Question

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 \( 2 \times 10^{-6} \, \text{N} \), then the value of \( x \) is approximately:

• A. \(1\)
• B. \(2.4\)
• C. \(1.4\)
• D. \(2\)

Hint:

The force between two parallel current-carrying wires is attractive if the currents are in the same direction and repulsive if they are in opposite directions. The force per unit length is directly proportional to the product of the currents and inversely proportional to the distance between the wires.

Answer 1

The Correct Option is C

Step 1: Problem Definition
Two parallel wires, each carrying a current of \( x \) A in the same direction, are separated by a distance of 0.20 m. The calculated force of attraction per meter between these wires is \( 2 \times 10^{-6} \, \text{N} \). The objective is to determine the value of \( x \).
Step 2: Applicable Formula
The formula for the force per unit length (\( F/L \)) between two parallel current-carrying wires is:\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}\]Key parameters:
\( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \) (permeability of free space),
\( I_1 = I_2 = x \) A (current magnitudes in both wires),
\( d = 0.20 \) m (inter-wire distance).
Step 3: Calculation and Simplification
Substituting the given values into the formula yields:\[2 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times x^2}{2 \pi \times 0.20}.\]Simplifying the expression leads to:\[2 \times 10^{-6} = \frac{4 \times 10^{-7} \times x^2}{0.40}.\]\[2 \times 10^{-6} = 10^{-7} \times x^2.\]Solving for \( x^2 \):\[x^2 = 2.\]Therefore, \( x = \sqrt{2} \approx 1.4 \).
Step 4: Option Comparison
The calculated value of \( x \) is approximately \(1.4\), which corresponds to option (C).Final Answer: The determined value for \( x \) is approximately \(1.4\).