# Force on wire due to current

1. Homework Statement
hi,im a bit confused by the following question.i think i have a fairly good idea of most of the solution but the actual wording of the question is stil throwing me.

Q: 3 parallel wires,arranged in the form of an equilateral triangle,of side length 10cm,each carry a current of 50 A in the same direction.Calculate the force on any one wire due to the other two.

2. Homework Equations
I used the equation F= 2*10^-7(I1)(I2)/d
a
and arranged wires as follows: *

b* c*

with wires going into page.

3. The Attempt at a Solution
i calculated the force on the top wire,a, due to b and c:
F=2*10^-7(50)(50)/0.1 = 5.0*10^-3 N in each case,along the line joining a to b and a to c

so this force F is attractive as the currents are in the same direction,yes?
what i am wondering is if i have to calculate the resultant force on a,that is,use for eg the parallelogram rule to find the resultant force, which would point in the direction straight down,or can i simply leave the ans. as is??

is there some sort of convention for expressing this direction or can i just say in the direction indicated on a diagram?

i used the cosine rule and found that the resultant force is (3^1/2)(5*10^-3)N in the direction pointing straight down so that the vector would bisect bc as shown in diagram,does this make sense?
thanks steve:rofl:

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ok that didnt work out,wires are meant to look like this:

..........a*

b* ............. c*

ignore dots