Magnetic Field From Displacement Current

Inside the gap the displacement current is spread uniformly across the plate area, so an Amperian loop of radius r encloses only the fraction of current passing through it. Applying the Ampere-Maxwell law to a loop of radius r < R gives B(2\pi r) = \mu_0 i_d (r^2/R^2), which rearranges to B = \frac{\mu_0 i_d r}{2\pi R^2}. Here i_d equals the conduction current 0.30 A, r = 0.04\ \text{m}, and R = 0.12\ \text{m}. Substituting, B = \frac{(2 \times 10^{-7})(0.30)(0.04)}{(0.12)^2} = \frac{2.4 \times 10^{-9}}{1.44 \times 10^{-2}} = 1.7 \times 10^{-7}\ \text{T}. The 5.0 \times 10^{-7}\ \text{T} option ignores the (r/R^2) weighting and uses the full-radius field. The 1.7 \times 10^{-6}\ \text{T} value is ten times too large from an exponent error. The 4.2 \times 10^{-8}\ \text{T} option mistakenly uses r/R rather than r/R^2 scaling. This mirrors the standard JEE treatment of the symmetric displacement-current field, where the changing electric flux is treated exactly like a uniform conduction current spread over the plate. Note that at the plate edge, where r = R, the formula collapses to the familiar B = \mu_0 i_d/(2\pi R), the same field a straight wire of that current would produce just outside the gap. As a magnitude check, fields from sub-ampere currents at centimetre distances should fall near 10^{-7}\ \text{T}, matching the answer.