Answer
(a) $2.57\times 10^{-5}\frac{N}{m}$
(b) same
Work Step by Step
(a) We know that
$\frac{F}{L}=\frac{\mu_{\circ}I_1I_2}{2\pi d}$
We plug in the known values to obtain:
$\frac{F}{L}=\frac{(4\pi \times 10^{-7})(2.75)(4.33)}{2\pi (0.0952)}$
$\frac{F}{L}=2.57\times 10^{-5}\frac{N}{m}$
(b) Yes these two forces are the same as they are an action-reaction pair according to Newton's third law.
