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?

The current needed is 1790.26 A

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;

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;

The current needed is calculated as;

Therefore, the current needed is 1790.26 A.

I = 1790.5 A

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

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explanation:

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