Two identical wires A and B, each of length $$l$$, carry the same current $$I$$. Wire A is bent into a circle of radius R and wire B is bent to form a square of side $$a$$. If $$B_A$$ and $$B_B$$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $$\frac{B_A}{B_B}$$ is

We are given two wires of the same material and the same length $$l$$. Both carry the same current $$I$$. Wire A is bent into a complete circle of radius $$R$$, whereas wire B is bent into a square of side $$a$$. We have to compare the magnetic fields produced at the geometrical centres of these two current-carrying loops.

First we deal with the circular loop. Because the whole length of wire becomes the circumference of the circle, we write

$$2\pi R = l \quad\Longrightarrow\quad R = \frac{l}{2\pi}.$$

The standard result for the magnetic field at the centre of a single-turn circular loop carrying current $$I$$ is stated as

$$B = \frac{\mu_0 I}{2R}.$$

Substituting the value of $$R$$ just obtained, we get

$$B_A = \frac{\mu_0 I}{2\left(\dfrac{l}{2\pi}\right)} = \frac{\mu_0 I}{2}\cdot \frac{2\pi}{l} = \frac{\mu_0 I \pi}{l}.$$

Now we turn to the square loop. Because the whole length of the wire forms the perimeter of the square, we have

$$4a = l \quad\Longrightarrow\quad a = \frac{l}{4}.$$

To find the magnetic field at the centre of a square loop, we look at the contribution from one side and then multiply by four. For a straight segment of finite length the magnetic field at a point lying on the perpendicular bisector of the segment is given by the Biot-Savart result

$$B = \frac{\mu_0 I}{4\pi r}\bigl(\sin\theta_1 + \sin\theta_2\bigr),$$

where $$r$$ is the perpendicular distance from the wire to the point and $$\theta_1,\theta_2$$ are the angles that the lines joining the point to the two ends of the segment make with the perpendicular.

For the centre of the square:

• The distance from the centre to the mid-point of any side is $$r = \dfrac{a}{2}.$$
• Each side extends equally on both sides of this perpendicular, so $$\theta_1 = \theta_2 = 45^\circ.$$ Hence $$\sin\theta_1 = \sin\theta_2 = \sin 45^\circ = \dfrac{1}{\sqrt{2}}.$$

Thus the field due to one side is

$$B_{\text{side}} = \frac{\mu_0 I}{4\pi\left(\dfrac{a}{2}\right)} \bigl(\sin 45^\circ + \sin 45^\circ\bigr) = \frac{\mu_0 I}{4\pi\left(\dfrac{a}{2}\right)} \bigl(\tfrac{1}{\sqrt{2}} + \tfrac{1}{\sqrt{2}}\bigr) = \frac{\mu_0 I}{4\pi\left(\dfrac{a}{2}\right)} \cdot \frac{2}{\sqrt{2}} = \frac{\mu_0 I}{2\pi\left(\dfrac{a}{2}\right)} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 I}{\pi a}\cdot\frac{1}{\sqrt{2}}.$$

Because there are four identical sides, the total field at the centre of the square loop is

$$B_B = 4\,B_{\text{side}} = 4\left(\frac{\mu_0 I}{\pi a}\cdot\frac{1}{\sqrt{2}}\right) = \frac{4\mu_0 I}{\pi a\sqrt{2}}.$$

We now substitute $$a = \dfrac{l}{4}$$ to express everything in terms of the common length $$l$$:

$$B_B = \frac{4\mu_0 I}{\pi\left(\dfrac{l}{4}\right)\sqrt{2}} = \frac{4\mu_0 I}{\dfrac{\pi l}{4}\sqrt{2}} = \frac{4\mu_0 I \times 4}{\pi l \sqrt{2}} = \frac{16\mu_0 I}{\pi l \sqrt{2}}.$$

We are now ready to take the ratio:

$$\frac{B_A}{B_B} = \frac{\dfrac{\mu_0 I \pi}{l}} {\dfrac{16\mu_0 I}{\pi l \sqrt{2}}} = \frac{\mu_0 I \pi}{l} \cdot \frac{\pi l \sqrt{2}}{16\mu_0 I} = \frac{\pi^2\sqrt{2}}{16}.$$

To put this in a simpler fractional form, we divide numerator and denominator by $$\sqrt{2}$$:

$$\frac{\pi^2\sqrt{2}}{16} = \frac{\pi^2}{16/\sqrt{2}} = \frac{\pi^2}{8\sqrt{2}}.$$

Therefore

$$\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}}.$$

Comparing with the given options, this matches Option B.

Hence, the correct answer is Option B.