In medical technology, Magnetic Resonance Imaging (MRI) produces detailed images of the interior of the human body. The cross section of a typical MRI is provided in the accompanying diagram. You are working with a team of physicians and physicists to learn more about the inner workings of the MRI. In one of your hospital rotations, a patient is placed into a MRI tunnel that is 0.40 m in diameter and 1.00 m long. A 200 A current creates an approximately uniform magnetic field inside the MRI tunnel. To carry such a large current, the wires (shown as cross sections, with current from the array of wires running out-of and into-the page, respectively) are cooled with liquid helium until they become superconducting. The head physicist informs you that the amount of magnetic energy stored in the MRI tunnel is on the order of 2.5 x 106 J. The other physicians are stumped as to calculating the number of turns of wire necessary to accomplish this storage of energy so they turn to you for your expert, detailed calculation.

Question

In medical technology, Magnetic Resonance Imaging (MRI) produces detailed images of the interior of the human body.  The cross section of a typical MRI is provided in the accompanying diagram. You are working with a team of physicians and physicists to learn more about the inner workings of the MRI.  In one of your hospital rotations, a patient is placed into a MRI tunnel that is 0.40 m in diameter and 1.00 m long. A 200 A current creates an approximately uniform magnetic field inside the MRI tunnel. To carry such a large current, the wires (shown as cross sections, with current from the array of wires running out-of and into-the page, respectively) are cooled with liquid helium until they become superconducting. The head physicist informs you that the amount of magnetic energy stored in the MRI tunnel is on the order of 2.5 x 106 J.  The other physicians are stumped as to calculating the number of turns of wire necessary to accomplish this storage of energy so they turn to you for your expert, detailed calculation.

anonymous · Answer

attachment

anonymous · Answer

@Michele_Laino @IrishBoy123

irishboy123 · Answer

try this approach... the inductance of a solenoid is $\large L = \mu \frac{N^2A}{l}$ http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/indsol.html#c1 and the energy required to create the magnetic field - from the energy stored in an inductor - is $\large E = {1\over 2} L I^2 \ = \large {1\over 2} .\mu \frac{N^2A}{l} . I^2$ http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html so $ N = \dfrac{1}{ I} \sqrt{\dfrac{2E l}{\mu A }}$ check algebra :p

anonymous · Answer

Hmm, that is interesting :P Thanks!

anonymous · Answer

@IrishBoy123 Here, is $\mu$ the magnetic constant for permeability of space? I read that most materials are relatively close to this value because they're para- or dia-magnetic.

anonymous · Answer

$\mu_0=4.0\pi \times 10^{-7} \frac{Tm}{A}$?