Question

The figure below shows wire 1 in cross section; the wire is long and straight, carries a current of 4.40 mA out of the page, and is at distance d1 = 2.40 cm from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance d2 = 5.40 cm from wire 1 and carries a current of 6.80 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?

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Answer 1

r = squareroot d^2_1 + d^2_2 cos theta = d_2/r The Force per unit length on wire 2 is F = mu_0 I_1 I_2/2 pi r The X - component is F_x = mu_0 I_1 I_2/2 pi r cos theta F_X = mu_0 I_1 I_2/2 pi d_2/r^2 = mu_0 I_1 I_2 d_2/2 pi (d^2_1 + d^2_2) F_X = 4 pi times 10^- 7 times 4.4 times 6.8 times 10^- 6 times 0.054/2 pi (0.024^2 + 0.054^2) \r\n Plugging in the numbers gives the x component of the magnetic force per unit length on wire 2 due to wire 1 is F_x = 9.2536 times 10^- 11 N/m

Plugging in the numbers gives the x component of the magnetic force per unit length on wire 2 due to wire 1 is

$F_x = 9.2536\times 10^{-11}\,N/m$