Displacement current in a capacitor

According to Gauss' law, the homogeous electric field inside the capacitor is $\begin{aligned} \oint \vec E \cdot d\vec A &= E \cdot \pi R^2 = \frac{Q}{\varepsilon_0} \\ \Rightarrow \quad E &= \frac{Q}{\varepsilon_0 \pi R^2} = \frac{Q_0 + I \cdot t}{\varepsilon_0 \pi R^2} \end{aligned}$ with the initial charge $Q_0$ at $t = 0$ . Inside the capacitor there is no electric current ( $\vec j = 0$ ), but there is temporal change $\dfrac{\partial E}{\partial t}$ of the electric field (displacement current) thereby inducing a magnetic field. Since the capacitor has a rotatianal symmetry, the magnetic field has the form $\vec B(\vec r, t) = B(r,t) \left( \begin{array}{c} -\sin \phi \\ \cos \phi \end{array} \right) = B(r,t) \vec e_\phi$ with the unit vector $\vec e_\phi \perp \vec r$ . Integrating Ampere's law over a circle with radius $r$ results $\begin{aligned} \int (\vec \nabla \times \vec B) \cdot d\vec A &= \oint \vec B \cdot d\vec l = B \cdot 2 \pi r \\ &= \mu_0 \varepsilon_0 \int \frac{\partial \vec E}{\partial t} \cdot d\vec A = \mu_0 \varepsilon_0 \frac{\partial E}{\partial t} \pi r^2 \\ \Rightarrow \quad B &= \frac{1}{2} \mu_0 \varepsilon_0 r \frac{\partial E}{\partial t} = \frac{\mu_0 I r}{2 \pi R^2} \end{aligned}$ Using the given numerical values, we get the result $B = \frac{4\pi\cdot 10^{-7} \cdot 1 \cdot 10^{-2}}{2 \pi \cdot 4 \cdot 10^{-4}} \,\text{T} = \frac{1}{2} 10^{-5} \,\text{T} = 5 \,\mathrm{\mu T}$