Point where magnetic field cancels between two current carrying wires.

[SherlockOhms]

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:
B_net = 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.

 

[SherlockOhms]

 

[haruspex]

B_net = 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?

 

[SherlockOhms]

Will both fields be pointing out of the page (in the positive k) in the region of 0-d?

 

[SherlockOhms]

Sorry! I meant into the page. In the negative k.

 

[haruspex]

Sorry! I meant into the page. In the negative k.

Yes, but I meant, what difference does it make to the equation?

 

[SherlockOhms]

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.

 

[haruspex]

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.