The answer is actually 10.4N

Step 1: The horizontal component of the force is given by
$F_x = F \cos \theta$

Step 2: Substitute $F = 12 \, \mathrm{N}$ and $\theta = 30^\circ$
$F_x = 12 \, N \cos 30^\circ$

Step 3: Use $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$
$F_x = 12 \, N \times 0.866$

Step 4: Simplify the product
$F_x = 10.392 \, N \approx 10.4 \, N$

Final answer: $\boxed{10.4 N}$