Question

A levitating train is three cars long (180 m) and has a mass of 100 metric tons (1 metric ton=1000 kg ). The current in the superconducting wires is about 500 kA, and even though the traditional design calls for many small coils of wire, assume for this problem that there is a 180-m-long wire carrying the current. Find the magnitude of the magnetic field needed to levitate the train.

Answer 1

`10.9\ \text{mT}`

Step-by-step explanation:

I = Current = 500 kA

L = Length of wire = 180 m

m = Mass of train =
`100* 1000=100000\ \text{kg}`

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

B = Magnetic field

The gravitational force and magnetic force will balance each other

`mg=BIL\\\Rightarrow B=(mg)/(IL)\\\Rightarrow B=(100000* 9.81)/(500* 10^3* 180)\\\Rightarrow B=0.0109\ \text{T}`

The magnitude of the magnetic field needed to levitate the train is
`10.9\ \text{mT}`