Wires Magnetic field question

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SUMMARY

The discussion focuses on calculating the magnetic field at the third vertex of an equilateral triangle formed by two long straight wires carrying a current of 2 A each, directed out of the paper. The magnetic field at the third vertex is determined using the formula B = μ₀I/(2πr), where μ₀ is the permeability of free space. Participants emphasize the importance of vector addition for magnetic fields, recommending the use of the right-hand rule to ascertain the direction of each wire's contribution. The final calculated magnetic field at the third vertex is 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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