A straight wire of length `pi//2` meter, is bent into a circular shape. If the wire were to carry a current of `5A`, calculate the magnetic field, due to it, before bending, at a point distance `0*01` times the radius of the circle formed from it. Also calculate the magnetic field, at the centre of the circular loop formed, for the same value of current.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Radius of the Circle
Given that the length of the wire is \( \frac{\pi}{2} \) meters, this length represents the circumference of the circular shape formed by the wire. The formula for the circumference \( C \) of a circle is given by:

\[ C = 2\pi r \]

Setting \( C = \frac{\pi}{2} \):

\[ \frac{\pi}{2} = 2\pi r \]

Now, solve for \( r \):

\[ r = \frac{\frac{\pi}{2}}{2\pi} = \frac{1}{4} \text{ meters} = 0.25 \text{ meters} \]

### Step 2: Calculate the Magnetic Field at a Point Outside the Wire
We need to calculate the magnetic field at a point that is \( 0.01 \) times the radius of the circle from the wire. First, calculate the distance from the center of the circle to this point:

\[ d = 0.01 \times r = 0.01 \times 0.25 = 0.0025 \text{ meters} \]

The magnetic field \( B \) at a distance \( d \) from a long straight wire carrying current \( I \) is given by:

\[ B = \frac{\mu_0 I}{2\pi d} \]

Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (the permeability of free space). Substituting the values:

\[ B = \frac{(4\pi \times 10^{-7}) \times 5}{2\pi \times 0.0025} \]

Simplifying this:

\[ B = \frac{(4 \times 10^{-7}) \times 5}{2 \times 0.0025} = \frac{20 \times 10^{-7}}{0.005} = \frac{20 \times 10^{-7}}{5 \times 10^{-3}} = 4 \times 10^{-4} \text{ Tesla} \]

### Step 3: Calculate the Magnetic Field at the Center of the Circular Loop
The magnetic field at the center of a circular loop carrying current \( I \) is given by:

\[ B = \frac{\mu_0 I}{4r} \]

Substituting the values:

\[ B = \frac{(4\pi \times 10^{-7}) \times 5}{4 \times 0.25} \]

Simplifying this:

\[ B = \frac{(4\pi \times 10^{-7}) \times 5}{1} = 20\pi \times 10^{-7} \text{ Tesla} \]

Calculating \( 20\pi \):

\[ B \approx 20 \times 3.14 \times 10^{-7} = 62.8 \times 10^{-7} \text{ Tesla} = 6.28 \times 10^{-6} \text{ Tesla} \]

### Final Answers
1. The magnetic field at a point \( 0.01 \) times the radius from the wire before bending is \( 4 \times 10^{-4} \) Tesla.
2. The magnetic field at the center of the circular loop is approximately \( 6.28 \times 10^{-6} \) Tesla.