Equilibrium Position of Bead on Accelerated Parabolic Wire

To find the equilibrium position of the bead with respect to the wire, we analyze the forces acting on the bead in the non-inertial frame of reference of the accelerating wire.

Let the acceleration of the wire be directed along the positive $x$-axis, $\vec{a} = a\hat{i}$. In the reference frame of the wire, the forces acting on the bead of mass $m$ are:

1. The force of gravity:
   $\vec{F}_g = -mg\hat{j}$
2. The pseudo force:
   $\vec{F}_p = -ma\hat{i}$
3. The normal force $\vec{N}$ exerted by the wire, which is perpendicular to the tangent of the parabolic curve at the position of the bead.

The equation of the parabolic wire is:
$y = kx^2$

The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative:
$\tan\theta = \frac{dy}{dx} = \frac{d}{dx}(kx^2) = 2kx$
where $\theta$ is the angle that the tangent to the wire makes with the horizontal ($x$-axis).

Since the wire is frictionless, the normal force $\vec{N}$ is perpendicular to the wire, making an angle $\theta$ with the vertical ($y$-axis). Resolving the normal force into horizontal and vertical components for the bead to be in equilibrium relative to the wire:
$N \cos\theta = mg$
$N \sin\theta = ma$

Dividing the two equations gives:
$\tan\theta = \frac{ma}{mg} = \frac{a}{g}$

Equating this to the geometric slope of the wire:
$2k|x| = \frac{a}{g}$

Solving for the distance of the bead from the $y$-axis, $|x|$:
$|x| = \frac{a}{2gk}$

Therefore, the correct option is B.