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) → \( 1:4 \)
The field at the centre of a circular coil of \( N \) turns and radius \( R \) is
\( \displaystyle B = \frac{\mu_0 N I}{2R} \)
The same wire is used, so its length is unchanged: \( N \times 2\pi R = \) constant, which gives \( R \propto 1/N \). Doubling the turns from 3 to 6 halves the radius.
\( \displaystyle B \propto \frac{N}{R} \propto N^{2} \)
\( \displaystyle \frac{B_1}{B_2} = \left(\frac{3}{6}\right)^{2} = \frac{1}{4} \)
So \( B_1 : B_2 = 1 : 4 \).