Question

Two current-carrying wires are parallel to each other. The current in one 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 . What is the change in the force between the wires?\

Answer 1

Answer:

The new force $F_N$ will be $\frac{6}{13}$ times the old force F. The change then will be $\Delta F=-\frac{7}{13}F$

Explanation:

The force between two current-carrying parallel wires is calculated with the formula:

$F=\frac{\mu_0I_1I_2\Delta L}{2\pi r}$

where r is the distance between them, $\Delta L$ a portion of length of the wires we consider, $I_1$ and $I_2$ their current intensity and $\mu_0=4\pi\times10^{-7}N/A^2$ 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):

$I_{N1}=2I_1\\I_{N2}=3I_2\\r_N=13r$

And the force then will be:

$F_N=\frac{\mu_0I_{N1}I_{N2}\Delta L}{2\pi r_N}=\frac{\mu_02I_13I_2\Delta L}{2\pi 13r}=\frac{6(\mu_0I_1I_2\Delta L)}{13(2\pi r)}=\frac{6}{13}F$

So the change will be:

$\Delta F=F_N-F=\frac{6}{13} F-F=(\frac{6}{13} -1)F=-\frac{7}{13}F$