Two identical wires X and Y, each of length L carry the same current I. Wire X is bent to form a square of side 'a' and wire Y is bent into a circle of radius R. If \( B_X \) and \( B_Y \) are the values of the magnetic field at the centres of the square and the circle respectively, the ratio \( B_X : B_Y \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{8\sqrt{2}}{\pi^2} \)
Both fields follow from the Biot-Savart law. For the square, each side is a finite straight wire; for the circle, the standard result for the centre of a loop is used.
Square: \( 4a = L \), so \( a = L/4 \). A finite straight wire at perpendicular distance \( d \) gives \( B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2) \). For one side, \( d = a/2 \) and each end is seen at \( 45^\circ \):
\( \displaystyle B_{\mathrm{side}} = \frac{\mu_0 I}{4\pi (a/2)}\left(\sin 45^\circ + \sin 45^\circ\right) = \frac{\sqrt{2}\,\mu_0 I}{2\pi a} \)
All four sides give a field in the same direction at the centre:
\( \displaystyle B_X = 4 \times \frac{\sqrt{2}\,\mu_0 I}{2\pi a} = \frac{2\sqrt{2}\,\mu_0 I}{\pi a} = \frac{8\sqrt{2}\,\mu_0 I}{\pi L} \)
Circle: \( 2\pi R = L \), so \( R = L/2\pi \).
\( \displaystyle B_Y = \frac{\mu_0 I}{2R} = \frac{\pi\,\mu_0 I}{L} \)
\( \displaystyle \frac{B_X}{B_Y} = \frac{8\sqrt{2}\,\mu_0 I/\pi L}{\pi\,\mu_0 I/L} = \frac{8\sqrt{2}}{\pi^{2}} \)