Question

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be:

• A. $n B$
• B. $n² B$
• C. $2nB$
• D. $2n² B$

Hint:

When wire is made into multiple turns, radius decreases and field increases as $n²$.

Answer 1

The Correct Option is B

The question tests your understanding of the magnetic field generated by a coil carrying current. Let us solve the problem step-by-step:

1. First, consider a wire bent into a circular loop of one turn carrying a current \(I\). The magnetic field at the center of this loop is given by the formula: \(B = \frac{\mu_0 I}{2R}\) where \(\mu_0\) is the permeability of free space and \(R\) is the radius of the loop.
2. According to the problem, this magnetic field is initially \(B\) when the wire is bent into one circle: \(B = \frac{\mu_0 I}{2R}\).
3. We then bend the wire into a new circular loop with \(n\) turns. The new magnetic field at the center of this coil with \(n\) turns is given by the formula: \(B' = n \cdot \frac{\mu_0 I}{2R}\).
4. Note, however, that bending the wire into \(n\) turns means that the total length of the wire in the coil remains the same. This implies that the radius changes to \(R/n\) to maintain the same wire length.
5. Substituting this modified radius into the equation for the magnetic field: \(B' = n \cdot \frac{\mu_0 I}{2(R/n)} = \frac{n^2 \cdot \mu_0 I}{2R}\).
6. Comparing this with the original magnetic field \(B = \frac{\mu_0 I}{2R}\), we find: \(B' = n^2 B\).

Thus, when the wire is bent into a circular loop with \(n\) turns, the magnetic field at the center becomes \(n^2\) times the original magnetic field. Therefore, the correct answer is \(n^2 B\).