Two identical circular wires P and Q each of radius R and carrying current I are kept in perpendicular planes such that they have a common center as shown in figure. Find the magnitude and direction of the net magnetic field at the common center of the two coils

$\frac{μ_0I}{\sqrt{2}R},β=45°$

The correct answer is Option (3) → $\frac{μ_0I}{\sqrt{2}R},β=45°$

The magnetic field at the center of the loop of radius R is,

$B_{loop}=\frac{μ_0I}{2R}$

$B_P=B_Q=\frac{μ_0I}{2R}$

Since, the $\vec B$ are perpendicular to each other,

$B_{net}=\sqrt{{B_P}^+{B_Q}^2}=\sqrt{\left(\frac{μ_0I}{2R}\right)^2+\left(\frac{μ_0I}{2R}\right)^2}$

$=\sqrt{2}.\frac{μ_0I}{2R}=\frac{μ_0I}{\sqrt{2}R}$

and, the net $\vec B$ is directed at a 45° angle to the z-axis and the x-axis, in the plane formed by these two axis.