A capacitor has circular plates of radius 4.0 cm…

Step 1. The small loop, $r = 2.0$ cm. The fraction is $\left(\dfrac{2.0}{4.0}\right)^2 = 0.25$, so $I_{\text{in}} = 0.25 \times 1.5\ \text{A} = 0.375\ \text{A}$.

Step 2. Its field. $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.375\ \text{A}}{0.020\ \text{m}} = 3.75\ \mu\text{T}$.

Step 3. The edge, $r = R$. All of $I_d$ is taken in: $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 1.5\ \text{A}}{0.040\ \text{m}} = 7.50\ \mu\text{T}$.

Step 4. The large loop, $r = 8.0$ cm. Again all of $I_d$ is taken in: $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 1.5\ \text{A}}{0.080\ \text{m}} = 3.75\ \mu\text{T}$.