Question

A coil having 9 turns carrying current produces magnetic field \( B_1 \) at the centre. Now that coil is rewound into 3 turns carrying same current. Then magnetic field at the centre \( B_2 \) is

• A. \( \frac{B_1}{9} \)
• B. \( 9B_1 \)
• C. \( 3B_1 \)
• D. \( \frac{B_1}{3} \)

Hint:

The field at the centre of a circular coil depends on the number of turns and the radius, and the radius changes when the same wire is rewound.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a coil of \(N\) turns and radius \(r\), the field at the centre is \(B = \dfrac{\mu_0 N I}{2r}\). The wire length is fixed, so rewinding changes both \(N\) and \(r\).

Step 2: Radius after Rewinding
Wire length \(L = N \times 2\pi r\), so \(r = \dfrac{L}{2\pi N}\). Thus \(r \propto \dfrac{1}{N}\).

Step 3: Calculation
Then \(B \propto \dfrac{N}{r} \propto N^2\). Therefore \(\dfrac{B_2}{B_1} = \left(\dfrac{3}{9}\right)^2 = \dfrac{1}{9}\).

Step 4: Compare the options
\(B_2 = \dfrac{B_1}{9}\), which matches the first option.

Final Answer:
The field at the centre becomes one-ninth of the original field. \[ \boxed{\text{(A) }\dfrac{B_1}{9}} \]