Magnets carrying very large currentsare used to produce the uniform, large-magnitude magnetic fieldsthat are required for magnetic resonance imaging (MRI). A typicalMRI magnet may be a solenoid that is 2.0 m long and 1.0 m in diameter,has a self-inductance of 4.4 H, and carries a current of 750 A. Anormal wire carrying that much current would dissipate a great dealof electrical power as heat, so most MRI magnets are made with coils of superconducting wire cooled by liquid helium at a temperature justunder its boiling point (4.2 K). After a current is established in the wire,the power supply is disconnected and the magnet leads are shorted togetherthrough a piece of superconductor so that the current flows withoutresistance as long as the liquid helium keeps the magnet cold.Under rare circumstances, a small segment of the magnet’s wiremay lose its superconducting properties and develop resistance. In thissegment, electrical energy is converted to thermal energy, which canboil off some of the liquid helium. More of the wire then warms upand loses its superconducting properties, thus dissipating even moreenergy as heat. Because the latent heat of vaporization of liquid heliumis quite low (20.9 kJ>kg), once the wire begins to warm up, all ofthe liquid helium may boil off rapidly. This event, called a quench, candamage the magnet. Also, a large volume of helium gas is generatedas the liquid helium boils off, causing an asphyxiation hazard, and theresulting rapid pressure buildup can lead to an explosion. You can seehow important it is to keep the wire resistance in an MRI magnet atzero and to have devices that detect a quench and shut down the currentimmediately. If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnet nonsuperconducting, how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter because the resistance will increase; (b) the time will be longer because the resistance will increase; (c) the time will be the same; (d) not enough information is given.
Magnets carrying very large currents
are used to produce the uniform, large-magnitude magnetic fields
that are required for magnetic resonance imaging (MRI). A typical
MRI magnet may be a solenoid that is 2.0 m long and 1.0 m in diameter,
has a self-inductance of 4.4 H, and carries a current of 750 A. A
normal wire carrying that much current would dissipate a great deal
of electrical power as heat, so most MRI magnets are made with coils of superconducting wire cooled by liquid helium at a temperature just
under its boiling point (4.2 K). After a current is established in the wire,
the power supply is disconnected and the magnet leads are shorted together
through a piece of superconductor so that the current flows without
resistance as long as the liquid helium keeps the magnet cold.
Under rare circumstances, a small segment of the magnet’s wire
may lose its superconducting properties and develop resistance. In this
segment, electrical energy is converted to thermal energy, which can
boil off some of the liquid helium. More of the wire then warms up
and loses its superconducting properties, thus dissipating even more
energy as heat. Because the latent heat of vaporization of liquid helium
is quite low (20.9 kJ>kg), once the wire begins to warm up, all of
the liquid helium may boil off rapidly. This event, called a quench, can
damage the magnet. Also, a large volume of helium gas is generated
as the liquid helium boils off, causing an asphyxiation hazard, and the
resulting rapid pressure buildup can lead to an explosion. You can see
how important it is to keep the wire resistance in an MRI magnet at
zero and to have devices that detect a quench and shut down the current
immediately. If part of the magnet develops resistance and liquid helium boils
away, rendering more and more of the magnet nonsuperconducting,
how will this quench affect the time for the current to drop to half of its
initial value? (a) The time will be shorter because the resistance will increase;
(b) the time will be longer because the resistance will increase;
(c) the time will be the same; (d) not enough information is given.
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