Two very long, parallel wires are separated by d = .035 m. The first wire carrie
ID: 2259934 • Letter: T
Question
Two very long, parallel wires are separated by d = .035 m. The first wire carries a current of I1 = .3A. The second wire carries a current of I2 = .3A as well.
Write an EXPRESSION of the minimal work per unit length needed to separate the wires from d to 2d.
Please note: the expression needs to include ln(2) within it, and any other natural logs (like ln(d)) cannot be used to answer this problem correctly.
*in the drawing of the problem, the current for BOTH I1 and I2 is going in the due north direction (same direction).
Explanation / Answer
Force between wires, F = mue*i1*i2*L/(2*pi*r)
Force per unit lentgh, F/L = mue*i1*i2*L/(2*pi*r)
work done per unit length = integral F*dr ( r is from d to 2*d)
= integral mue*i1*i2/(2*pi*r)*dr ( r is from d to 2*d)
= mue*i1*i2/(2*pi) * ln(r) ( r is from d to 2*d)
= mue*i1*i2/(2*pi)*ln(2*d/d)
= mue*i1*i2/(2*pi)*ln(2)
=4*pi*10^-7*0.3*0.3/(2*pi) * ln(2)
= 1.2476*10^-8 J
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