Question

In the figure, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3. In unit-vector notation, what is the net magnetic force per meter of wire length on wire 4?

Answer 1

Using superposition of forces, F4= F14 + F24 + F34. With θ=45°, the situation is as shown on the right.

The components of F4 are given by

F4x =−F43 − F42cosθ

=μ$_0$$i^2$/2πa -μ$_0$$i^2$cos45/2$\sqrt{2}$πa

=-3μ$_0$$i^2$/4πa

and F4y = F41 −F42​sinθ

=μ$_0$$i^2$/2πa -μ$_0$$i^2$vsin45/2$\sqrt{2}$πa

=μ$_0$$i^2$/4πa

Thus,

F4 =(F4$x^{2}$+F4$y^2$)$^1^/^2$

=[(-3μ$_0$$i^2$/4πa )$^1^/^2$+(μ$_0$$i^2$/4πa)$^1^/^2$]$^1^/^2$

=$\sqrt{10}$μ$_0$$i^2$/4πa

=($\sqrt{10}$(4π×$10^-^7$)×7.5$0^2$)/4π(0.135)

​=1.317×1$0^-^4$ N/m

and F4 makes an angle ϕ with the positive x axis, where ϕ=ta$n^-^1$( F4x, F4)

=tan−1(-1/3)=162°

In unit-vector notation, we have

F4=(1.317×1$0^-^4$)[cos 162i+sin 162j]

=(-1.252×1$0^-^4$)i+(0.406×1$0^-^4$)j

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