MHTCET 19 Apr 2026 Shift 1 Paper

Concept:The magnetic field on the axis of a circular coil depends on its radius, current, and the distance of the point from the centre.Explanation:For a single-turn coil of radius $R$ carrying current $I$, the axial magnetic field at distance $x$ from the centre is:$$B = \frac{\mu_0 I R^2}{2\left(R^2 + x^2\right)^{3/2}}$$For the bigger coil $Y$: $R_Y = 2r$ and $x_Y = d$.$$B_y = \frac{\mu_0 I (2r)^2}{2\left((2r)^2 + d^2\right)^{3/2}} = \frac{2\mu_0 I r^2}{\left(d^2 + 4r^2\right)^{3/2}}$$For the smaller coil $X$: $R_X = r$ and $x_X = \frac{d}{2}$.$$B_x = \frac{\mu_0 I r^2}{2\left(r^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}}$$Since $r^2 + \frac{d^2}{4} = \frac{d^2 + 4r^2}{4}$, we get:$$B_x = \frac{\mu_0 I r^2}{2 \cdot \frac{\left(d^2 + 4r^2\right)^{3/2}}{8}} = \frac{4\mu_0 I r^2}{\left(d^2 + 4r^2\right)^{3/2}}$$Comparing $B_x$ with $B_y$:$$B_x = 2 B_y$$Answer:$B_x = 2 B_y$Therefore, the correct option is C.