The electric current in a circular coil of 3 turns produces a magnetic field $B_1$ at its centre. The coil is unwound and is rewound into a circular coil of 6 turns. Now on passing the same current, the magnetic field produced is $B_2$ at its centre. The ratio $B_1: B_2$ is

1 : 4

The correct answer is Option (3) → 1 : 4

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.

Given the same wire is used: total length constant. If $N$ is doubled, the length per turn halves, so radius halves:

$R \propto \frac{1}{N}$

Thus magnetic field becomes

$B \propto \frac{N}{R} \propto \frac{N}{\frac{1}{N}} = N^2$

Hence, $B \propto N^2$

So,

$\frac{B_1}{B_2} = \left(\frac{N_1}{N_2}\right)^2 = \left(\frac{3}{6}\right)^2 = \frac{1}{4}$

Answer: $B_1 : B_2 = 1 : 4$