Question

A wire of length L carrying a current of I ampere is initially bent in the form of a circular coil of one turn. If the same wire is bent into a circular coil of two turns without any change in the current, then the magnetic field at the center of the coil:

• A. No change in the value
• B. Becomes twice to initial value
• C. Reduces to half of initial value
• D. Becomes four times to initial value

Hint:

For a fixed wire length, $B \propto n^2$.

Answer 1

The Correct Option is D

Step 1: Field at the centre of a coil.
A circular coil of $n$ turns and radius $R$ carrying current $I$ gives a central field \[ B = \frac{n\,\mu_0 I}{2R}. \] The wire length $L$ is fixed, so I will write $R$ in terms of $L$ for each case.
Step 2: One turn.
The whole wire makes one loop, so $L = 2\pi R_1$, giving $R_1 = \frac{L}{2\pi}$. Then \[ B_1 = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\pi \mu_0 I}{L}. \]
Step 3: Two turns.
Now the wire wraps twice, so $L = 2(2\pi R_2) = 4\pi R_2$, giving $R_2 = \frac{L}{4\pi}$.
Step 4: Field for the two-turn coil.
With $n = 2$, \[ B_2 = \frac{2\,\mu_0 I}{2R_2} = \frac{2\,\mu_0 I}{2(L/4\pi)} = \frac{4\pi \mu_0 I}{L}. \]
Step 5: Compare the two fields.
Dividing, $\frac{B_2}{B_1} = \frac{4\pi \mu_0 I / L}{\pi \mu_0 I / L} = 4$. So $B_2 = 4B_1$.
Step 6: Conclusion.
The field at the centre becomes four times the original value. \[ \boxed{4 \times \text{initial}} \]