Question

Two identical wires, carrying equal currents are bent into circular coils A and B with 2 and 3 turns respectively. The ratio of the magnetic fields at the centres of the coils A and B is

• A. 4:9
• B. 2:3
• C. 9:4
• D. 3:2

Hint:

If a wire of fixed length is bent into $N$ turns, the magnetic field at the center is proportional to $N^2$. $B \propto N^2$.

Answer 1

The Correct Option is A

Step 1: Relation between N and r:
Let the length of the wire be $L$. For coil A with $N_A = 2$ turns of radius $r_A$: $L = N_A (2\pi r_A) \implies r_A = \frac{L}{2\pi N_A} \propto \frac{1}{N_A}$. For coil B with $N_B = 3$ turns of radius $r_B$: $r_B \propto \frac{1}{N_B}$.
Step 2: Magnetic Field Formula:
Magnetic field at the center of a coil with $N$ turns: $B = \frac{\mu_0 N I}{2r}$. Since $I$ is same, $B \propto \frac{N}{r}$. Substituting $r \propto \frac{1}{N}$, we get $B \propto \frac{N}{1/N} \propto N^2$.
Step 3: Calculate Ratio:
\[ \frac{B_A}{B_B} = \left( \frac{N_A}{N_B} \right)^2 \] \[ \frac{B_A}{B_B} = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]