Question

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is \( B \). It is then bent into a circular coil of \( n \) turns. The magnetic field at the centre of this coil of \( n \) turns will be

• A. \( 2Bn^2 \)
• B. \( n^2B \)
• C. \( n^3B \)
• D. \( nB \)

Hint:

When forming a coil with \( n \) turns using the same wire, the magnetic field at the center multiplies by \( n \), not \( n^2 \), provided the radius remains unchanged.

Answer 1

The Correct Option is D

Magnetic field at the center of a circular loop carrying current is given by: \[ B = \frac{\mu_0 I}{2R} \] If the same wire is bent into \( n \) turns (i.
e.
, a coil), the magnetic field at the center becomes: \[ B_n = n \cdot B \] Hence, the magnetic field becomes \( n \) times the original field \( B \).