The magnetic field at the centre of a square loop of side 5 cm carryi…

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

Each side of the square is a finite straight wire. Its field at a point at perpendicular distance \( d \) is given by the Biot-Savart law:

    \( \displaystyle B = \frac{\mu_0 I}{4\pi d}\,(\sin\theta_1 + \sin\theta_2) \)

At the centre P, \( d = a/2 \), and each end of a side is seen at \( 45^\circ \). By the right-hand rule all four sides give fields in the same direction (perpendicular to the plane of the loop), so the four contributions add.

    \( a = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
    \( d = a/2 = 0.025\,\mathrm{m} \)
    \( I = 5\,\mathrm{A} \)

    \( \displaystyle B_{\text{side}} = \frac{10^{-7} \times 5}{0.025}\,(\sin 45^\circ + \sin 45^\circ) = 2\times10^{-5} \times \sqrt{2}\,\mathrm{T} \)

    \( B = 4\,B_{\text{side}} = 8\sqrt{2}\times10^{-5}\,\mathrm{T} \)