An aluminum wire with mass/unit length of p = 5*10^-4 kg/m lies on the ground, running east-west. The wire feels the earth's magnetic field, which has magnitude B = 5*10^-5 T. Assume the wire is in Columbia, where the Earth's magnetic field runs due north, parallel to ground. A current flows in the wire, in the west-to-east direction as shown. How large would the current have to be to lift the cable off the ground? Recall, g=9.8 m/s^2.
To understand how to calculate the current required to lift the cable off the ground, we need to consider the force exerted on the wire due to the interaction between the magnetic field and the current flowing through the wire.
The force experienced by a current-carrying wire in a magnetic field is given by the equation:
πΉ = πΌπ΅πΏsinΞΈ
where πΉ is the force, πΌ is the current, π΅ is the magnetic field, πΏ is the length of the wire, and ΞΈ is the angle between the direction of the current and the magnetic field.
In this scenario, the wire is lying on the ground and running east-west, while the Earth's magnetic field runs due north. Since the wire is parallel to the ground, the angle ΞΈ between the current and the magnetic field is 90 degrees.
The force required to lift the wire off the ground is equal to the weight of the wire, which can be calculated using the equation:
π = πππΏ
where π is the weight, π is the mass per unit length of the wire, π is the acceleration due to gravity, and πΏ is the length of the wire.
To determine the current required to lift the wire, we equate the force due to the magnetic field to the weight of the wire:
πΌπ΅πΏsinΞΈ = πππΏ
Simplifying the equation, we get:
πΌ = (ππ) / (π΅sinΞΈ)
Substituting the given values into the equation:
πΌ = ((5*10^-4 kg/m) * 9.8 m/s^2) / ((5*10^-5 T) * sin(90Β°))
Simplifying further:
πΌ = (4.9*10^-3 N) / (5*10^-5 T)
πΌ β 98 A
Therefore, the current required to lift the cable off the ground would be approximately 98 Amperes.