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
Answer & explanation
Correct answer: option 3
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: $B_1 : B_2 = 1 : 4$