A current I flows through a thin wire shaped as regular polygon of n sides which can be inscribed in a circle of radius R. The magnetic fidl induction at the center of polygon due to one side of the polygon is

Options

A. `(mu_0I)/(piR) (tanpi/n)`
B. `(mu_0I)/(4piR) tanpi/n`
C. `(mu_0I)/(2piR) (tanpi/n)`
D. `(mu_0I)/(2piR) (cospi/n)`

Explanation

To find the magnetic field induction at the center of a regular polygon with \( n \) sides, inscribed in a circle of radius \( R \), due to one side of the polygon, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Geometry**:
- A regular polygon with \( n \) sides is inscribed in a circle of radius \( R \). The center of the polygon coincides with the center of the circle.

2. **Calculate the Angle Subtended by One Side**:
- The total angle around a point is \( 360^\circ \) or \( 2\pi \) radians. Therefore, the angle subtended by one side of the polygon at the center is:
\[ \theta = \frac{2\pi}{n} \]

3. **Determine the Perpendicular Distance**:
- When we bisect the angle \( \theta \), we get two angles of \( \frac{\theta}{2} = \frac{\pi}{n} \). The perpendicular distance \( D \) from the center to the midpoint of one side is given by:
\[ D = R \cos\left(\frac{\pi}{n}\right) \]

4. **Use the Formula for Magnetic Field due to a Finite Wire**:
- The magnetic field \( B \) at a distance \( D \) from a straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{4\pi D} \left( \sin \theta_1 + \sin \theta_2 \right) \]
- Here, \( \theta_1 \) and \( \theta_2 \) are the angles from the wire to the endpoints of the segment, which in this case are both \( \frac{\pi}{n} \).

5. **Substitute Values**:
- Since \( \sin \theta_1 = \sin \theta_2 = \sin\left(\frac{\pi}{n}\right) \), we have:
\[ B = \frac{\mu_0 I}{4\pi D} \left( 2 \sin\left(\frac{\pi}{n}\right) \right) \]
- Substituting \( D = R \cos\left(\frac{\pi}{n}\right) \):
\[ B = \frac{\mu_0 I}{4\pi R \cos\left(\frac{\pi}{n}\right)} \left( 2 \sin\left(\frac{\pi}{n}\right) \right) \]

6. **Simplify the Expression**:
- This simplifies to:
\[ B = \frac{\mu_0 I}{2\pi R} \cdot \frac{\sin\left(\frac{\pi}{n}\right)}{\cos\left(\frac{\pi}{n}\right)} \]
- Which can be expressed as:
\[ B = \frac{\mu_0 I}{2\pi R} \tan\left(\frac{\pi}{n}\right) \]

7. **Direction of the Magnetic Field**:
- Using the right-hand thumb rule, the direction of the magnetic field due to the current flowing in the wire will be inside the plane of the polygon.

### Final Result:
The magnetic field induction at the center of the polygon due to one side is:
\[ B = \frac{\mu_0 I}{2\pi R} \tan\left(\frac{\pi}{n}\right) \]