Calculate the extension of a spring between current-carrying wires

[jisbon]

Homework Statement

Consider two current-carrying wires of length 1 m each and are held in
equilibrium by a spring which has an unextended length of 0.75m
constant of 0.005 N/m. Each wire carries a current of 20 A in opposite directions. Calculate the extension of the spring. Assume that there is no friction
between the wires and the floor and the wires are rigid.

Relevant Equations

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From what I understand, the force between two current-carrying wires can be calculated as:
$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi R} $$

Doesn't this mean:

$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi 1^{2}}=\dfrac {1}{2}kx^{2} $$ ?

$$\dfrac {\left( 4\pi \times 10^{-7}\right) \left( 20\right) ^{2}}{2\pi \left( 0.75\right) }=\dfrac {1}{2}\left( 0.005\right) x^{2} $$

turns out my answer is incorrect, with the correct being 0.0208m

What is my error here?
Cheers

 

[TSny]

From what I understand, the force between two current-carrying wires can be calculated as:
$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi R} $$

This formula is only accurate if the distance between the two wires is much smaller than the length of the wires, which is not the case here. But it appears that they want you to ignore this and go ahead and use this formula. Otherwise, it would be much more difficult.

Doesn't this mean:

$$\dfrac {F}{L}=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi 1^{2}}=\dfrac {1}{2}kx^{2} $$ ?

Note that $\frac{1}{2}kx^{2}$ does not represent a force (or a force per unit length).

$$\dfrac {\left( 4\pi \times 10^{-7}\right) \left( 20\right) ^{2}}{2\pi \left( 0.75\right) } $$

Will the distance between the wires be 0.75 m after the spring has stretched some?

 

[jisbon]

This formula is only accurate if the distance between the two wires is much smaller than the length of the wires, which is not the case here. But it appears that they want you to ignore this and go ahead and use this formula. Otherwise, it would be much more difficult.

Note that $\frac{1}{2}kx^{2}$ does not represent a force (or a force per unit length).

Will the distance between the wires be 0.75 m after the spring has stretched some?

Nope it will be more than 0.75.
And yes, force = kx, my bad
So, will it be:
$$kx=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi \left( x+0.75\right) }$$
Answer still seems to be wrong

 

[TSny]

$$kx=\dfrac {\mu _{0}I_{1}I_{2}}{2\pi \left( x+0.75\right) }$$
Answer still seems to be wrong

This should give you the "correct" answer. Check your work.

 

[jisbon]

This should give you the "correct" answer. Check your work.

Seems to be a calculation error from my part. Thanks :)