AIIMS 2019 Physics Inductance MCQ Question

The inductance $L$ of a toroid can be calculated using the formula $L = \frac{{\mu_0 N^2 A}}{{2\pi r}}$, where $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{H/m}$), $N$ is the number of turns, $A$ is the cross-sectional area in square meters, and $r$ is the radius in meters. Substituting $N = 500$, $A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2$, and $r = 40 \, \text{cm} = 0.4 \, \text{m}$ into the formula yields an inductance of approximately $125 \, \mu\text{H}$.

Options B, C, and D are incorrect because they either miscalculate the inductance or do not apply the correct formula for a toroidal inductor. For instance, option B (250 μH) may arise from an error in the area or radius, while option C (0.00248 μH) and option D (zero) do not reflect the physical properties of the toroid as defined by the given parameters.