A parallel plate capacitor with circular plates of radius $R$ is being charged as shown in the figure. Let $B(r)$ be the induced magnetic field at a distance $r$ from the central axis between the plates. Assuming $d \ll R$, the ratio $\frac{B(2d)}{B(d)}$, while charging, is
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Inside the plates ($r \lt R$), the induced magnetic field grows linearly with $r$ ($B \propto r$).
Outside the plates ($r \gt R$), it falls off as $1/r$ ($B \propto 1/r$).
Always check if the given distances are inside or outside the radius $R$!
Step 1: Recall where the field comes from. While the capacitor charges, the changing electric field between the plates acts like a current, the displacement current, and that creates a magnetic field circling the axis.
Step 2: Use the Ampere-Maxwell loop law inside the plates. For a circle of radius $r$ inside the plates ($r \lt R$), the enclosed displacement current grows with the area, so \[ B(r) \cdot 2\pi r = \mu_0 I_d \frac{\pi r^2}{\pi R^2}. \]
Step 3: Simplify the relation. Cancelling and solving gives \[ B(r) = \frac{\mu_0 I_d \, r}{2\pi R^2}. \] So inside the plates, $B$ is directly proportional to $r$.
Step 4: Confirm both points are inside. Since the gap satisfies $d \ll R$, both $r = d$ and $r = 2d$ are well within the plate radius, so the simple proportional rule applies to each.
Step 5: Form the ratio. Because $B$ scales with $r$, \[ \frac{B(2d)}{B(d)} = \frac{2d}{d} = 2. \]
Step 6: State the answer. The field at twice the radius is simply twice as strong. \[ \boxed{\dfrac{B(2d)}{B(d)} = 2} \]
