One of the two identical conducting wires of length $$L$$ is bent in the form of a circular loop and the other one into a circular coil of $$N$$ identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop $$(B_L)$$ to that at the centre of the coil $$(B_C)$$, i.e. $$\frac{B_L}{B_C}$$ will be:

Let the current flowing through both wires be $$I$$ and let the total length of each wire be $$L$$. Both wires are identical, so they have the same length.

First we take the wire that is bent into a single circular loop. If the radius of this loop is $$R_1$$, then its circumference must equal the length of the wire. Therefore, using the basic relation “circumference = $$2\pi\times\text{radius}$$”, we have

$$2\pi R_1 = L \;\;\Longrightarrow\;\; R_1 = \frac{L}{2\pi}.$$

The magnetic field at the centre of a single circular loop carrying current $$I$$ is given by the standard formula

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

Substituting $$R = R_1$$, we obtain the field at the centre of this loop:

$$B_L = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2\left(\dfrac{L}{2\pi}\right)} = \frac{\mu_0 I}{2}\cdot\frac{2\pi}{L} = \frac{\mu_0 I\pi}{L}.$$

Now consider the second wire that is wound into a circular coil of $$N$$ identical turns. Let the radius of each turn be $$R_2$$. Because there are $$N$$ turns, the total length consumed is $$N$$ times the circumference of one turn. Hence

$$N\,(2\pi R_2) = L \;\;\Longrightarrow\;\; R_2 = \frac{L}{2\pi N}.$$

The magnetic field at the centre of a circular coil having $$N$$ turns is $$N$$ times the field produced by one turn (all turns are coaxial and their fields add). Therefore, using the same single-turn formula but multiplied by $$N$$, we have

$$B_C = N\left(\frac{\mu_0 I}{2R_2}\right) = \frac{\mu_0 N I}{2R_2}.$$

Substituting $$R_2 = \dfrac{L}{2\pi N}$$ into the above expression gives

$$B_C = \frac{\mu_0 N I}{2\left(\dfrac{L}{2\pi N}\right)} = \frac{\mu_0 N I}{2}\cdot\frac{2\pi N}{L} = \frac{\mu_0 I\pi N^2}{L}.$$

We now form the required ratio $$\dfrac{B_L}{B_C}$$:

$$\frac{B_L}{B_C} = \frac{\dfrac{\mu_0 I\pi}{L}} {\dfrac{\mu_0 I\pi N^2}{L}} = \frac{\mu_0 I\pi}{L}\cdot\frac{L}{\mu_0 I\pi N^2} = \frac{1}{N^2}.$$

Hence, the correct answer is Option A.