Two parallel wires are separated by 5.95 cm, each carrying 2.5 A of current in the same direction.What is the magnitude of the force per unit length between the wires? (unit of N/m)
Question
Two parallel wires are separated by 5.95 cm, each carrying 2.5 A of current in the same direction.What is the magnitude of the force per unit length between the wires? (unit of N/m)
Solution
The force per unit length (F/L) between two parallel wires carrying current can be calculated using Ampere's law, which states:
F/L = μ0 * I1 * I2 / (2πr)
where:
- μ0 is the permeability of free space, which is 4π x 10^-7 T m/A
- I1 and I2 are the currents in the two wires
- r is the distance between the wires
Given:
- I1 = I2 = 2.5 A (since both wires carry the same current)
- r = 5.95 cm = 0.0595 m (converted from cm to m)
Substituting these values into the formula, we get:
F/L = 4π x 10^-7 T m/A * (2.5 A)^2 / (2π * 0.0595 m)
Solving this equation will give us the force per unit length between the wires.
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