A 4.0 cm long straight wire carrying a current of 8A is placed perpendicular to an uniform magnetic field of strength 0.15 T. The magnetic force on the wire is _________ mN.

The magnetic force $$F$$ on a straight conductor of length $$l$$ carrying current $$I$$ in a uniform magnetic field of magnitude $$B$$ is given by the formula

$$F = I\,l\,B\,\sin\theta$$,

where $$\theta$$ is the angle between the direction of current and the magnetic field.

Here the wire is placed perpendicular to the field, so $$\theta = 90^{\circ}$$ and $$\sin 90^{\circ} = 1$$. Thus,

$$F = I\,l\,B$$ $$-(1)$$

Convert the length from centimetres to metres:
$$l = 4.0\ \text{cm} = 4.0 \times 10^{-2}\ \text{m} = 0.04\ \text{m}$$

Substitute $$I = 8\ \text{A}$$, $$l = 0.04\ \text{m}$$ and $$B = 0.15\ \text{T}$$ into $$(1)$$:

$$F = 8 \times 0.04 \times 0.15$$

First multiply $$8 \times 0.04 = 0.32$$.
Then multiply $$0.32 \times 0.15 = 0.048$$.

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

To express the force in milli-newtons (1 mN = $$10^{-3}$$ N):
$$0.048\ \text{N} = 0.048 \times 10^{3}\ \text{mN} = 48\ \text{mN}$$.

Therefore, the magnetic force acting on the wire is $$\mathbf{48\ mN}$$.