A solid wire with a radius of 10 cm carries a 5.0 A current that is...

Using Ampere's circuital law, the magnetic field at a distance $r$ from the axis of a solid cylindrical wire of radius $R$ carrying current $I$ is given by $B = \dfrac{\mu_0 I}{2 \pi r}$ for $r \ge R$.

Here, the point is at a distance $r = 10 \text{ cm} = 0.1 \text{ m}$, which is exactly on the surface of the wire ($R = 10 \text{ cm}$).

Substituting the given values:

$B = \dfrac{4 \pi \times 10^{-7} \times 5.0}{2 \pi \times 0.1}$

$B = \dfrac{2 \times 10^{-7} \times 5.0}{0.1}$

$B = 100 \times 10^{-7} \text{ T}$

$B = 10 \times 10^{-6} \text{ T} = 10 \mu\text{T}$