Final answer:
To levitate a conductive bar in a magnetic field, the magnetic force on the bar must equal the gravitational force. By setting these forces equal and solving for current, the minimum current needed is found to be approximately 56.8 mA for the given parameters.
Step-by-step explanation:
To determine the minimum current that must be passed through a thin conductive bar to levitate it in a magnetic field, we must equate the magnetic force with the gravitational force acting on the bar.
The magnetic force Fm on a current-carrying conductor in a magnetic field is given by Fm = I * L * B * sin(θ), where I is the current, L is the length of the conductor, B is the magnetic field strength, and θ is the angle between the conductor and the magnetic field. In this case, the bar is perpendicular to the magnetic field, so θ = 90° (or π/2 radians), and sin(θ) = 1, simplifying the formula to Fm = I * L * B.
Gravitational force Fg on the bar is given by Fg = m * g, where m is the mass and g is the acceleration due to gravity (approximated as 9.81 m/s2).
For levitation, these two forces must be equal: Fm = Fg. Thus, I * L * B = m * g, and solving for I gives us: I = (m * g) / (L * B).
Plugging in the given values, m = 0.067 kg, L = 2.45 m, B = 0.477 T, and g = 9.81 m/s2, we get:
I = (0.067 kg * 9.81 m/s2) / (2.45 m * 0.477 T) = 0.0568 A
Therefore, the minimum current that needs to be passed through the bar for levitation is approximately 0.0568 A, or 56.8 mA.