The magnetic field at the centre of a square loop of side 5 cm carrying a current of 5 A, is:

$8\sqrt{2} × 10^{-5} T$

The correct answer is Option (4) → $8\sqrt{2} × 10^{-5} T$

For a square loop, the magnetic field at the centre due to one side is:

$B_{one} = \frac{\mu_0 I}{4 \pi r} (\sin 45^\circ + \sin 45^\circ)$

Side $a = 5 \text{ cm} = 0.05 \text{ m}$

Distance of centre from one side:

$r = \frac{a}{2} = 0.025 \text{ m}$

Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$,

$B_{one} = \frac{\mu_0 I}{4 \pi r} \times \sqrt{2}$

$B = 4 B_{one}$$

Substituting $\mu_0 = 4\pi \times 10^{-7}$, $I = 5\text{A}$, and $r = 0.025\text{m}$,

$B = 4 \times \frac{4\pi \times 10^{-7} \times 5}{4\pi \times 0.025} \times \sqrt{2}$

$B = 8\sqrt{2} \times 10^{-5} \text{ T}$

Hence, the correct answer is: $8\sqrt{2} \times 10^{-5} \text{ T}$