Answer:
The new force [tex]F_N[/tex] will be [tex]\frac{6}{13}[/tex] times the old force F. The change then will be [tex]\Delta F=-\frac{7}{13}F[/tex]
Explanation:
The force between two current-carrying parallel wires is calculated with the formula:
[tex]F=\frac{\mu_0I_1I_2\Delta L}{2\pi r}[/tex]
where r is the distance between them, [tex]\Delta L[/tex] a portion of length of the wires we consider, [tex]I_1[/tex] and [tex]I_2[/tex] their current intensity and [tex]\mu_0=4\pi\times10^{-7}N/A^2[/tex] the vacuum permeability.
If the current in one wire is increased by a factor of 2, the current in the other is increased by a factor of 3 , and the distance between the wires is decreased by a factor of 13, then we would have a new situation N where (considering the previous variables as an initial situation):
[tex]I_{N1}=2I_1\\I_{N2}=3I_2\\r_N=13r\\[/tex]
And the force then will be:
[tex]F_N=\frac{\mu_0I_{N1}I_{N2}\Delta L}{2\pi r_N}=\frac{\mu_02I_13I_2\Delta L}{2\pi 13r}=\frac{6(\mu_0I_1I_2\Delta L)}{13(2\pi r)}=\frac{6}{13}F[/tex]
So the change will be:
[tex]\Delta F=F_N-F=\frac{6}{13} F-F=(\frac{6}{13} -1)F=-\frac{7}{13}F[/tex]
