Question

Chapter 21, Problem 83 Two parallel rods are each 0.52 m in length. They are attached at their centers to a spring that is initially neither stretched nor compressed. The spring has a spring constant of 220 N/m. When 1100 A of current is in each rod in the same direction, the spring is observed to be compressed by 2.4 cm. Treat the rods as long, straight wires and find the separation between them when the current is present. Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work

Answer 1

Using Force balance in horizontal direction:

Force on one of the wires due to other wire = Force due to compression of spring

Fm = Fs

Using Hooke's Law: Fs = k*x

Since both wires have current in same direction so force between them will be attractive, which will be:

Fm = $\mu_0$ *I1*I2*L/(2*pi*d)

So,

$\mu_0$*I1*I2*L/(2*pi*d) = k*x

d = separation between wires = $\mu_0$ *I1*I2*L/(2*pi*k*x)

Using given values:

I1 = I2 = 1100 Amp

L = length of wire = 0.52 m

k = spring constant = 220 N/m

x = compression of spring = 2.4 cm = 0.024 m

So,

d = 4*pi*10^-7*1100*1100*0.52/(2*pi*220*0.024)

d = 0.0238 m = 2.38*10^-2 m = 2.38 cm

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