KCET 2026 Physics Solved Paper

Concept: The magnetic field at an axial point of a circular current loop is directed along the loop’s axis, and the net field is found by vector addition of the fields from both loops.Explanation:For a circular loop of radius $R$ carrying current $I$, the magnetic field at an axial distance $x$ is given by$$B = \frac{\mu_0 I R^2}{2\left(R^2 + x^2\right)^{3/2}}.$$From the geometry of the figure, each loop’s centre is at a distance $x = \sqrt{3}R$ from point $P$.Substituting $x = \sqrt{3}R$:$$B_1 = B_2 = \frac{\mu_0 I R^2}{2\left(R^2 + 3R^2\right)^{3/2}} = \frac{\mu_0 I R^2}{2\left(4R^2\right)^{3/2}}.$$Since $\left(4R^2\right)^{3/2} = 8R^3$, we get$$B_1 = B_2 = \frac{\mu_0 I R^2}{2 \cdot 8R^3} = \frac{\mu_0 I}{16R}.$$Take the axis of the first loop along the $x$-axis. Then its field is$$\vec{B}_1 = \frac{\mu_0 I}{16R}\,\hat{i}.$$The second loop’s axis is inclined at $45^\circ$ to the first, so its field also makes $45^\circ$ with the $x$-axis:$$\vec{B}_2 = \frac{\mu_0 I}{16R}\left(\cos 45^\circ\,\hat{i} + \sin 45^\circ\,\hat{j}\right) = \frac{\mu_0 I}{16R}\left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right).$$Adding the two fields:$$\vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I}{16R}\left[\left(1 + \frac{1}{\sqrt{2}}\right)\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right].$$Simplify the coefficients:$$\vec{B}_{\text{net}} = \frac{\mu_0 I}{16\sqrt{2}R}\left[(\sqrt{2}+1)\,\hat{i} + \hat{j}\right].$$Answer: The resultant magnetic field at $P$ is$$\boxed{\frac{\mu_0 I}{16\sqrt{2}R}\left[(\sqrt{2}+1)\,\hat{i} + \hat{j}\right]}$$So, the correct option is A.