Getting less total B-field from two infinitely long parallel wires

[adamaero]

Homework Statement

Two infinitely long parallel wires separated by 1 m are at 6 m above ground. Each carries a 60-Hz AC current of 100 amperes (rms), but in opposing directions. Find the magnetic flux density B at a test location on the ground which is at an equal distance from the wires.

Relevant Equations

B = u*i/(2*pi*R)

I got 4.7*sin(377t) micro-Tesla for each wire, but where are they getting the total B-field equaling 0.78 micro-Tesla?
Is it some hidden trigonometry?

 

[TSny]

Welcome to PF!
Did you calculate the vertical component of B₁? You can use similar triangles as suggested in the solution.

 

[adamaero]

Did you calculate the vertical component of B₁?

I don't know the dimensions. All I see is the h (6) and d (1).
Sqrt(6^2+1/2^2) used to calculate B.

 

[phyzguy]

I don't know the dimensions. All I see is the h (6) and d (1).
Sqrt(6^2+1/2^2) used to calculate B.

You have d and h. What other dimensions do you think you need to know? Each B field vector is perpendicular to its corresponding radius vector.

 

[adamaero]

You have d and h. What other dimensions do you think you need to know?

I don't know.

Each B field vector is perpendicular to its corresponding radius vector.

So how does that help? I suppose I can guess terms, trial and error: sqrt(j^2+k^2) = 0.78.

...I hate playing this game between inquiry based learning and beating around the bush. When the semester actually starts, I can't spend a week on one problem like this. Quizzes are more frequent than that...too little too late...
...ya, ya, it's been only three days...but it'll probably be a week of volleying back and forth until enough help comes or the approach is uncovered from all the piecemeal comments...

 

[phyzguy]

I don't know.

So how does that help? I suppose I can guess terms, trial and error: sqrt(j^2+k^2) = 0.78.

I guess I don't understand what you think is missing. You've calculated the magnitude of the B field vectors, and you know their directions. Do you know how to do vector addition?

 

[adamaero]

I guess I don't understand what you think is missing. You've calculated the magnitude of the B field vectors, and you know their directions. Do you know how to do vector addition?

4.7+4.7 = 0.78
No, I do not.

 

[phyzguy]

You need to go back and learn how to do vector addition, otherwise there is no hope of solving problems like this. Basically, each component of the two vectors needs to be added independently.

 

[adamaero]

So the issue isn't me knowing vector addition...
I suppose one should only use this site as a starting point to define the question, then after one reply, ask it on allaboutcircuits...

Welcome to PF!
Did you calculate the vertical component of B₁? You can use similar triangles as suggested in the solution.

How are the vertical components calculated? h = 6 m...
B1 = u*i/(2*pi*R)
B1 = 2*10^(-7)/6 = 3.33e-8

 

[phyzguy]

As it says in the problem statement, you have two vectors, and the resultant magnetic field is the sum of the two (B = B1 + B2). B1 is the magnetic field due to wire 1, and B2 is the magnetic field due to wire 2. The total field is the vector sum of these 2, since Maxwell's equations are linear. Can you write out B1 and B2 in vector form?

By vector form, I mean like this:
\vec{B1} = B1_x \vec{e_x} + B1_y \vec{e_y} \vec{B2} = B2_x \vec{e_x} + B2_y \vec{e_y}