Answer:
F = μi²l/πa
Explanation:
The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by
F = Bilsinθ
The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa
So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa
the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa
Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90
So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and
F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa
Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)
The resultant force F is thus
F = √(F₁² + F₂² + 2F₁F₂cos60°)
F = √(F₁² + F₂² + 2F₁F₂ × 0.5)
F = √(F₁² + F₂² + 2F₁F₂) (since F₁ = F₂)
F = √(2F₁² + 2F₁²) = √(4F₁²)
F = 2F₁
F = 2μi²l/2πa
F = μi²l/πa
