Two parallel wires which carry currents I1 = 69 A and I2. The top wire is held in position, the bottom wire is prevented from moving sideways but can slide up and down without friction. If the wires have a mass of 5.3 g per metre of length calculate current I2 such that the lower wire levitates at a position 4.8 cm below the top wire.

To determine the current I2 required for the lower wire to levitate at a position 4.8 cm below the top wire, we can use the principle of magnetic force between two parallel wires.

The magnetic force between two parallel wires is given by the equation:

F = (μ₀ * I1 * I2 * L) / (2π * d)

Where:
F is the magnetic force between the wires,
μ₀ is the permeability of free space (4π × 10^-7 T·m/A),
I1 is the current in the top wire,
I2 is the current in the bottom wire,
L is the length of the wires,
and d is the distance between the wires.

In this case, the wires are parallel, and the magnetic force is responsible for levitating the lower wire.

Given:
I1 = 69 A
L = 1 m (length of wire)
m = 5.3 g = 5.3 × 10^-3 kg (mass per meter of wire)
d = 4.8 cm = 0.048 m (distance between the wires)

First, we can calculate the weight of the lower wire using its mass per meter:

Weight = m * g * L

Where:
g is the acceleration due to gravity (9.81 m/s²)

Weight = (5.3 × 10^-3 kg) * (9.81 m/s²) * (1 m) = 5.3 × 10^-2 N

To achieve levitation, the magnetic force should balance the weight. Therefore, we have:

F = Weight

Substituting the given values:

(μ₀ * I1 * I2 * L) / (2π * d) = 5.3 × 10^-2 N

Rearranging the equation, we can solve for I2:

I2 = (5.3 × 10^-2 N * 2π * d) / (μ₀ * I1 * L)

Substituting the given values:

I2 = (5.3 × 10^-2 N * 2π * 0.048 m) / (4π × 10^-7 T·m/A * 69 A * 1 m)

Simplifying:

I2 = (5.3 × 10^-2 N * 0.096π m) / (276π × 10^-6 T·m²/A) = (5.3 × 0.096) / (276 × 10^-6) A

I2 ≈ 0.183 A

Therefore, the current I2 required for the lower wire to levitate at a position 4.8 cm below the top wire is approximately 0.183 A.

To solve this problem, we will use the concept of magnetic forces between current-carrying wires. The magnetic force between two parallel wires is given by the equation:

F = (𝜇₀ * I₁ * I₂ * L)/(2πd)

Where:
F is the magnetic force between the wires,
𝜇₀ is the permeability of free space (4π × 10⁻⁷ T*m/A),
I₁ and I₂ are the currents in the first and second wires respectively,
L is the length of the wires, and
d is the distance between the wires.

In this case, we are given I₁ = 69 A, L = 1 m, and d = 4.8 cm = 0.048 m. We need to find I₂, the current flowing through the lower wire such that it levitates at a position 4.8 cm below the top wire.

First, let's calculate the magnetic force on the lower wire that opposes its weight. For the wire to levitate, the magnetic force should be equal and opposite to the weight of the wire. The weight of the wire is given by:

W = m * g

Where m is the mass per unit length of the wire and g is the acceleration due to gravity. In this case, m = 5.3 g/m = 5.3 × 10⁻³ kg/m and g = 9.8 m/s².

So, W = (5.3 × 10⁻³ kg/m) * (9.8 m/s²) = 5.194 × 10⁻² N/m.

Now, we can equate the magnetic force and the weight of the wire:

F = W

Substituting the values, we get:

(𝜇₀ * I₁ * I₂ * L)/(2πd) = 5.194 × 10⁻² N/m

Simplifying the equation:

I₂ = (2πd * W)/(𝜇₀ * I₁ * L)

Substituting the given values:

I₂ = (2π * 0.048 m * 5.194 × 10⁻² N/m) / (4π × 10⁻⁷ T*m/A * 69 A * 1 m)

Simplifying further:

I₂ = 0.002748 A ≈ 2.748 A

Therefore, the current I₂ flowing through the lower wire should be approximately 2.748 A for it to levitate at a position 4.8 cm below the top wire.