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 1.50-mm-diameter superconducting wire.

What current is needed?

Answer 1

I = 1790.5 A

Step-by-step explanation:

The magnetic field due to a solenoid is given by the following formula:

B = μ₀NI/L

where,

B = Magnetic Field Required = 1.5 T

μ₀ = 4π x 10⁻⁷ T/A.m

L = length of Solenoid = 1.8 m

I = Current needed = ?

N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200

Therefore,

1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m

I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)

I = 1790.5 A

Answer 2

The current needed is 1790.26 A

Step-by-step explanation:

Given;

magnitude of magnetic field, B = 1.5 T

length of the solenoid, L = 1.8 m

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

The magnetic field is given by;

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

Where;

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

I is current in the solenoid

N is the number of turns, calculated as;

`N = (Length \ of\ solenoid)/(diameter \ of \ wire) \\\\N = (1.8)/(1.5*10^(-3)) =1200 \ turns`

The current needed is calculated as;

`I = (BL)/(\mu_o N) \\\\I = (1.5 *1.8)/(4\pi *10^(-7) *1200) \\\\I = 1790.26 \ A`

Therefore, the current needed is 1790.26 A.