Displacement Current, Maxwell's Equations & Production question from JEE Main 2026 (21 Jan, Shift 2)

Answer: 1800

1. $J_c = \sigma E_0$ and $J_d = \varepsilon_0 \omega E_0$ (maximum values), so $\dfrac{J_c}{J_d} = \dfrac{\sigma}{\omega \varepsilon_0}$.
2. $\varepsilon_0 = \dfrac{1}{4\pi \times 9 \times 10^{9}}$, $\omega = 2\pi \times 10^{8}$, so $\omega\varepsilon_0 = \dfrac{10^{8}}{18 \times 10^{9}} = \dfrac{1}{180}$.
3. $\dfrac{J_c}{J_d} = 10 \times 180 = 1800$.
4. Answer: 1800.