Mutual inductance coil line

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Homework Statement



Find the mutual inductance between a coil with N loops and an infinite straight line with current I on its axis.

Homework Equations



M12 = M21 = flux 1/I 2 = flux 2/ I 1

emf = -d/dt flux

i = emf /R

R = eta*l/S

The Attempt at a Solution



I have done this problem before using a infinite straight line and a toroidial solenoid with current however here it is a "coil with N loops" so I am unsure about how it changes it.

The field generated by the line is:

mu_0 * I2 /(2*pi*r)

Is the flux through the coil the same as through a toroidial solenoid:

flux = mu_0 * N * I2 * h * ln(R2/R1)/(2*pi) ?

Thanks
 

Answers and Replies

  • #2
rude man
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Homework Statement



Find the mutual inductance between a coil with N loops and an infinite straight line with current I on its axis.

Homework Equations



M12 = M21 = flux 1/I 2 = flux 2/ I 1

Not quite right. What about the no. of turns in the coil?
The field generated by the line is:

mu_0 * I2 /(2*pi*r)

Yes, but see below.
Is the flux through the coil the same as through a toroidial solenoid:

flux = mu_0 * N * I2 * h * ln(R2/R1)/(2*pi) ?

Thanks

No. See below:

What is the B field due to the wire? What is its direction? To what extent does the wire's B field couple into the coil windings?

Refresher:

let wire = 1
coil = 2

Then M21 = N2ø21/i1
ø21 = flux in the coil due to the wire current i1
=
∫∫B21*dA2
B21 is the B vector inside the coil due to the wire current
dA2 is a vector element of area pointing along the normal of each coil winding (i.e. the coil's axis).
 
  • #3
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Ok, so the flux which I need to find would just be

B = mu_0 * I2 /(2*pi*r)
A = pi*r^2?
 
  • #4
rude man
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Ok, so the flux which I need to find would just be

B = mu_0 * I2 /(2*pi*r)
A = pi*r^2?

You need to think of the direction of the B field induced by the wire and how it can develop flux thru the coil.

Your B for the wire is OK in magnitude but you need to think direction.
What is the dot-product of B due to the wire and area of the coil? Look at what I posted before.
 
  • #5
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They would be perpendicular so the dot product in the integral would reduce to the magnitude no? Sorry I am having a hard time with this for some reason.
 
  • #6
rude man
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They would be perpendicular so the dot product in the integral would reduce to the magnitude no? Sorry I am having a hard time with this for some reason.

What is the dot product of any two mutually perpendicular vectors?
 
  • #7
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The dot product of perpendicular vectors is 0, so the flux is 0 and then M would be zero but that doesn't seem right.

So then they are parallel instead but then they would reduce to what I had before for B and A but that isn't right either gah
 
  • #8
rude man
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The dot product of perpendicular vectors is 0, so the flux is 0 and then M would be zero but that doesn't seem right.

And why do you say "that doesn't seem right"?
 
  • #9
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Hmm, just because I haven't encountered that before not for any physical reason
 
  • #10
STEMucator
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Suppose the straight wire (label it #1) is hooked up to a corresponding potential difference. Let there also be a variable resistor or something of that nature to cause the current in the straight wire to be changing with respect to time (if we change the resistance we change the current). The way I see it, the current in the straight wire has a magnetic field around it, which can be found with the corresponding right hand rule.

This magnetic field (if placed inside the coil on its radial axis) should cause a changing magnetic flux through the coil as the current increases (or decreases). This will obviously induce an emf and current in the coil. Label the coil #2.

The mutual inductance of the coil (#2) with respect to the straight wire (#1) is given by:

$$M_{21} = \frac{N_2 \Phi_{21}}{i_1}$$

Where ##N_2 \Phi_{21}## is the flux linkage. Calculating it you should find:

$$N_2 \Phi_{21} = N_2 B_1A_2$$

The magnetic field is given by ##B_1 = \frac{\mu_0 i_1}{2 \pi R_1}## and the area of the coil is given by ##A_2 = \pi R^2_2##. Dividing by ##i_1## would give you the mutual inductance.

EDIT: Also forgot to mention the dot product is actually zero, didn't notice that until now sorry. The mutual inductance is indeed zero.
 
Last edited:
  • #11
rude man
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Hmm, just because I haven't encountered that before not for any physical reason

Well, Faraday and Maxwell didn't lie! :smile:

Do you think you understand the situation now? Let me know if there are lingering questions ...
 
  • #12
rude man
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This magnetic field (if placed inside the coil on its radial axis) should cause a changing magnetic flux through the coil as the current increases (or decreases). This will obviously induce an emf and current in the coil. Label the coil #2.

You're not paying attention to the direction of the B field set up by the wire. That B field is concentric with the coil. But in order for an emf to be induced in the coil the B field needs to have an axial component. So the wire B field, being perpendicular to the coil normal (axis) will not induce an emf in the coil, and the mutual inductance is zero.
 
  • #13
STEMucator
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You're not paying attention to the direction of the B field set up by the wire. That B field is concentric with the coil. But in order for an emf to be induced in the coil the B field needs to have an axial component. So the wire B field, being perpendicular to the coil normal (axis) will not induce an emf in the coil, and the mutual inductance is zero.

Yes sorry, I was scribbling my drawing a little too quickly and didn't notice I messed up my RHR.

My apologies.
 
  • #14
rude man
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EDIT: Also forgot to mention the dot product is actually zero, didn't notice that until now sorry. The mutual inductance is indeed zero.

OK, sorry, didn't catch the edit when I wrote my reply to your post.
rm
 

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