What contradiction is found by using Ampere's circuital law to obtain the magnetic field during the charging of a capacitor?

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(N/A) When applying Ampere's circuital law to a loop enclosing a charging capacitor, we encounter a contradiction.
Consider a loop of radius $r$ outside the capacitor plates as shown in figure $(a)$. Applying Ampere's circuital law gives $\oint \vec{B} \cdot d\vec{l} = \mu_0 i(t)$, which yields a non-zero magnetic field $B = \frac{\mu_0 i(t)}{2\pi r}$.
Now, consider a surface bounded by the same loop but passing through the region between the capacitor plates, as shown in figures $(b)$ and $(c)$. Since there is no conduction current flowing between the plates, the current enclosed by this surface is zero $(\sum I = 0)$.
Applying Ampere's circuital law to this surface gives $\oint \vec{B} \cdot d\vec{l} = \mu_0 (0) = 0$, which implies $B = 0$.
This is a contradiction because the magnetic field at the same point $P$ cannot be both non-zero and zero. This inconsistency led Maxwell to propose the existence of displacement current $i_d = \epsilon_0 \frac{d\Phi_E}{dt}$ to complete the law.

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Similar Questions

Match List-$I$ with List-$II$:
List-$I$ List-$II$
$A$. Gauss's Law in Electrostatics $I$. $\oint \vec{E} \cdot d \vec{l} = -\frac{d \phi_B}{d t}$
$B$. Faraday's Law $II$. $\oint \vec{B} \cdot d \vec{A} = 0$
$C$. Gauss's Law in Magnetism $III$. $\oint \vec{B} \cdot d \vec{l} = \mu_0 i_C + \mu_0 \epsilon_0 \frac{d \phi_E}{d t}$
$D$. Ampere-Maxwell Law $IV$. $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$

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