Magnetic field at the centre of an $N$-turn circular coil

The magnetic field at the centre of a circular coil is $B = \frac{\mu_0 N I}{2R}$ where:

• $N$ = number of turns
• $I$ = current
• $R$ = radius of the coil

Initial coil

Initially, the coil has $N_1 = 2$ turns and produces field $B_1$: $B_1 = \frac{\mu_0 N_1 I}{2R_1} = \frac{\mu_0 (2) I}{2R_1}$

After rewinding

The wire is the same, so its total length remains constant.

Length of a circular coil of $N$ turns: $L = N \cdot 2\pi R$

Since the wire is unchanged, $N_1 2\pi R_1 = N_2 2\pi R_2$ $N_1 R_1 = N_2 R_2$

Given $N_1 = 2$ and $N_2 = 5$, $2R_1 = 5R_2$ $R_2 = \frac{2}{5}R_1$

Field in the new coil

Now, $B_2 = \frac{\mu_0 N_2 I}{2R_2} = \frac{\mu_0 (5) I}{2R_2}$

Take the ratio

$\frac{B_2}{B_1} = \frac{\mu_0 N_2 I}{2R_2} \cdot \frac{2R_1}{\mu_0 N_1 I}$ $\frac{B_2}{B_1} = \frac{N_2 R_1}{N_1 R_2}$

Substitute $N_1=2$, $N_2=5$, and $R_2 = \frac{2}{5}R_1$: $\frac{B_2}{B_1} = \frac{5R_1}{2 \cdot \left(\frac{2}{5}R_1\right)}$ $\frac{B_2}{B_1} = \frac{5}{\frac{4}{5}} = \frac{25}{4}$

Correct option

$\boxed{\frac{B_2}{B_1} = \frac{25}{4}}$

So the correct answer is Option B.