A 2.4 m long wire with a weight of 0.127 N/m is suspended directly above an infinitely long straight wire. The top wire carries a current of 29 A, and the bottom wire carries a current of 37 A. The permeability of free space is \( 1.25664 \times 10^{-6} \) N/A\(^2\). Determine the distance of separation between the wires to ensure that the top wire is held in place by magnetic repulsion.

To calculate the distance of separation between the wires needed to ensure that the top wire is held in place by magnetic repulsion, we need to set up an equation where the magnetic force between the wires equals the gravitational force acting on the top wire.

The formula for the magnetic force per unit length between two long straight wires carrying currents I1 and I2 separated by a distance r is given by the Ampère's force law:

\[ F_m = \frac{\mu_0 I_1 I_2}{2 \pi r} \]

where: - \( F_m \) is the force per unit length between the wires (N/m) - \( \mu_0 \) is the permeability of free space (N/A²) - \( I_1 \) and \( I_2 \) are the currents in the wires (A) - \( r \) is the distance of separation between the wires (m)

The gravitational force per unit length acting on the top wire is given by its weight per unit length (w), which is 0.127 N/m.

To balance the gravitational force and hold the wire in place, these forces must be equal:

\[ F_m = w \] \[ \frac{\mu_0 I_1 I_2}{2 \pi r} = w \]

We can solve this equation for \( r \):

\[ r = \frac{\mu_0 I_1 I_2}{2 \pi w} \]

Let's plug in the values:

\[ I_1 = 29 \text{ A} \] \[ I_2 = 37 \text{ A} \] \[ w = 0.127 \text{ N/m} \] \[ \mu_0 = 1.25664 \times 10^{-6} \text{ N/A}^2 \]

\[ r = \frac{(1.25664 \times 10^{-6} \text{ N/A}^2)(29 \text{ A})(37 \text{ A})}{2 \pi (0.127 \text{ N/m})} \]

Now, calculate \( r \):

\[ r = \frac{(1.25664 \times 10^{-6})(29)(37)}{2 \pi (0.127)} \]

\[ r ≈ \frac{1.3568 \times 10^{-3}}{0.79796} \]

\[ r ≈ 1.701 \times 10^{-3} \text{ m} \]

So, the required distance of separation between the wires to balance the gravitational force with the magnetic repulsion is approximately 1.701 mm.

Extra: The formula used to calculate the magnetic force between two wires comes from the Biot-Savart Law, which describes how currents produce magnetic fields. Ampère's force law is a direct application of this, representing the interaction between two magnetic fields created by the currents in each wire.

The force between two parallel current-carrying wires is attractive if the currents are in the same direction, and repulsive if they are in the opposite directions. In this problem, we are working under the assumption that the currents are in opposite directions (which wasn't stated but implied), as we are using the magnetic repulsion to counteract gravity.

Permeability of free space (\( \mu_0 \)) is a constant that represents the capability of vacuum to support the formation of a magnetic field. The value of \( \mu_0 \) is key in calculating magnetic forces and fields in a vacuum or air which is considered as having similar magnetic properties.

The calculations done are significant in understanding forces between wires in electrical circuits, which is not only a matter of theoretical physics but also practical in designing electrical distribution systems. Safe distances need to be maintained to ensure forces between wires do not cause them to move undesirably, which could lead to short-circuits or other safety hazards.