Two long, straight, parallel wires carry current in the same direction. If the wires are 24 cm apart and carry currents of 2.0 A and 4.0 A, respectively, find the force per unit length on each wire?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/wirfor.html
The above site will provide the formula
To find the force per unit length on each wire, we can use the formula for the magnetic force between two parallel current-carrying wires:
F = (μ₀ * I₁ * I₂ * L) / (2πd)
where F is the force between the wires, μ₀ is the permeability of free space (4π × 10^-7 T m/A), I₁ and I₂ are the currents in the wires, L is the length of the wire segment, and d is the distance between the wires.
In this case, we need to calculate the force per unit length, so we divide the above formula by the length (L) of the wire segment:
F/l = (μ₀ * I₁ * I₂) / (2πd)
Given that the wires are 24 cm (or 0.24 m) apart and carry currents of 2.0 A and 4.0 A, respectively, we can substitute these values into the formula:
F/l = (4π × 10^-7 T m/A) * (2.0 A) * (4.0 A) / (2π * 0.24 m)
Simplifying the equation:
F/l = (8π × 10^-7 T m²) / (2π * 0.24 m)
We can cancel out the π and m in the denominator and multiply 8 and 10^-7:
F/l = (8 * 10^-7 T m) / (0.48 m)
Finally, we can divide 8 * 10^-7 by 0.48 to find the force per unit length on each wire:
F/l ≈ 1.67 × 10^-6 N/m
Therefore, the force per unit length on each wire is approximately 1.67 x 10^-6 Newtons per meter.