1. A 30 V battery is connected to a resistor that is rated at 250 with a mass of

Question

1. A 30 V battery is connected to a resistor that is rated at 250 with a mass of 0.5 grams and a length of 1 cm. What magnetic field strength will cause this resistor to "levitate" in the Earth's gravitational field? Neglect the weight of any connections between the resistor and the battery. 2. A levitating train is three cars long (180 m) and has a mass of 100 metric tons (1 metric ton = 1000 kg). The current that runs in the superconducting wires is about 500 kiloamps 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 this current. Find the size of the magnetic field needed to levitate the train. 1. A 30 V battery is connected to a resistor that is rated at 250 with a mass of 0.5 grams and a length of 1 cm. What magnetic field strength will cause this resistor to "levitate" in the Earth's gravitational field? Neglect the weight of any connections between the resistor and the battery. 2. A levitating train is three cars long (180 m) and has a mass of 100 metric tons (1 metric ton = 1000 kg). The current that runs in the superconducting wires is about 500 kiloamps 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 this current. Find the size of the magnetic field needed to levitate the train. 1. A 30 V battery is connected to a resistor that is rated at 250 with a mass of 0.5 grams and a length of 1 cm. What magnetic field strength will cause this resistor to "levitate" in the Earth's gravitational field? Neglect the weight of any connections between the resistor and the battery. 2. A levitating train is three cars long (180 m) and has a mass of 100 metric tons (1 metric ton = 1000 kg). The current that runs in the superconducting wires is about 500 kiloamps 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 this current. Find the size of the magnetic field needed to levitate the train.

Explanation / Answer

force due to magnetic field F= B i L

force on wire F = mg

i = V/ R = 30/250=0.12 A

BiL=mg

B=(mg)/(iL)

B=(0.5*10-3 kg * 9.8 ) / (0.12*0.01m)

B=4.083 T

2)

B=(mg)/(iL)

B=(100*1000*9.8)/(500*1000*180)

B=0.010 T

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