In a Magnetic Resonance Imaging machine, there are several superconducting solenoid coils that provide the uniform magnetic field inside the bore. Let's look at just one of them and ask why they are made from superconducting material rather than ordinary copper wire such as used in your house.
Assume that a coil is made with 2000 windings of current-carrying wire. The bore has a radius of 0.5 m, and the coil is 0.3 m long.
(a) What current, in amperes, is flowing in the wire to get a magnetic field of 2.0 T at the center of the solenoid?
(b) If we assume the wire is made of copper and is not a superconductor, we can calculate the resistance of the wire to carrying that much current. The resistance R (measured in units of Ohms) can be found if we know the resistivity ρ (measured in units of Ohms times meters), the cross-sectional area of the wire a, and the length of the wire L: R = ρ * L/a. Assume that the wire is 1 mm^2 and the resistivity is 16.78 × 10^-9 Ohm * m. What is the electrical resistance of the coil?
(c) Ohm's law states that the voltage applied to a wire is equal to the current times the resistance: V = I * R. What voltage would be required to drive this much current through the copper wire?
(d) The power required to do this is P (measured in watts) and is the current times the voltage, P = V * I. What power is required to drive this current? How many 60-watt light bulbs is this equivalent to?
(e) Now assume that the magnetic coil is made of superconducting wire such as niobium-titanium. Now that the resistivity ρ = 0, what is the 1. current, 2. voltage, and 3. power used now?