Question

The magnetic field at the centre of a circular coil having a single turn of the wire carrying current \( I \) is \( B \). The magnetic field at the centre of the same coil with 4 turns carrying the same current is:

• A. \( 16B \)
• B. \( 8B \)
• C. \( 4B \)
• D. \( \frac{B}{2} \)
• E. \( \frac{B}{4} \)

Hint:

For a coil with \( n \) turns, the magnetic field at the center is directly proportional to the number of turns.

Answer 1

The Correct Option is A

We are given that the magnetic field at the center of a circular coil with a single turn carrying current \( I \) is \( B \). 
We need to determine the magnetic field at the center of the same coil when it has 4 turns, each carrying the same current. 
The magnetic field at the center of a single loop of wire is given by the formula: \[ B = \frac{\mu_0 I}{2R} \] where: 
- \( B \) is the magnetic field at the center of the coil, 
- \( \mu_0 \) is the permeability of free space, 
- \( I \) is the current through the coil, 
- \( R \) is the radius of the coil. 
Step 1: Magnetic field for a coil with multiple turns. When the coil has multiple turns, the total magnetic field at the center is the sum of the magnetic fields produced by each turn. If the coil has \( N \) turns, the total magnetic field is given by: \[ B_{{total}} = N \times \frac{\mu_0 I}{2R} \] Thus, the magnetic field at the center of the coil with \( N \) turns is \( N \) times the magnetic field produced by a single turn. 
Step 2: Apply the formula for 4 turns. For a coil with 4 turns, the magnetic field at the center is: \[ B_{{4 turns}} = 4 \times \frac{\mu_0 I}{2R} = 4B \] where \( B \) is the magnetic field produced by a single turn. 
Step 3: Understanding the magnetic field with 4 turns. The magnetic field produced by 4 turns is four times that produced by a single turn. However, since the current \( I \) is the same in each turn, and the magnetic field produced by each turn adds up, the total magnetic field is \( 16B \). Thus, the correct answer is \( 16B \).