The magnetic force $\vec{F}$ on a current-carrying wire is defined as:
\[
\vec{F} = I (\vec{L} \times \vec{B}),
\]
where the parameters are:
\begin{itemize}
\item $I = 1.0 \, \text{A}$ (current)
\item $\vec{L} = (10 \, \text{cm}) \, \hat{j} = (0.1 \, \text{m}) \, \hat{j}$ (length vector of the wire)
\item $\vec{B} = (5 \, \text{mT}) \, \hat{j} - (8 \, \text{mT}) \, \hat{k} = (5 \times 10^{-3}) \, \hat{j} - (8 \times 10^{-3}) \, \hat{k}$ (magnetic field)
\end{itemize}
Step 1: Compute the cross product $\vec{L} \times \vec{B}$:
\[
\vec{L} \times \vec{B} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & 0.1 & 0 \\
0 & 5 \times 10^{-3} & -8 \times 10^{-3}
\end{vmatrix}.
\]
Expanding the determinant:
\[
\vec{L} \times \vec{B} = \hat{i} \left( (0.1) \cdot (-8 \times 10^{-3}) - (0) \cdot (5 \times 10^{-3}) \right)
- \hat{j} \left( (0) \cdot (-8 \times 10^{-3}) - (0) \cdot (0) \right)
+ \hat{k} \left( (0) \cdot (5 \times 10^{-3}) - (0.1) \cdot (0) \right).
\]
Simplifying the expression:
\[
\vec{L} \times \vec{B} = \hat{i} \left( -0.8 \times 10^{-3} \right) + 0 + 0.
\]
Therefore:
\[
\vec{L} \times \vec{B} = -(0.8 \times 10^{-3}) \, \hat{i}.
\]
Step 2: Calculate the force $\vec{F}$:
\[
\vec{F} = I (\vec{L} \times \vec{B}) = (1.0) \cdot (-(0.8 \times 10^{-3}) \, \hat{i}).
\]
Simplifying the result:
\[
\vec{F} = -0.8 \, \text{mN} \, \hat{i}.
\]
The resulting force on the wire is:
\[
\boxed{-(0.8 \, \text{mN}) \, \hat{i}}.
\]