Question

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

• A. It becomes half of its original value \( \left(\frac{B}{2}\right) \)
• B. It remains unchanged \( (B) \)
• C. It becomes twice the original value \( (2B) \)
• D. It becomes four times the original value \( (4B) \)

Hint:

For a circular coil, the magnetic field at the center is given by \( B=\frac{\mu_0 n I}{2R} \), hence it is directly proportional to the number of turns. If the number of turns is doubled while other factors remain constant, the magnetic field also doubles.

Answer 1

The Correct Option is A

The magnetic field at the center of a circular coil carrying current is given by \( B = \frac{\mu_0 I N}{2R} \).

Here, \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( N \) is the number of turns, and \( R \) is the radius of the coil.

Let the initial magnetic field be \( B_0 = \frac{\mu_0 I N_0}{2R} \).

If the number of turns is doubled, then \( N = 2N_0 \).

The new magnetic field becomes \( B = \frac{\mu_0 I (2N_0)}{2R} = 2B_0 \).

Thus, the magnetic field at the center becomes twice the original value.