Physics - Magnetism Question with Solution | TestHub

 An attractive force will act normal to  $X$ towards $Y$ because the direction of the currents in the wires is the same.

Force exerted on length $l$ due to the magnetic field is given as $F=F_{XY}=F_{YX}=I_1{l}_1 B_{12}$

$\Rightarrow F=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}{l}_1$

Given here, $I_1=2 \mathrm{A}, I_2= 3 \mathrm{A}, l_1=5 \mathrm{m}, r=5 \mathrm{cm}$.

Putting the values, we have

$F=\frac{4\pi \times {10}^{-7}\times 2\times 3}{2\pi \times 5\times {10}^{-2}}\times 50\times {10}^{-2}$

$=1.2\times {10}^{-5} \mathrm{N}$ directed towards wire $X$.