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 :

Solution

Physics sir · 4 steps

Both wires carry the same x, so the force goes as x²: set it equal and take a square root.

1. Force per length. Two long parallel wires carrying I₁ and I₂ at separation d attract each other with this force per metre when the currents are in the same direction.

   F/L = (μ₀ I₁ I₂)/(2π d)

2. Put in numbers. Here I₁ = I₂ = x and μ₀/2π = 2 × 10⁻⁷ T m A⁻¹. So F/L = (2 × 10⁻⁷ x²)/0.20 = 10⁻⁶ x² N m⁻¹.
3. Solve for x. Set 10⁻⁶ x² = 2 × 10⁻⁶. This gives x² = 2, so x = √2 ≈ 1.41 A.
4. So. The current in each wire is about 1.4 A, which is option (c).

The idea to keep. When both wires carry the same current, the force per metre depends on I². Always take the square root at the end, and use μ₀/2π = 2 × 10⁻⁷ to save time.

Answer from the NTA final answer key. Solution by PrepWiser.