Two long straight aluminum wires, each of diameter 0.42 mm, carry the same current but in opposite directions. They are suspended by 0.50-m-long strings as shown in Fig. 28–66. If the suspension strings make an angle of 3.0° with the vertical and are hanging freely, what is the current in the wires?

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1. Identify the forces acting on the wires: The tension in the strings (T), the gravitational force (mg), and the magnetic force (Fm) due to the currents in the wires. Since the wires carry currents in opposite directions, the magnetic force will be repulsive.
2. Set up the equilibrium conditions for the forces. Since the wires are in equilibrium, the sum of the forces in the horizontal direction (x-axis) must be zero, and the sum of the forces in the vertical direction (y-axis) must also be zero.
3. Calculate the magnetic force (Fm) using the formula for the magnetic force between two parallel currents: Fm = \(\frac{\mu_0 I^2 L}{2\pi d}\), where \(\mu_0\) is the permeability of free space (4\(\pi\) x 10^-7 T·m/A), I is the current, L is the length of the wires within the magnetic field, and d is the separation distance between the wires.
4. Use trigonometry to find the separation distance (d) between the wires. Since the angle with the vertical is 3.0°, the horizontal distance (d) can be calculated as d = 2Lsin(θ), where L is the length of the string and θ is the angle.
5. Solve the equilibrium equations to find the current (I). Use the equation from step 3 and substitute the expression for d from step 4. Solve for I to find the current in the wires.