Four identical wires of length L each are bent into four different planar figures: an equilateral triangle, a square, a regular hexagon and a circle. The loops are kept far away from each other and made to carry equal current I. The magnetic field intensity at the centre of the figures are recorded as `B_(T),B_(S),B_(H)` and `B_(C)`. Choose the correct option(s):

Options

A. `B_(T)=(27mu_(0)I)/(2piL)`
B. `B_(H)=(6 sqrt(3)mu_(0)I)/(piL)`
C. `B_(S)=(8sqrt(2)mu_(0)I)/(piL)`
D. `B_(T)gtB_(S)gtB_(H)gtB_(C)`

Explanation

To solve the problem, we need to calculate the magnetic field intensity at the center of each of the four shapes formed by bending the wires: an equilateral triangle, a square, a regular hexagon, and a circle.

### Step-by-step Solution:

1. **Identify the Length of Each Side:**
Each wire has a total length \( L \). For each shape, we need to determine the length of each side:
- **Equilateral Triangle:** Each side \( = \frac{L}{3} \)
- **Square:** Each side \( = \frac{L}{4} \)
- **Regular Hexagon:** Each side \( = \frac{L}{6} \)
- **Circle:** The circumference \( = L \), so the radius \( R = \frac{L}{2\pi} \)

2. **Calculate the Magnetic Field for Each Shape:**
The magnetic field \( B \) at the center of a loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{4\pi R} \cdot \text{(sum of contributions from each segment)} \]

- **For the Equilateral Triangle:**
Each segment contributes equally. The distance from the center to each side is \( \frac{L}{6\sqrt{3}} \). The angle subtended by each side at the center is \( 60^\circ \).
\[ B_T = \frac{3 \cdot \mu_0 I}{4\pi \left(\frac{L}{6\sqrt{3}}\right)} \cdot \sin(60^\circ) = \frac{27 \mu_0 I}{2\pi L} \]

- **For the Square:**
Each side contributes equally. The distance from the center to each side is \( \frac{L}{8} \). The angle subtended by each side at the center is \( 90^\circ \).
\[ B_S = \frac{4 \cdot \mu_0 I}{4\pi \left(\frac{L}{8}\right)} \cdot \sin(45^\circ) = \frac{8\sqrt{2} \mu_0 I}{\pi L} \]

- **For the Regular Hexagon:**
Each side contributes equally. The distance from the center to each side is \( \frac{L}{12} \). The angle subtended by each side at the center is \( 60^\circ \).
\[ B_H = \frac{6 \cdot \mu_0 I}{4\pi \left(\frac{L}{12}\right)} \cdot \sin(30^\circ) = \frac{6\sqrt{3} \mu_0 I}{\pi L} \]

- **For the Circle:**
The magnetic field at the center of a circular loop is given by:
\[ B_C = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2 \cdot \frac{L}{2\pi}} = \frac{\mu_0 I \pi}{L} \]

3. **Comparing the Magnetic Fields:**
Now we compare the magnitudes of the magnetic fields:
- \( B_T = \frac{27 \mu_0 I}{2\pi L} \)
- \( B_S = \frac{8\sqrt{2} \mu_0 I}{\pi L} \)
- \( B_H = \frac{6\sqrt{3} \mu_0 I}{\pi L} \)
- \( B_C = \frac{\mu_0 I \pi}{L} \)

### Conclusion:
After calculating the magnetic fields, we can rank them based on their values:
- \( B_T > B_S > B_H > B_C \)