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 magnetic field produced by a coil is directly proportional to the number of turns. If the number of turns is reduced, the magnetic field decreases by the same factor.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
When a coil is rewound, its total length remains constant. This change in the number of turns affects the radius of the coil, both of which contribute to the magnetic field at the centre.
Step 2: Key Formula or Approach:
1. Magnetic field at the centre: \( B = \frac{\mu_0 N I}{2R} \).
2. Length of wire: \( l = N \cdot 2\pi R \) (Constant).
Step 3: Detailed Explanation:
Initially: \( N_1 = 9 \), field is \( B_1 \). Radius is \( R_1 \).
Finally: \( N_2 = 3 \), field is \( B_2 \). Radius is \( R_2 \).
From length conservation: \( 9 \cdot 2\pi R_1 = 3 \cdot 2\pi R_2 \Rightarrow R_2 = 3R_1 \).
Magnetic field ratio:
\[ \frac{B_2}{B_1} = \frac{N_2/R_2}{N_1/R_1} = \frac{3/(3R_1)}{9/R_1} = \frac{1}{9} \]
\[ B_2 = \frac{B_1}{9} \]
Step 4: Final Answer:
The new magnetic field is \( B_1/9 \).