1. Use the magnetic force formula on a current-carrying wire

For a straight wire in a magnetic field,

$F = BIL\sin\theta$

where:

• $B = 0.15\,\text{T}$
• $I = 8\,\text{A}$
• $L = 4.0\,\text{cm} = 0.040\,\text{m}$
• $\theta = 90^\circ$ since the wire is perpendicular to the field

So,

$\sin 90^\circ = 1$

2. Substitute the values

$F = 0.15 \times 8 \times 0.040 \times 1$

$F = 0.048\,\text{N}$

3. Convert to milli-newton

Since

$1\,\text{N} = 1000\,\text{mN}$

therefore,

$0.048\,\text{N} = 48\,\text{mN}$

4. Final answer

The magnetic force on the wire is

$\boxed{48\,\text{mN}}$

So the required integer answer is:

$\boxed{48}$

5. Comparison with stored correct answer

Stored correct answer = $48$

Our derived answer = $48$

They match.