In one of your hospital rotations, a patient is placed into a MRI tunnel that is

Question

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. 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 10^6 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.

Explanation / Answer

Magnetic energy

E=(1/2)LI2

2.5*106=(1/2)*L*2002

L=125 H

Inductance of a solenoid is given by

L=uoN2A/l

125=(4pi*10-7)*N2*(pi*0.42/4)/1

N=28135 turns

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