Off center hole in a current-carrying wire - magnetic field?

  • #1
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Homework Statement


A long copper rod 8 cm in diameter has an off-center cylindrical hole 4 cm in diameter down its full length. This conductor carries a current of 900 amps flowing in the direction “into the paper.” What is the direction, and strength in tesla, of the magnetic field at the point P that lies on the axis of the outer cylinder?

(To clarify, point P is in the center of the big cylinder, and the small cut-out cylinder is tangent to the edge of the big cylinder and touches point P at the top).

Homework Equations


B(r) = mu*I / 2pi*r

The Attempt at a Solution



I used superposition of currents, saying that the cut out cylinder is equivalent to a cylinder carrying 900 A out of the page superposed on the big cylinder carrying 900 A into the page.

The magnetic field due to the small superposed cylinder is mu*900 / (2pi*0.02) = 9*10^11 T.

I didn't know how to handle the magnetic field from the big cylinder, however, since point P is at the center, which means r=0. Any ideas?
 

Answers and Replies

  • #2
haruspex
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B(r) = mu*I / 2pi*r
That is only valid outside the wire. The situation inside is analogous to the gravitational field inside the Earth.
 
  • #3
gneill
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You might want to think about the proposed current in the smaller cylinder. Consider the current density for the actual current in the compromised cylinder.

For the magnetic field due to the large (intact) cylinder you might get away with a hand-wavy symmetry argument, or, look up "The field inside a current carrying wire". You should find a suitable derivation. It depends on both the wire's overall radius and the radial position from the center of the wire.


Edit: Ah. @haruspex beat me to the post!
 
  • #4
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I tried this again soon after posting with a slightly different strategy:

I put B in terms of current density, so B(r) = mu*J*r/2. J = 900/(pi * [0.04^2 - 0.02^2]) = 750,000/pi. Thus I treat the small cut out as J=-750,000/pi.


B due to the big cylinder = 0 since r=0. B due to the small cylinder = mu*Jr/2 = 3*10^11 tesla. Does this sound right?
 
  • #5
gneill
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I tried this again soon after posting with a slightly different strategy:

I put B in terms of current density, so B(r) = mu*J*r/2. J = 900/(pi * [0.04^2 - 0.02^2]) = 750,000/pi. Thus I treat the small cut out as J=-750,000/pi.
How do you justify the current magnitudes being the same? Shouldn't the smaller cylinder be carrying what is "cut out" from the larger one if the current density were the same throughout?
edit: never mind. You're using current density! D'uh! Move along, nothing to see here...
B due to the big cylinder = 0 since r=0. B due to the small cylinder = mu*Jr/2 = 3*10^11 tesla. Does this sound right?
That's an awful lot of teslas. The order of magnitude looks much too large to me. What value are you using for μ?
 
  • #6
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How do you justify the current magnitudes being the same? Shouldn't the smaller cylinder be carrying what is "cut out" from the larger one if the current density were the same throughout?
edit: never mind. You're using current density! D'uh! Move along, nothing to see here...

That's an awful lot of teslas. The order of magnitude looks much too large to me. What value are you using for μ?


Oops - it actually comes out to 3*10^-3 tesla (30 gauss). That makes more sense
 
  • #7
gneill
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:smile:
 

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