Answer:
F/L = μ₀I₁I₂/2πr
when a force of 2 x 10⁻⁷ N is produced between two conductors 1 m apart from each other and having a cross-sectional radius 1 m. The current on each conductor in this situation is 1 Ampere
Explanation:
Consider two conductors of same length parallel to each other. The Magnetic field due to current in 1st conductor is given by Ampere's Law as:
B₁ = μ₀I₁/2πr
where,
B₁ = Magnetic Field due to Conductor 1
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I₁ = Current through conductor 1
r = radius of cross section of conductor
Now, the force of the conductors on each other is given by:
F₁₂ = B₁I₂L
where,
F₁₂ = Force of conductor 1 on conductor 2
B₁ = Magnetic Field of Conductor 1
I₂ = Current through conductor 2
L = Length
Therefore,
F₁₂ = μ₀I₁I₂L/2πr
The force of conductor 2 on conductor will also have same magnitude but opposite direction:
F₁₂ = |- F₂₁| = F
<u>F/L = μ₀I₁I₂/2πr</u>
F/L = μ₀I₁I₂L/2πr
Now, to define S.I unit of current (Ampere) we substitute values:
F/(1 m) = (4π x 10⁻⁷ N/A²)(1 A)(1 A)/2π(1 m)
F = 2 x 10⁻⁷ N
<u>So, when a force of 2 x 10⁻⁷ N is produced between two conductors 1 m apart from each other and having a cross-sectional radius 1 m. The current on each conductor in this situation is 1 Ampere </u>
