Question

Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of 1.35 cm, experiencing a force per unit length of \( 4.76 \times 10^{-2} \, \text{N/m} \).

Hint:

The force between two parallel wires is proportional to the product of the currents and inversely proportional to the distance between them.

Answer 1

Step 1: Formula for the force between two parallel wires.
The force per unit length \( F/L \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) is given by: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r} \] where \( \mu_0 \) is the permeability of free space, \( r \) is the distance between the wires, and \( I_1 \) and \( I_2 \) are the currents in the wires.
Step 2: Given Data.
The force per unit length is \( 4.76 \times 10^{-2} \, \text{N/m} \), the distance between the wires is \( r = 1.35 \, \text{cm} = 0.0135 \, \text{m} \), and the currents in the wires are equal, so \( I_1 = I_2 = I \).
Step 3: Solve for the current.
Substituting the known values into the formula: \[ \frac{4.76 \times 10^{-2}}{1} = \frac{(4 \pi \times 10^{-7}) I^2}{2 \pi \times 0.0135} \] Simplifying: \[ 4.76 \times 10^{-2} = \frac{2 \times 10^{-7} I^2}{0.0135} \] \[ I^2 = \frac{4.76 \times 10^{-2} \times 0.0135}{2 \times 10^{-7}} \] \[ I = 2.35 \, \text{A} \]