Part A A levitating train is three cars long (180 m) and has a mass of 110 metri

Question

Part A

A levitating train is three cars long (180 m) and has a mass of 110 metric tons (1 metric ton = 1000 kg). The current in the superconducting wires is about 300 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. (Express your answer to two significant figures.)

Part B

A small 40-turn current loop with a 2.00-cm diameter is suspended in a region with a magnetic field of 7000 G, with the plane of the loop parallel with the magnetic field direction.

What is the current in the loop if the torque exerted by the magnetic field on the loop is 7.00×

105 Nm? (Express your answer to two significant figures.)

Explanation / Answer

part A)

length , L = 180 m

mass , m = 110 * 1000 kg

I = 300 *1000 A

let the magnetic field is B

as F = B *I*L

m * g = B *I*L

110 * 1000 * 9.8 = B * 300 *1000 * 180

B = 0.0199 T

the magnetic field needed is 0.0199 T

part B)

N = 40

radius =2/2 = 1 cm

magntic field , B = 7000 G = 0.7 T

Now , let the current in the loop is I

Torque exerted = N *B * A * I

7 *10^-5 = 40 * 0.7 * pi * 0.01^2 * I

I = 7.9 *10^-3 A

the current in the coil is 7.9 *10^-3 A

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