A regular polygon of n sides is formed by bending a wire of total length `2 pi r` which carries a current i. (a) Find the magnetic field B at the centre of the polygon. (b) By letting `n rarr oo` , deduce the expression for the magnetic field at the centre of a circular current.

Explanation

To solve the problem, we will break it down into two parts:

### Part (a): Finding the Magnetic Field \( B \) at the Centre of the Polygon

1. **Determine the Side Length of the Polygon**:
The total length of the wire is given as \( 2 \pi r \). Since the wire is bent into a regular polygon with \( n \) sides, the length of each side \( l \) can be calculated as:
\[ l = \frac{2 \pi r}{n} \]

2. **Identify the Geometry**:
The polygon can be inscribed in a circle of radius \( r \). The angle subtended by each side at the center of the polygon is:
\[ \theta = \frac{2 \pi}{n} \]

3. **Calculate the Perpendicular Distance**:
For each side of the polygon, we need to find the perpendicular distance \( a \) from the center to the midpoint of the side. The angle at the center for each side is \( \frac{\theta}{2} = \frac{\pi}{n} \). Using basic trigonometry:
\[ a = r \cos\left(\frac{\pi}{n}\right) \]

4. **Magnetic Field due to One Side**:
The magnetic field \( dB \) at the center due to one side of the wire can be expressed using the Biot-Savart Law:
\[ dB = \frac{\mu_0 I}{4 \pi a} \cdot l \sin\left(\frac{\pi}{n}\right) \]
Here, \( l = \frac{2 \pi r}{n} \) and \( a = r \cos\left(\frac{\pi}{n}\right) \).

5. **Total Magnetic Field**:
Since there are \( n \) sides, the total magnetic field \( B \) at the center is:
\[ B = n \cdot dB = n \cdot \frac{\mu_0 I}{4 \pi r \cos\left(\frac{\pi}{n}\right)} \cdot \frac{2 \pi r}{n} \sin\left(\frac{\pi}{n}\right) \]
Simplifying this gives:
\[ B = \frac{\mu_0 I}{2 r} \cdot \frac{\sin\left(\frac{\pi}{n}\right)}{\cos\left(\frac{\pi}{n}\right)} \]
Using the identity \( \tan\left(\frac{\pi}{n}\right) = \frac{\sin\left(\frac{\pi}{n}\right)}{\cos\left(\frac{\pi}{n}\right)} \), we can write:
\[ B = \frac{\mu_0 I}{2 r} \tan\left(\frac{\pi}{n}\right) \]

### Part (b): Deducing the Expression for the Magnetic Field at the Centre of a Circular Current

1. **Let \( n \to \infty \)**:
As \( n \) approaches infinity, the angle \( \frac{\pi}{n} \) approaches zero. In this limit, we can use the small angle approximation:
\[ \tan\left(\frac{\pi}{n}\right) \approx \frac{\pi}{n} \]

2. **Substituting in the Expression**:
Substituting this approximation into the expression for \( B \):
\[ B = \frac{\mu_0 I}{2 r} \cdot \frac{\pi}{n} \]
As \( n \) approaches infinity, the expression simplifies to:
\[ B = \frac{\mu_0 I}{2 r} \]

Thus, the magnetic field at the center of a circular current carrying loop is:
\[ B = \frac{\mu_0 I}{2 r} \]