mu_○N^2I\pi r^2}{2r}$$

View attachment 340010
$$E_{max}=N^2πr^2Bω=N×0.603V$$
$$<E_{avg.}>=<N^2πr^2Bω\sin⁡(ωt)>=E_{max.}<sin⁡(ωt)>=0$$
$$I_{max.}=N×0.0603A$$
Average Power loss=$$\frac{<(E_{ind}^2)>}{R}=N^2×0.018W$$

My answers have an extra $N$ or $N^2$ multiplied with the given correct answers to the question.

The correct solutions are : $0.603$ ,$0$ ,$0.0603$, $0.018$

 

[Charles Link]

The coils are in an external magnetic field. If there is a current through them it will generate its own magnetic field, but I think you need to read the problem more carefully.

 

[Aurelius120]

The coils are in an external magnetic field. If there is a current through them it will generate its own magnetic field, but I think you need to read the problem more carefully.

The entire coil of 20 turns is rotated in magnetic field, total flux linked is $N\phi$. EMF induced per turn is $\frac{\Delta(N\phi)}{\Delta t}$ Total EMF in the coil is $N\frac{\Delta(N\phi)}{\Delta t}$
So where am I wrong?

 

[Charles Link]

If the coils are generating the field, then the field is proportional to $N$ for a single bunch of very closely grouped coils, i.e. the flux per coil is $N$ times the flux created by a single coil.

That case is not what you have here. Instead the field comes from some external source, and the flux linked per coil is just $A B \cos{\theta}$.

 

[TSny]

View attachment 340057
The entire coil of 20 turns is rotated in magnetic field, total flux linked is $N\phi$.

Yes, assuming $\phi$ is the flux through one turn of the coil.

EMF induced per turn is $\frac{\Delta(N\phi)}{\Delta t}$

No, $\frac{\Delta(N\phi)}{\Delta t}$ is the total EMF induced in the coil. It is not the EMF per turn.

 

[Aurelius120]

Does this also hold for solenoid?
@Charles Link
Ok I think I get it. When current passes through the coil each turn produces its own magnetic field which contributes to the net flux of every other turn.
But here each turn got the same magnetic field and therefore same flux linked with it.

Though here are some doubts:

If the coils are generating the field, then the field is proportional to $N$ for a single bunch of very closely grouped coils, i.e. the flux per coil is $N$ times the flux created by a single coil.

1. Shouldn't the flux linked with each coil be $(N-1)$ times the flux per turn?(since the magnetic field of a turn cannot affect itself or apply Lorentz Force on its own electrons)
2. How can the flux linked with each turn and subsequently the emf induced per turn be the same? (For turn at an end of the coil all flux will be incoming or outgoing while for a turn in the middle the net flux will be zero as outgoing flux will cancel incoming flux and similarly each turn will have a different net flux associated with it)

That case is not what you have here. Instead the field comes from some external source, and the flux linked per coil is just $A B \cos{\theta}$.

 

[Charles Link]

Shouldn't the flux linked with each coil be (N−1) times the flux per turn?(since the magnetic field of a turn cannot affect itself or apply Lorentz Force on its own electrons)

No, it is indeed $N$.

For a solenoid, near the ends there will be less flux for an air core, (some of the flux lines enter and exit near the ends, rather than at the ends etc.), but approximately the same flux everywhere for an iron core, such as in a transformer. In any case, the magnetic field inside the solenoid points in the + z-direction =(i.e. for positive current), it points forward=you get the strongest field in the middle of the solenoid, and it does not cancel in any case. The flux lines go along the inside of the solenoid in one direction. (You reverse the direction of current and the magnetic field then points the other way).

 

[Charles Link]

a follow-on: You may find it of interest that the self-inductance of a single circular loop gets complicated by the inductance coming from the field interacting with the electrons of the wire. I did the calculation for the case where that part can be neglected, (just considering the flux that goes through the empty face of the loop and not through the wire), and got that $L=\mu_o R$. (It involves a little calculus to get this result).[Edit: See post 14. This result is incorrect.]

I found with a google though that this simple result isn't taught in the textbooks because the diameter of the wire winds up being part of the formula, and needs to be taken into account. See http://www.enigmatic-consulting.com.../More_elaborate_circular_coil_derivation.html

This latest isn't needed to solve your homework problem, but I thought you might find it of interest.

 

[hutchphd]

There are no "flux linkages" salient to this answer. This is a moving wire in a given external field. A "linkage" would imply that the current in one coil somehow affected that in another. So your initial equation does not apply, as previously explained by others N² implies two interactions with each coil (source+reciever) and the source is external.. Also the average of $sin^2$ is ½ sothe energy calculation is numerically wrong I believe.


-

I did the calculation for the case where that part can be neglected and got that L=μoR. (It involves a little calculus to get this result).

I'm a little confused by the result for L as R gets large. Is the self-inductance of a large loop arbitrarilly large (maybe my intuition fails here)

 

[Charles Link]

I'm a little confused by the result for L as R gets large. Is the self-inductance of a large loop arbitrarilly large (maybe my intuition fails here)

The area is getting larger as $R$ increases. The flux through the face of the circular loop of radius $R$, surprisingly enough is $\mu_o R I$, assuming I computed it correctly. I used Biot-Savart and started for one segment $dl$ on the ring, and did the integral for $d \Phi=A \, dB =dl \, I (\mu_o/4 \pi) 2 \int\limits_{0}^{2R } \cos{\theta} (r \theta /r^2) \, dr$ where $\theta=\arccos(r/2R)$. The integral does a sweep at radius $r$ from the segment $dl$ of width $dr$ that subtends an angle of $2 \theta$, in order to find the flux through the face. (The $\cos{\theta}$ comes from the sine term that you get in the vector cross-product). Edit: Scratch that=I think I might have miscalculated it. I need to integrate over $\theta$ as well. Let me try again...What I'm getting for a second attempt: $d \Phi=A \, dB =dl \, I (\mu_o/4 \pi) 2 \int\limits_{0}^{2R } \int\limits_{0}^{\theta_o} (\cos{\theta} /r^2)r \, dr \, d \theta$, where $\theta_o=\arccos(r/2R)$ , this time I think I set up the integral correctly=it's a double integral=the result diverges. I'll need to check it again, but this could be why the textbooks often use approximate results for this case=e.g. using the field in the center of the loop for the approximate average value, instead of trying to do a more precise calculation.

[Edit: Putting in a radius of $a$ for the wire, (making $a$ the lower limit of the $dr$ integral), I'm now getting $L \approx \mu_o R \ln(1.5 R/a)$ which is a result similar to the result that was computed in the link of post 12, my initial result is incorrect].

In any case, all this is extra work, but the OP might find it of interest.

 

[Charles Link]

I would like to make one additional comment on this: It seems for a single loop, if we make the wire arbitrarily thin, the integral diverges when we calculate the magnetic flux through the loop. We don't encounter that same problem when we consider the magnetic flux through a solenoid, where in this case, we can also make the wire arbitrarily thin, but we spread out the wire in the z-direction when we consider $n$ turns per unit length, so that the mathematics works for the solenoid, unlike for the single loop.

We can compute the $B$ field for a long solenoid and get a uniform $B=\mu_o n I$ in the z direction, with $n=N/l$, so that the flux links are $\Phi=\mu_o N^2 A I/l$.