Two long, straight parallel wires separated by a distance of 5.0\text{ cm} in air carry eq… — JAMB UTME

Two long, straight parallel wires separated by a distance of $5.0\text{ cm}$ in air carry equal currents in opposite directions. If the repulsive force per unit length between the wires is $1.6 \times 10^{-3}\text{ N/m}$ , determine the magnitude of the current flowing through each wire in amperes. (Take $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Answer: 20 A

Answer

The magnitude of the current flowing through each wire is

20\text{ A}

Using the parallel conductor force formula

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

, we substitute

\frac{F}{L} = 1.6 \times 10^{-3}\text{ N/m}

d = 0.05\text{ m}

, and

\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}

. Simplifying yields

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2

, giving

I^2 = 400

and

I = 20\text{ A}

Step-by-Step Solution

Recall the expression for force per unit length between two current-carrying parallel wires

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

The magnetic field generated by one wire exerts a magnetic force on the current in the adjacent wire.

Convert distance to meters and substitute all given values into the formula

d = 0.05\text{ m}

, leading to

1.6 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (0.05)}

Standard SI unit for distance is meters, necessary for dimensional consistency.

Simplify the equation and compute the current magnitude

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2 \implies I^2 = 400 \implies I = 20\text{ A}

Solving the quadratic term gives the scalar current magnitude in amperes.

Key Concept

Force per unit length between parallel current-carrying conductors

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