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Consider a clinical MRI (magnetic resonance imaging) superconducting magnet that has a diameter of 1.27 m, length of 1.03 m, and a uniform magnetic field of 3.79 T.  a) Determine the energy density of the magnetic field.  J/m3J/m3  b) Determine the total energy in the solenoid.  J

Question

Consider a clinical MRI (magnetic resonance imaging) superconducting magnet that has a diameter of 1.27 m, length of 1.03 m, and a uniform magnetic field of 3.79 T.  a) Determine the energy density of the magnetic field.  J/m3J/m3  b) Determine the total energy in the solenoid.  J

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Solution

a) The energy density (u) of a magnetic field is given by the formula:

u = B² / (2μ)

where B is the magnetic field strength and μ is the permeability of free space. The permeability of free space (μ) is a constant, approximately 4π x 10^-7 T m/A.

So, substituting the given values:

u = (3.79 T)² / (2 * 4π * 10^-7 T m/A) u = 14.37 T² / (8π * 10^-7 T m/A) u = 1.8 x 10^6 J/m³

b) The total energy (U) in the solenoid can be found by multiplying the energy density by the volume of the solenoid. The volume (V) of a cylinder is given by the formula:

V = πr²h

where r is the radius and h is the height. Given that the diameter is 1.27 m, the radius is 1.27 m / 2 = 0.635 m. The height is the length of the solenoid, 1.03 m.

So, substituting these values:

V = π * (0.635 m)² * 1.03 m V = 1.33 m³

Then, the total energy is:

U = u * V U = 1.8 x 10^6 J/m³ * 1.33 m³ U = 2.4 x 10^6 J

This problem has been solved

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