The electric current in a circular coil of 2 turns produces a magnetic induction $$B_1$$ at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction $$B_2$$ at its centre. The ratio of $$\dfrac{B_2}{B_1}$$ is:

A circular coil of 2 turns is unwound and rewound into 5 turns. We need to find $$\frac{B_2}{B_1}$$.

Relate the radii using the total wire length.

The total length of wire remains constant:

$$L = n_1 \times 2\pi r_1 = n_2 \times 2\pi r_2$$ $$2 \times 2\pi r_1 = 5 \times 2\pi r_2$$ $$r_2 = \frac{2r_1}{5}$$

Write the magnetic field at the centre of each coil.

$$B = \frac{\mu_0 n I}{2r}$$ $$B_1 = \frac{\mu_0 \times 2 \times I}{2r_1} = \frac{\mu_0 I}{r_1}$$ $$B_2 = \frac{\mu_0 \times 5 \times I}{2r_2} = \frac{5\mu_0 I}{2 \times \frac{2r_1}{5}} = \frac{5\mu_0 I}{\frac{4r_1}{5}} = \frac{25\mu_0 I}{4r_1}$$

Calculate the ratio.

$$\frac{B_2}{B_1} = \frac{\frac{25\mu_0 I}{4r_1}}{\frac{\mu_0 I}{r_1}} = \frac{25}{4}$$

The correct answer is Option B: $$\dfrac{25}{4}$$.