Question

Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral triangle of side a, and oriented perpendicular to the triangle.
Find an expression for the magnitude of the force on each wire.

Express your answer in terms of the variables I, l, a, and appropriate constants.

Answer 1

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

Answer 2

Answer:

$F_{1}=\frac{\mu _{0}I_{1}I_{2}}{2\pi a}l+\frac{\mu _{0}I_{1}I_{3}}{2\pi a}l=\frac{\mu _{0}I_{1}}{2\pi a}l(I_{2}+I_{3})$

Explanation:

To solve this problem we can use the expression for the force between two wires with currents I1 and I2, at a distance of r.

$F_{1}=F_{2}=\frac{\mu _{0}I_{1}I_{2}}{2\pi r}L$

where L is the length of the wires. Wires feel the same force and the currents are in the same direction.

For our problem the three wires the force is the same. By taking into account that the distance among them is a we have

$F_{1}=F_{2}=F_{3}$

and the force on on of the wire is the sum of the effects of the others

$F_{1}=\frac{\mu _{0}I_{1}I_{2}}{2\pi a}l+\frac{\mu _{0}I_{1}I_{3}}{2\pi a}l=\frac{\mu _{0}I_{1}}{2\pi a}l(I_{2}+I_{3})$

where we have taken r=a (wires are equidistant in a equilateral triangle)and L=l for our problem

I hope this is useful for you

regards