Wires Magnetic field question

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To calculate the magnetic field at the third vertex of an equilateral triangle formed by two long straight wires carrying 2 A out of the paper, the magnetic field from each wire must be determined using the formula B = μ0*I/(2π*r). The contributions from both wires cannot be simply added as scalars; instead, vector addition must be applied, taking into account the angles formed by the triangle. The right-hand rule helps to establish the direction of the magnetic field produced by each wire. The final magnetic field at the third vertex is calculated to be 1.7 x 10^-5 T.
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Two long straight wires piece on the plane of the paper at vertices of an equilateral triangle. They each carry 2 A, out of the paper. The magnetic field at the third vertex has a magnetic of...?

The third vertex is the top of the equilateral triangle. I don't know what equation to use to figure this one out. I know that the magnetic field is eqaul to u0*I/2*pi*r...so do I figure out the magnetic field for each of the straight wires and add together?
(The answer should be 1.7*10^-5 T.)
 
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the magnetic field is a vector so you can't just add them if it is an equalateral triangle it should be easy to find the angles thus use geometry to find the magnetic field at the 3rd vertex if you know the current they each carry then you can sub into the magnetic field equation and hopefully this should give you the answer

hope this helps
 
Use the right-hand rule to find each current's contribution to the total field.
(the B-field encircles its source current). Always add vectors tail-to-tip first.
 
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