Question

In the figure below, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 5.50 cm. Each wire carries 19.0 A, and all the currents are out of the page. In unit-vector notation, what is the net magnetic force per meter of wire length on wire 1? (Substitute numeric values, do not use variables.) F with arrow L = Incorrect: Your answer is incorrect. N/m

Answer 1

We'll figure out the force each of wires 2, 3 and 4 exerts on 1, and then add to obtain the total force.

Pair 1-2:

The wires carry the same current I in the same direction. Hence, the magnetic force between the two wires is attractive. It has magnitude μ_₀I²/(2πa) per unit length. Set wire 1 at the origin and take x to the right, y up, and z out of the page (it seems like you have this part under control); then, the force is in the positive direction. F_₂ = μ_₀I²/(2πa) i.

Pair 1-3:

Same analysis, only the force is in the negative y direction. F_₃ = -μ_₀I²/(2πa) j.

Pair 1-4:

The two are now a distance a√2 apart, and the attractive force is in the direction i - j. A unit vector in this direction is 1/√2 (i - j). Hence, F_₄ = μ_₀I²/(2πa√2) * (1/√2)(i - j) = μ_₀I²/(4πa) (i - j).

Hence, the total force per unit length is:

F₂ + F₃ + F₄
= μ_₀I²/(2πa) i - μ_₀I²/(2πa) j + μ_₀I²/(4πa) (i - j)
= 3μ_₀I²/(4πa) (i - j)
= 3(10^-7)(19.0)²/(0.055)(i - j) N/m

=1.96 x 10^-3 (i-j)N/m