Situated in a horizontal plane, a circular loop of radius 4.0 cm...

The magnitude of the magnetic field on the axis of a circular loop is given by:

$B = \dfrac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}$

Given values are:
$I = 5.0 \text{ A}$
$R = 4.0 \text{ cm} = 0.04 \text{ m}$
$z = 3.0 \text{ cm} = 0.03 \text{ m}$
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$

Substituting the values into the formula:

$B = \dfrac{4\pi \times 10^{-7} \times 5.0 \times (0.04)^2}{2((0.04)^2 + (0.03)^2)^{3/2}}$

$B = \dfrac{4\pi \times 10^{-7} \times 5.0 \times 1.6 \times 10^{-3}}{2(1.6 \times 10^{-3} + 0.9 \times 10^{-3})^{3/2}}$

$B = \dfrac{32\pi \times 10^{-10}}{2(2.5 \times 10^{-3})^{3/2}}$

$B = \dfrac{32\pi \times 10^{-10}}{2(125 \times 10^{-6})}$

$B = \dfrac{32\pi \times 10^{-10}}{250 \times 10^{-6}} = 1.28\pi \times 10^{-5} \text{ T}$

Using $\pi \approx 3.14$:

$B \approx 1.28 \times 3.14 \times 10^{-5} \text{ T} \approx 4.0 \times 10^{-5} \text{ T}$

According to the right-hand thumb rule, since the current is in the clockwise direction when viewed from above, the magnetic field lines enter the loop from above and exit from below. Thus, the direction of the magnetic field at any point on the axis is downwards.