A long horizontal wire carries a current of 48A . A second wire, made of 2.6mm diameter copper wire and parallel to the first, is kept in suspension magnetically 15cm below(Figure 1) .

Determine the magnitude of the current in the lower wire.

To determine the magnitude of the current in the lower wire, we can use Ampere's law.

Ampere's law states that the magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

First, let's find the magnetic field produced by the current in the long horizontal wire at a distance of 15 cm below it. We can use the formula for the magnetic field around a straight current-carrying wire:

B = (μ₀ * I) / (2π * r)

Where:
B is the magnetic field
μ₀ is the permeability of free space (constant value: 4π * 10^(-7) T·m/A)
I is the current
r is the distance from the wire

Plugging in the given values:
I = 48 A
r = 15 cm = 0.15 m

B = (4π * 10^(-7) T·m/A * 48 A) / (2π * 0.15 m)
B ≈ 0.512 T (Tesla)

Now, the magnetic field created by the upper wire is acting downwards. In order to suspend the lower wire magnetically, the magnetic field due to the lower wire should be equal in magnitude but in the opposite direction.

Next, we can find the magnetic field produced by the lower wire at the location of the upper wire. Again, we can use the same formula:

B = (μ₀ * I) / (2π * r)

Where:
B is the magnetic field
μ₀ is the permeability of free space (constant value: 4π * 10^(-7) T·m/A)
I is the current (which is what we are trying to find)
r is the distance from the wire, which is the same as before: 0.15 m

Setting the two magnetic field equations equal to each other and solving for I, we have:

0.512 T = (4π * 10^(-7) T·m/A * I) / (2π * 0.15 m)

Simplifying the equation:

0.512 T = (2 * 10^(-7) T·m/A * I) / (0.15 m)

Cross-multiplying and solving for I:

0.512 T * 0.15 m = 2 * 10^(-7) T·m/A * I

0.0768 T·m = 2 * 10^(-7) T·m/A * I

I ≈ (0.0768 T·m) / (2 * 10^(-7) T·m/A)

I ≈ 384 A

Therefore, the magnitude of the current in the lower wire is approximately 384 A.