A current of  $$5\,A$$  exists in a square loop of side  $$\frac{1}{\sqrt{2}}\,m.$$ Then the magnitude of the magnetic field  $$B$$ at the centre of the square loop will be $$p \times10^{-6}\,T,$$ where value of $$p$$  is  $$\underline{\hspace{2cm}}.$$ $$\left[\mu_0=4\pi\times10^{-7}\,TmA^{-1}\right] $$

The magnetic field at the centre of a square loop carrying current can be found by applying the Biot-Savart law to one side and then using symmetry. Given that I = 5 A, a = 1/√2 m and μ₀ = 4π×10⁻⁷ T m A⁻¹, consider a single straight segment of the loop. For a finite straight wire the field at a perpendicular distance d, subtending angles θ₁ and θ₂ at the observation point, is $$B_{\text{wire}} = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2).$$ In our square loop the distance from the centre to each side is d = a/2, and each side subtends equal angles θ₁ = θ₂ = 45° at the centre. Substituting these values gives $$B_1 = \frac{\mu_0 I}{4\pi\,(a/2)}\bigl(\sin45°+\sin45°\bigr) = \frac{\mu_0 I}{2\pi a}\times\frac{2}{\sqrt2} = \frac{\sqrt2\,\mu_0 I}{2\pi a}.$$ By symmetry all four sides contribute equally and their fields add in the same direction, so the total field is $$B = 4B_1 = 4\;\frac{\sqrt2\,\mu_0 I}{2\pi a} = \frac{2\sqrt2\,\mu_0 I}{\pi a}.$$ Substituting I = 5 A, a = 1/√2 m and μ₀ = 4π×10⁻⁷ T m A⁻¹ yields $$B = \frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi\times(1/\sqrt2)}.$$ The π factors cancel, giving $$B = \frac{2\sqrt2\times4\times5\times10^{-7}}{1/\sqrt2} = 40\sqrt2\times10^{-7}\times\sqrt2 = 40\times2\times10^{-7} = 8\times10^{-6}\text{ T}.$$ Therefore, the magnetic field at the centre of the square loop is $$B = 8\times10^{-6}\text{ T},$$ and the required value of p is 8.