Solution :
The direction of current is into the page.
The magnetic field [tex]$B_1$[/tex] due to the wire is perpendicular to AP and the magnetic field [tex]$B_2$[/tex] due to the wire is perpendicular to BP. Thus, the net magnetic field at point P is towards the right.
The net magnetic field at point P is
[tex]$B_{net} = B_1 \cos \theta + B_2 \cos \theta$[/tex]
The magnetic field is
[tex]$B_1=B_2=\frac{\mu_0I}{2 \pi L}$[/tex]
The angle θ is
[tex]$\theta = \cos^{-1}\frac{h}{L}$[/tex]
The net magnetic field at point P is
[tex]$B_{net}=\left(\frac{\mu_0I}{2 \pi L}\right)\left(\frac{h}{L}\right)+\left(\frac{\mu_0I}{2 \pi L}\right)\left(\frac{h}{L}\right)$[/tex]
[tex]$=\frac{\mu_0 Ih}{\pi L^2}$[/tex]
a). The direction of the current for both the wires is from north to south. The net magnetic field is towards the right when we are facing south.
[tex]$B_{net}=\frac{\mu_0 Ih}{\pi L^2}$[/tex]
[tex]$=\frac{\mu_0I L}{\pi \left(\sqrt{h^2+(d/2)^2\right)^2}}$[/tex]
[tex]$=\frac{4 \pi \times 10^{-7} \times 100 \times 15}{\pi \left(\sqrt{15^2+(3/2)^2\right)^2}}$[/tex]
[tex]$=2.64 \times 10^{-6} \ T$[/tex]
The direction is towards the west.
b). Since, both the magnetic field are equal and opposite, so no magnetic field is produced.
