Question

What happens to the magnetic field at the center of a circular coil if the number of turns is doubled while keeping the current constant?

• A. It becomes half \( \left(\frac{B}{2}\right) \)
• B. It remains the same
• C. It doubles \( (2B) \)
• D. It becomes four times \( (4B) \)

Hint:

For a circular current loop, \[ B = \frac{\mu_0 n I}{2R} \] Magnetic field increases directly with the number of turns and current, and decreases with increasing radius.

Answer 1

The Correct Option is C

Concept: The magnetic field at the center of a circular coil carrying current is given by: \[ B = \frac{\mu_0 n I}{2R} \] where
• \( \mu_0 \) = permeability of free space
• \( n \) = number of turns of the coil
• \( I \) = current
• \( R \) = radius of the coil

Step 1: Observe the proportionality. From the formula, \[ B \propto n \] Thus, the magnetic field is directly proportional to the number of turns.

Step 2: Apply the given condition. If the number of turns is doubled: \[ n \rightarrow 2n \] Substituting into the relation, \[ B' = \frac{\mu_0 (2n) I}{2R} = 2B \]

Step 3: Interpret the result. Hence, the magnetic field at the center of the circular coil becomes twice its original value. \[ \boxed{B' = 2B} \]