Two horizontal parallel wires which carry currents I1 = 45 A (top)and I2 (bottom). 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 9.2 g per metre of length, calculate current I2 such that the lower wire levitates at a position 5.6 cm below the top wire.
To solve this problem, we can use the principle of magnetic levitation. The magnetic field generated by the current in the top wire interacts with the current in the bottom wire to create an upward magnetic force that balances the weight of the bottom wire, causing it to levitate.
Let's break down the steps to find the current I2:
Step 1: Calculate the magnetic field created by the top wire at the position of the bottom wire.
The magnetic field due to a long, straight wire at a distance r from the wire is given by the formula:
B = (μ₀ * I) / (2πr)
where B is the magnetic field, μ₀ is the permeability of free space (4π * 10^(-7) T·m/A), I is the current, and r is the distance from the wire.
In this case, the distance between the wires is 5.6 cm, which is equal to 0.056 m. The current in the top wire is given as I1 = 45 A. Using the formula, we can calculate the magnetic field at the position of the bottom wire.
B = (4π * 10^(-7) T·m/A * 45 A) / (2π * 0.056 m)
B = 0.045 / 0.112
B = 0.4017 T (to four significant figures)
Step 2: Calculate the magnetic force on the bottom wire.
The magnetic force felt by a current-carrying wire in a magnetic field is given by the formula:
F = BIL
where F is the force, B is the magnetic field, I is the current, and L is the length of the wire in the magnetic field.
In this case, the length of the wires is not given, but we are given that the mass of the wires is 9.2 g per metre of length. We can assume a unit length of 1 meter. So the force on 1 meter of wire is equal to its weight.
F = mg
where F is the force (weight) of the wire, m is the mass of the wire per unit length, and g is the acceleration due to gravity (9.8 m/s²).
In this case, the force is equal to the weight of the wire, as it is levitating.
F = (9.2 g/m * 1 m) * 9.8 m/s²
F = 90.16 x 10^(-3) kg * 9.8 m/s²
F = 0.8821 N (to four significant figures)
Step 3: Equate the magnetic force on the bottom wire to the weight of the wire and solve for I2.
F = BIL
0.8821 N = 0.4017 T * I2 * 1 m
I2 = (0.8821 N) / (0.4017 T)
I2 ≈ 2.194 A (to three significant figures)
Therefore, the current I2 that would cause the lower wire to levitate at a position 5.6 cm below the top wire is approximately 2.194 A.