Magnetic resonance imaging often need magnetic fields of a strength of around 1.50 T. The solenoid is 1.80 meters long and 75.0 cm in diameter. It is tightly wound with a single layer of 2.00 mm diameter superconducting wire.

Required:

What current is needed?

Answer:The current needed is

2387.32 AExplanation:Given;

strength of the magnetic field, B = 1.5 T

length of the solenoid, L = 18 m

diameter of the solenoid, D = 75 cm = 0.75 m

diameter of the superconducting wire, d = 2 mm = 0.002 m

The number of turns of the solenoid is calculated as;

[tex]N = \frac{length \ of \ solenoid}{diameter \ of \ wire } = \frac{1.8}{0.002} = 900 \ turns[/tex]

The magnetic field strength is given by;

[tex]B = \frac{\mu_0 NI}{L} \\\\[/tex]

Where;

I is the current needed

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

[tex]I = \frac{BL}{\mu_0 N} =\frac{1.5 \times 1.8}{4\pi \times 10^{-7} \ \times 900} \\\\I = 2387.32 \ A[/tex]

Therefore, the current needed is

2387.32 A