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 2.20-mm-diameter superconducting wire.

Required:
What current is needed?

Answer 1

The current needed is 2625.72 A.

Step-by-step explanation:

Given;

magnetic field strength, B = 1.5 T

length of the solenoid, L = 1.8 m

diameter of the solenoid, d = 75 cm

The number of turns is given by;

`N = (Length \ of solenoid )/(diameter \ of wire)\\\\N = (1.8)/(2.2*10^(-3)) = 818.18 \ turns`

The magnetic field is given by;

`B = (\mu_o NI)/(l)`

where;

I is the current needed

`I = (Bl)/(\mu_o N) \\\\I = ((1.5)(1.8))/((4\pi*10^(-7))(818.18)) \\\\I = 2625.72 \ A`

Therefore, the current needed is 2625.72 A.