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: Understanding the Problem
Two parallel wires carry current \( x \) A in the same direction and are 0.20 m apart. The force of attraction per meter between the wires is \( 2 \times 10^{-6} \, \text{N} \). We need to find the value of \( x \).
Step 2: Using the Formula for Force Between Two Current-Carrying Wires
The force per unit length (\( F/L \)) between two parallel current-carrying wires is given by: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d} \] where:
\( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \) (permeability of free space),
\( I_1 = I_2 = x \) A (currents in the wires),
\( d = 0.20 \) m (distance between the wires).
Step 3: Substituting the Given Values
Substitute the values into the formula: \[ 2 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times x^2}{2 \pi \times 0.20}. \] Simplify the equation: \[ 2 \times 10^{-6} = \frac{4 \times 10^{-7} \times x^2}{0.40}. \] \[ 2 \times 10^{-6} = 10^{-7} \times x^2. \] \[ x^2 = 2. \] \[ x = \sqrt{2} \approx 1.4. \]
Step 4: Matching with the Options
The closest option to our calculated value is (C) \(1.4\). Final Answer: The value of \( x \) is approximately \(1.4\).