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Self inductance of a coaxial system of cylinders

  1. Mar 27, 2017 #1
    1. The problem statement, all variables and given/known data

    A thin hollow cylinder of radius a is surrounded co-axially by another hollow cylinder of radius b, where b>a. An electric current I flows through them (I is into the plane of paper (x) in outer cylinder and coming out of plane of paper (.) in inner cylinder). Find the:

    a) Self inductance per unit length.

    b) Magnitude of the pressure exerted on each cylinder and state whether the force on each cylinder is tending to burst apart or to collapse the cylinder.

    2. Relevant equations





    3. The attempt at a solution

    The magnetic field inside the inner cylinder and outside and outer cylinder will be zero. Magnetic field will exist only in a<r<b region. So magnetic field at a distance r will be B=μoI/2πr.
    We then take a rectangular strip of length l and width dr. Flux through it will be:


    Integrating this from a to b, we get total flux through that region.

    Self inductance/unit length will be Φ/(i*l) which can be easily calculated.

    In the first part, I don't understand why we are calculating flux through a rectangular strip and then integrating it. Why aren't we taking total flux through a circle of radius r and thickness dr?

    In the second part, all I can think is to calculate force on each of the two cylinders. For calculating force, we need net B on the cylinder. Can anyone tell me how to do it?

    Kindly help.

    Edit: I got the second part. It was similar to calculate electrostatic pressure at a point. Please tell why we have assumed a rectangular strip in first part. Thanks.
    Last edited: Mar 27, 2017
  2. jcsd
  3. Mar 27, 2017 #2


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    Is your circle of radius r and thickness dr concentric with the cylinders? If so, what is the magnetic flux through it?
  4. Mar 27, 2017 #3
    I assumed a concentric circle and then calculated dΦ=B.A=(μoI/2πr)*2πrdr.
    And then integrated from a to b to get μoI(b-a)
    This seems wrong.. but idk why?
  5. Mar 27, 2017 #4


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    The magnetic flux is not just dΦ = B dA. It is ##d\Phi = \vec{B} \cdot \hat{n} ~dA## where ##\hat{n}## is the normal to the area element ##dA##. What is the normal for your choice of dA? What is the direction of the magnetic field? What does that make the dot product?
  6. Mar 27, 2017 #5
    Okay.. so the magnetic field is downwards(if we take the left portion of figure) and area vector is outside the plane of paper? giving me net dot product to be 0?
  7. Mar 27, 2017 #6


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    Yes. There is no magnetic flux through the area element that you propose. What about the rectangular strip?
  8. Mar 27, 2017 #7
    In rectangular strip, how will we take area vector? I assume it will be in the direction of magnetic field. But how do we see that?

    Is there a rule to determine the direction of area vector? If so, then please tell. I always get confused in such cases.
  9. Mar 27, 2017 #8


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    This is a good question. Say you have a closed loop in the plane of the screen. You can orient the loop in one of two ways.
    1. Pick the sense of line integration first, "counterclockwise" or "clockwise". Then the normal to the surface follows from the right hand rule, out of the screen if you picked counterclockwise or into the screen if you picked clockwise.
    2. Pick the direction of the normal first, "out of" or "into" the screen. Then the sense of integration follows from the right hand rule, counterclockwise if you picked out of or clockwise if you picked into.

    Note that this rule has nothing to do with the direction of the B-field. The rule relates circulation (sense of line integration) with direction (normal to the area element.)
  10. Mar 27, 2017 #9
    Thanks. But I still can't figure how to apply this for the rectangular strip sorry.
  11. Mar 27, 2017 #10


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    1. Draw the strip of length L and width dr.
    2. Orient it.
    3. Find the magnetic flux dΦ through the strip. Note that if dr is small enough, the B-field can be assumed to be constant over the strip's surface.
    4. Add all such contributions dΦ continuously from the inner radius to the outer to find the total flux.
  12. Mar 27, 2017 #11

    rude man

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    You can also look at this from an energy viewpoint:
    1/2 Li2 = mag. field energy
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