Ampere's loop inside capacitor 
The figure is a view of one plate of a parallel-plate capacitor from within the capacitor. 
In the question, we are required to rank the 4 paths (a, b, c and d) according to the value of $\oint B\cdot ds$. 
I thought that all 4 paths are tie because in this case $\oint B\cdot ds=\mu_0 (\epsilon_0 \frac{d\Phi_E}{dt})$, where the value of $\frac{d\Phi_E}{dt}$ is the same for all paths.
But the given answer says that b, c, d tie and a is the smallest one. Why?
 A: The reason for your puzzlement is the way that you are writing Ampere's law. Write it like this:
$$ \oint \vec{B} \cdot d\vec{s} = \mu_0 \int ( \vec{J} + \epsilon_0 \frac{\partial \vec{E}}{\partial t}) \cdot d\vec{A} ,$$
where this is the vacuum version and the integral on the right is a surface integral that includes the current density and the displacement current density.
This surface integral is not evaluated over the whole capacitor plate, it is evaluated over a surface that is bounded by the loop over which you evaluate the line integral of the B -field on the left hand side.
Thus within the capacitor plates, $\partial \vec{E}/\partial t$ is uniform and parallel to $d\vec{A}$
$$ \oint \vec{B} \cdot d\vec{s} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} \int dA, $$
where $\int dA$ is just the enclosed area between the plates. Outside the plates
$$ \oint \vec{B} \cdot d\vec{s} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} A = \mu_0 \epsilon_0 \frac{d \Phi_E}{dt}$$
where $A$ is the total area of a plate and $\Phi_E$ is the total flux of electric field.
Thus for an ideal capacitor with uniform E-field (and uniform rate of change of E-field and no current density)  the line integral of the B-field around a loop will be proportional to the area enclosed by the loop until the loop encloses the entire area of the capacitor; at which point it is constant because there is no (changing) E-field beyond this.
