Two identical circular current loops carrying equal currents are placed with their axes inclined at \(45^\circ\) to each other as shown in the figure. The resultant magnetic field at P is

The magnetic field at a point on the axis of a circular current loop is given by:

\(B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\)

where \(R\) is the radius of the loop and \(x\) is the distance from the center to the point.

For both loops, the distance from the center to point P is \(x = \sqrt{3}R\). Substituting this value:

\(B_1 = B_2 = \dfrac{\mu_0 I R^2}{2(R^2 + (\sqrt{3}R)^2)^{3/2}} = \dfrac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \dfrac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \dfrac{\mu_0 I R^2}{2(8R^3)} = \dfrac{\mu_0 I}{16R}\)

Let the magnetic field due to the first loop be \(\vec{B}_1\). Based on the geometry, its axis is along the x-axis, so the field is directed along the positive x-axis:

\(\vec{B}_1 = \dfrac{\mu_0 I}{16R} \hat{i}\)

The second loop is inclined such that its axis makes an angle of \(45^\circ\) with the first loop's axis. The magnetic field \(\vec{B}_2\) due to this loop is directed along its axis towards P. Resolving this field into components gives:

\(\vec{B}_2 = \dfrac{\mu_0 I}{16R} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = \dfrac{\mu_0 I}{16R} \left( \dfrac{1}{\sqrt{2}} \hat{i} + \dfrac{1}{\sqrt{2}} \hat{j} \right)\)

The resultant magnetic field at P is the vector sum of the fields from both loops:

\(\vec{B}_{net} = \vec{B}_1 + \vec{B}_2\)

\(\vec{B}_{net} = \dfrac{\mu_0 I}{16R} \hat{i} + \dfrac{\mu_0 I}{16R} \left( \dfrac{1}{\sqrt{2}} \hat{i} + \dfrac{1}{\sqrt{2}} \hat{j} \right)\)

\(\vec{B}_{net} = \dfrac{\mu_0 I}{16R} \left( \left( 1 + \dfrac{1}{\sqrt{2}} \right) \hat{i} + \dfrac{1}{\sqrt{2}} \hat{j} \right)\)

\(\vec{B}_{net} = \dfrac{\mu_0 I}{16R} \left( \dfrac{\sqrt{2} + 1}{\sqrt{2}} \hat{i} + \dfrac{1}{\sqrt{2}} \hat{j} \right)\)

\(\vec{B}_{net} = \dfrac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}]\)