Two long wires are placed parallel to each other with a distance 1 m between them. The wires carry equal current 1 A in the same direction. The magnetic field intensity at a point P which is equidistant from the wires is `1.6xx10^(-7)T`. Then, the distance of this point from any one of the wires is _________ m.
[Permeability of vacuum, `mu_(0)=4pixx10^(-7)T-m//A`]

Explanation

To solve the problem, we need to find the distance \( x \) from point P to either of the wires. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have two long parallel wires separated by a distance of 1 meter, each carrying a current of 1 A in the same direction. The point P is equidistant from both wires.

### Step 2: Define the Distance
Let the distance from point P to either wire be \( x \). Since the total distance between the two wires is 1 meter, we can express this as:
\[ x + x = 1 \quad \text{or} \quad 2x = 1 \]
Thus, \( x = 0.5 \) m. However, we need to find the magnetic field intensity at point P.

### Step 3: Magnetic Field Due to a Long Straight Wire
The magnetic field intensity \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 4: Calculate the Total Magnetic Field at Point P
Since point P is equidistant from both wires, the magnetic field contributions from both wires at point P will add up. Therefore, the total magnetic field \( B \) at point P is:
\[ B = B_1 + B_2 = \frac{\mu_0 I}{2 \pi x} + \frac{\mu_0 I}{2 \pi x} = \frac{2 \mu_0 I}{2 \pi x} = \frac{\mu_0 I}{\pi x} \]

### Step 5: Substitute the Given Values
We know:
- \( B = 1.6 \times 10^{-7} \, T \)
- \( I = 1 \, A \)
- \( \mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A \)

Substituting these values into the equation:
\[ 1.6 \times 10^{-7} = \frac{(4 \pi \times 10^{-7}) \cdot 1}{\pi x} \]

### Step 6: Simplify the Equation
The \( \pi \) cancels out:
\[ 1.6 \times 10^{-7} = \frac{4 \times 10^{-7}}{x} \]

### Step 7: Solve for \( x \)
Rearranging gives:
\[ x = \frac{4 \times 10^{-7}}{1.6 \times 10^{-7}} \]
Calculating this:
\[ x = \frac{4}{1.6} = 2.5 \, m \]

### Final Answer
The distance of point P from either of the wires is:
\[ \boxed{2.5 \, m} \]