The two wires in a DC adapter cable are 5 mm apart and carry a current of 4 A.If the cable is 1.5 m long,calculate the force between the wires and state if this force is attractive or not. A)9.6 $\times$ 10^-4 N,attractive B)9.6 $\times$ 10^-7 N,attractive C)1.1 $\times$ 10^-8 N,attractive D)9.6 $\times$ 10^-4 N,repulsive E)1.1 $\times$ 10^-8 N,repulsive

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The force between the wires can be calculated using the formula F = μ0I1I2/(2πd), where μ0 is the permeability of free space (4π × 10⁻⁷ N/A²), I1 and I2 are the currents in the wires, and d is the distance between the wires. Plugging in the given values, we get F = (4π × 10⁻⁷ × 4 × 4)/(2π × 0.005) = 9.6 × 10⁻⁴ N. The force is repulsive because the currents are flowing in the same direction and the wires are parallel, causing a magnetic field that pushes them apart. Therefore, the correct answer is D.

The Two Wires in a DC Adapter Cable Are 5 ×\times×

The Two Wires in a DC Adapter Cable Are 5 $\times$