Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is `2xx10^(-6)` N, then the value of x is approximately :

Options

A. 1
B. 2.4
C. 1.4
D. 2

Explanation

To solve the problem, we will use the formula for the force per unit length between two parallel current-carrying wires. The force per unit length \( F/L \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( d \) is given by:

\[ \frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d} \]

Where:
- \( \mu_0 \) is the permeability of free space, which is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).
- \( I_1 \) and \( I_2 \) are the currents in the wires.
- \( d \) is the distance between the wires.

Given that both wires carry the same current \( x \) A, we can rewrite the equation as:

\[ \frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{x^2}{d} \]

We are given:
- \( \frac{F}{L} = 2 \times 10^{-6} \, \text{N/m} \)
- \( d = 0.20 \, \text{m} \)

Now, substituting the values into the equation:

\[ 2 \times 10^{-6} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{x^2}{0.20} \]

Simplifying the left side:

\[ 2 \times 10^{-6} = 2 \times 10^{-7} \cdot \frac{x^2}{0.20} \]

Now, multiply both sides by \( 0.20 \):

\[ 0.20 \cdot 2 \times 10^{-6} = 2 \times 10^{-7} x^2 \]

Calculating the left side:

\[ 4 \times 10^{-7} = 2 \times 10^{-7} x^2 \]

Now, divide both sides by \( 2 \times 10^{-7} \):

\[ \frac{4 \times 10^{-7}}{2 \times 10^{-7}} = x^2 \]

This simplifies to:

\[ 2 = x^2 \]

Taking the square root of both sides:

\[ x = \sqrt{2} \]

Calculating \( \sqrt{2} \):

\[ x \approx 1.41 \, \text{A} \]

Thus, the value of \( x \) is approximately \( 1.41 \, \text{A} \).