A staright wire of length `(pi^(2))` meter is carrying a current of `2A` and the magnetic field due to it is measured at a point distant `1 cm` from it. If the wire is to be bent into a circles and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be

Options

A. `50:1`
B. `1:50`
C. `100:1`
D. `1:100`

Answer

To solve the problem, we need to calculate the magnetic field produced by a straight wire and then compare it to the magnetic field produced by a circular wire of the same length carrying the same current.

### Step-by-Step Solution:

1. **Calculate the Magnetic Field due to the Straight Wire (B1)**:
The formula for the magnetic field (B) at a distance (r) from a long straight wire carrying current (I) is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2I}{r} \]
Here, \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\), \(I = 2 \, \text{A}\), and \(r = 0.01 \, \text{m}\) (1 cm).

Substituting the values:
\[ B_1 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 2}{0.01} \]
Simplifying:
\[ B_1 = 10^{-7} \cdot \frac{4}{0.01} = 10^{-7} \cdot 400 = 4 \times 10^{-5} \, \text{T} \]

2. **Determine the Length of the Wire and Radius of the Circle**:
The length of the wire is given as \(\pi^2\) meters. When bent into a circle, the circumference (C) of the circle is equal to the length of the wire:
\[ C = 2\pi r = \pi^2 \]
Solving for the radius (r):
\[ r = \frac{\pi^2}{2\pi} = \frac{\pi}{2} \, \text{m} \]

3. **Calculate the Magnetic Field at the Center of the Circular Wire (B2)**:
The formula for the magnetic field at the center of a circular loop carrying current (I) is:
\[ B = \frac{\mu_0 I}{2r} \]
Substituting the values:
\[ B_2 = \frac{4\pi \times 10^{-7} \cdot 2}{2 \cdot \frac{\pi}{2}} = \frac{4\pi \times 10^{-7} \cdot 2}{\pi} = 8 \times 10^{-7} \, \text{T} \]

4. **Calculate the Ratio of the Magnetic Fields (B2/B1)**:
Now we find the ratio of the magnetic field at the center of the circular wire to the magnetic field due to the straight wire:
\[ \text{Ratio} = \frac{B_2}{B_1} = \frac{8 \times 10^{-7}}{4 \times 10^{-5}} = \frac{8}{400} = \frac{1}{50} \]

Thus, the ratio of the magnetic field at the center of the circular wire to that obtained in the first case is:
\[ \text{Ratio} = 1 : 50 \]