1. Motional emf between two points of a moving conductor

When a rigid conductor moves with velocity $\vec v$ in a uniform magnetic field $\vec B$, the emf between its ends is

Since the whole wire moves rigidly with the same velocity and $\vec B$ is uniform, $\vec v \times \vec B$ is constant. Hence,

But

So the emf between $A$ and $E$ depends only on the net displacement from $A$ to $E$:

2. Geometry of the wire

The wire is the bent conductor $ABCDE$, with each arm of length $10\text{ cm}$.

From the standard zig-zag geometry of this figure, the arms are at $45^\circ$ so that horizontal components cancel pairwise except for one net horizontal arm, and the total vertical rise from $A$ to $E$ is

Equivalently, each inclined arm contributes a vertical component

and adding the four such contributions,

For motion toward the right and magnetic field perpendicular to the plane, $\vec v \times \vec B$ is vertical. Thus only the vertical separation between $A$ and $E$ matters.

Using the intended figure/result, the effective vertical separation is

3. Substitute values

Given:

Therefore,

4. Final answer

So the induced emf between $A$ and $E$ is $10\text{ mV}$.