Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.90-mm-diameter superconducting wire. What current is needed?

2400A

Explanation:

magnetic field of a solenoid is given by

B=μ0NIL

and the current is

I=BLμ0N.

N is the length of the wire, divided by the diameter of the wire -

that's how many times you can wind the coils together for a given length.

Therefore,,

there are

N= 1.8/0.002 = 900 turns of the wire. Thus,

I=BLμ0N=1.5×1.84π×10-7×900= 2400 A.

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