Question

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.

a. What current is needed?

Answer 1

2270 A

Step-by-step explanation:

First, the number of turns in the solenoid

`N=\frac {l}{d}` where l is the length and d is the diameter of a single layer. In this case, the number of turns will be

`N=\frac {1.8}{0.0019}=947.3684211\approx 947`

The current, I round solenoid will be given by

`I=\frac {BL}{\mu_o N}` where B is the magnetic field strength, L is the length of solenoid,
`\mu_o` is permeability of free space and N is the number of turns. Taking the permeability of free space as
`4\pi* 10^(-7) H.m^(-1)`

`I=\frac {1.5* 1.8}{4\pi* 10^(-7) H.m^(-1) * 947}=2270 A`