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 × 10^–6 N, then the value of x is approximately:

• A. 1
• B. 2.4
• C. 1.4
• D. 2

Answer 1

The Correct Option is C

To solve this question, we will use the formula for the magnetic force per unit length between two parallel current-carrying wires. The formula is given by:

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

where:

• \(F/L\) is the force per unit length between the wires (given as \(2 \times 10^{-6} \, \text{N/m}\))
• \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7} \, \text{Tm/A}\)
• \(I_1\) and \(I_2\) are the currents in the wires
• \(d\) is the distance between the wires (given as 0.20 m)

Since both wires carry the same current \(I\), the equation becomes:

\[\frac{2 \times 10^{-6}}{1} = \frac{(4\pi \times 10^{-7}) \cdot I^2}{2\pi \cdot 0.20}\]

We can simplify this equation to solve for \(I\):

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

Multiplying both sides by \(0.20\) gives:

\[(2 \times 10^{-6}) \times 0.20 = 2 \times 10^{-7} \cdot I^2\]

Simplifying, we get:

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

Divide both sides by \(2 \times 10^{-7}\):

\[2 = I^2\]

Taking the square root of both sides yields:

\[I = \sqrt{2} \approx 1.414 \, \text{A}\]

Therefore, the current \(I\) is approximately 1.4 A. Hence, the correct answer is 1.4.