Point where magnetic field cancels between two current carrying wires.

  • #1
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Homework Statement


Problem statement along with relevant diagram attached with picture below.


Homework Equations


The Biot-Savart Law for a long current carrying wire.
B = (μ)(I)/2(pi)(d)
Where d is the perpendicular distance between the wire and the point at which the field is being calculated. μ is the permeability of free space.

The Attempt at a Solution


(a) Point at which field is null:
Net magnetic field = 0.
Thus, the sum of the magnetic fields at a certain point is 0. (One will point into the page and one will point out of the page)
Thus:
Bnet = 0 ⇔ (μ)(I1)/2(pi)(ρ) = (μ)(I1/2)/2(pi)(d - ρ)
Calculate ρ. I think it works out to be 2d/3.
Could someone tell me if this is correct?
Also, for part (b) what adjustment should be made? Assuming I did (a) correctly. Thanks.
 

Answers and Replies

  • #3
haruspex
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Bnet = 0 ⇔ (μ)(I1)/2(pi)(ρ) = (μ)(I1/2)/2(pi)(d - ρ)
Calculate ρ. I think it works out to be 2d/3.
Sounds right.
Also, for part (b) what adjustment should be made?
What change do you think it makes to the field from I2?
 
  • #4
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Will both fields be pointing out of the page (in the positive k) in the region of 0-d?
 
  • #5
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Sorry! I meant into the page. In the negative k.
 
  • #6
haruspex
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Sorry! I meant into the page. In the negative k.
Yes, but I meant, what difference does it make to the equation?
 
  • #7
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Well. instead of having:
(I)(μ)/(2)(pi)(ρ) - (I/2)(μ)/(2)(pi)(d - ρ) = 0

We'll now have:
(I)(μ)/(2)(pi)(ρ) + (I/2)(μ)/(2)(pi)(d - ρ) = 0
I think the answer works out to be ρ = 2d.
 
  • #8
haruspex
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Well. instead of having:
(I)(μ)/(2)(pi)(ρ) - (I/2)(μ)/(2)(pi)(d - ρ) = 0

We'll now have:
(I)(μ)/(2)(pi)(ρ) + (I/2)(μ)/(2)(pi)(d - ρ) = 0
I think the answer works out to be ρ = 2d.
Looks right.
 

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