Calculate the force F on the wire shown in the figure, given the current I = 1A, magnetic field B = 2T, and the parabolic shape of the wire described by y2 = 2x, with AC = 2m.

The magnetic field is uniform and into the page, with magnitude $B = 2 \, \text{T}$.
The current flowing through the wire is $I = 1 \, \text{A}$.
The wire segment follows the path of the parabola $y^2 = 2x$ from point C to point A.
The vertical distance between C and A is given as $AC = 2 \, \text{m}$.
Since the parabola is symmetric about the x-axis and the points C and A lie on a vertical line (same x-coordinate), if the y-coordinate of C is $y_C$ and A is $y_A$, then $y_C = -y_A$.
Given $AC = |y_C - y_A| = |y_C - (-y_C)| = |2y_C| = 2 \, \text{m}$.
Therefore, $y_C = 1 \, \text{m}$ and $y_A = -1 \, \text{m}$.
Now, find the x-coordinates using $y^2 = 2x$:
For point C: $x_C = \frac{y_C^2}{2} = \frac{(1)^2}{2} = 0.5 \, \text{m}$. So, $C = (0.5, 1)$.
For point A: $x_A = \frac{y_A^2}{2} = \frac{(-1)^2}{2} = 0.5 \, \text{m}$. So, $A = (0.5, -1)$.
The current flows from C to A. The effective length vector $\vec{L}_{eff}$ is the displacement vector from C to A:

$$\vec{L}_{eff} = (x_A - x_C)\hat{i} + (y_A - y_C)\hat{j}$$ $$\vec{L}_{eff} = (0.5 - 0.5)\hat{i} + (-1 - 1)\hat{j}$$ $$\vec{L}_{eff} = 0\hat{i} - 2\hat{j} = -2\hat{j} \, \text{m}$$

The magnetic field $\vec{B}$ is into the page. If $\hat{i}$ is along x and $\hat{j}$ is along y, then $\hat{k}$ is out of the page. So, into the page is $-\hat{k}$.

$$\vec{B} = -2\hat{k} \, \text{T}$$

For a current-carrying wire in a uniform magnetic field, the magnetic force is given by:

$$\vec{F} = I (\vec{L}_{eff} \times \vec{B})$$

Substitute the values:

$$\vec{F} = (1 \, \text{A}) ((-2\hat{j} \, \text{m}) \times (-2\hat{k} \, \text{T}))$$ $$\vec{F} = (1) ((-2) \times (-2)) (\hat{j} \times \hat{k}) \, \text{N}$$

Recall the cross product identity $\hat{j} \times \hat{k} = \hat{i}$:

$$\vec{F} = 4 (\hat{i}) \, \text{N}$$ $$\vec{F} = 4\hat{i} \, \text{N}$$

The magnitude of the force is $4 \, \text{N}$.