detailed solution

Correct option is E

By symmetry, each of the wires  and  carries a current of 5A.  As the separation between them is 1cm , the magnetic force per unit length of DC is
  $\begin{array}{l}\frac{\mathrm{dF}}{\mathrm{dl}}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1{\mathrm{i}}_2}{2\mathrm{\pi d}}=\frac{\left(2\times {10}^{-7}{\mathrm{TmA}}^{-1}\left)\right(5\mathrm{A}\left)\right(5\mathrm{A})\right.}{1\times {10}^{-2}\mathrm{m}}\\ =5\times {10}^{-4}\mathrm{TA}=5\times {10}^{-4}{\mathrm{Nm}}^{-1}.\end{array}$
 
 

detailed solution

Correct answer is 5

By symmetry, each of the wires  and  carries a current of 5A.  As the separation between them is 1cm , the magnetic force per unit length of DC is
  $\begin{array}{l}\frac{\mathrm{dF}}{\mathrm{dl}}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1{\mathrm{i}}_2}{2\mathrm{\pi d}}=\frac{\left(2\times {10}^{-7}{\mathrm{TmA}}^{-1}\left)\right(5\mathrm{A}\left)\right(5\mathrm{A})\right.}{1\times {10}^{-2}\mathrm{m}}\\ =5\times {10}^{-4}\mathrm{TA}=5\times {10}^{-4}{\mathrm{Nm}}^{-1}.\end{array}$