Question

A levitating train is three cars long (210 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 210-m-long, straight wire carrying the current beneath the train. A perpendicular magnetic field on the track levitates the train.

Find the magnitude of the magnetic field needed to levitate the train.

= _____T

Answer 1

The magnitude of the magnetic field needed to levitate a three-car long train with a mass of 100 metric tons and a current of 500 kA in the superconducting wires is approximately 0.0093 teslas (T).

Step-by-step explanation:

To find the magnitude of the magnetic field needed to levitate the train, we first need to determine the force that the magnetic field must exert to counteract gravity and keep the train afloat.

This force is equivalent to the weight of the train, which can be calculated using the formula F = m × g, where m is the mass and g is the acceleration due to gravity (9.8 m/s2).

The mass of the train is given as 100 metric tons, which is 100,000 kg (since 1 metric ton = 1000 kg).

The weight of the train is thus F = 100,000 kg × 9.8 m/s2 = 980,000 N.

This force must be provided by the magnetic force from the interaction of the current in the wire and the magnetic field, according to the formula F = I × L × B, { where I is the current, L is the length of the conductor within the magnetic field, and B is the magnetic field strength }.

We already have the current (I = 500 kA = 500,000 A) and the length of the wire (L = 210 m), which is the same as the length of the train.

Solve for B: 980,000 N = 500,000 A × 210 m × B

B = 980,000 N / (500,000 A × 210 m)

B ≈ 0.0093 T

The magnitude of the magnetic field needed to levitate the train is approximately 0.0093 teslas (T).