Question

The number of turns is increased by \( n \) times in a current-carrying circular coil. Then the magnetic field at the center of the coil is:

• A. Increased by \( n \) times
• B. Decreased by \( n \) times
• C. Increased by \( n^2 \) times
• D. Unchanged

Hint:

The magnetic field at the center of a coil is directly proportional to the number of turns squared, assuming current and radius are constant.

Answer 1

The Correct Option is C

The magnetic field at the center of a current-carrying circular coil is given by the formula: \[ B = \frac{\mu_0 n I}{2 R} \]
where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length,
- \( I \) is the current through the coil, and
- \( R \) is the radius of the coil.
When the number of turns is increased by a factor of \( n \), the total number of turns becomes \( n \times N \) (where \( N \) is the initial number of turns). As a result, the magnetic field at the center increases by a factor of \( n^2 \), because the magnetic field is proportional to the square of the number of turns in the coil.
Thus, the magnetic field increases by \( n^2 \) times.