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 point outside a long straight current-carrying wire is given by

$B = \dfrac{\mu_0 I}{2 \pi r}$

Given that the point is at a distance $r = 20 \text{ cm} = 0.2 \text{ m}$ from the axis, which is greater than the radius of the wire ($R = 10 \text{ cm}$), the point lies outside the wire.

Substituting the given values $I = 5.0 \text{ A}$ and $r = 0.2 \text{ m}$:

$B = \dfrac{4\pi \times 10^{-7} \times 5.0}{2 \pi \times 0.2}$

$B = \dfrac{2 \times 10^{-7} \times 5.0}{0.2}$

$B = \dfrac{10 \times 10^{-7}}{0.2}$

$B = 50 \times 10^{-7} \text{ T}$

$B = 5.0 \times 10^{-6} \text{ T} = 5.0 \mu\text{T}$