Question

A thin conductive bar of mass 0.071 kg and of length 1.04 m is to be levitated using a magnetic field of magnitude 0.511 T. What is the minimum current that can be passed through the bar to achieve the levitation?

Answer 1

To levitate a conductive bar using a magnetic field, the magnetic force must be equal to the gravitational force. Using the equation ILB = mg where L is the length of the bar, B is the magnetic field, and g is the acceleration due to gravity, we calculate the minimum current required for levitation to be approximately 1.35 A.

Step-by-step explanation:

The student has asked how to levitate a thin conductive bar using a magnetic field. In order to levitate the bar, the magnetic force must equal the gravitational force. The magnetic force can be described as F = ILB sin(θ), where I is the current, L is the length of the bar, and B is the magnetic field. Since the bar and the magnetic field are perpendicular to each other, θ = 90° and sin(θ) = 1. The gravitational force is Fg = mg, where m is the mass of the bar and g is the acceleration due to gravity.

Setting the magnetic force equal to the gravitational force gives us ILB = mg. Solving for I, the minimum current needed to levitate the bar is I = mg/(LB). Plugging in the values (m = 0.071 kg, L = 1.04 m, B = 0.511 T, and g = 9.81 m/s2), we find that the minimum current I is approximately 1.35 A.