The magnetic field of a proton is approximately like that of a circular current

Question

The magnetic field of a proton is approximately like that of a circular current loop 0.650 Times 10^-15 m in radius carrying 1.05 Times 10^4 A. An MRI machine needs to be able to manipulate these fields. To see why an MRI utilizes iron to increase the magnetic field created by a coil calculate the current needed in a 400 loop-per-meter circular coil 0.660 m in radius to create a 1.20 T field (typical of an MRI instrument) at its center with no iron insert. What is the field at the center of a proton? Notice how it compares to the field we used in the previous calculation.

Explanation / Answer

a.) Magnetic field inside a solenoid (coil) B = o  nI where n is the number of turns per unit length

and I is current.

1.2 = 4 x 10-7 x 400 x I

I = 2387.324 Amperes

So, 2387.324 Amperes of current is needed

b.) Magnetic field at the centre of a circular current carrying loop (as in the case of proton) is B = o I / 2R

Bo = 4 x 10-7 x 1.05 x 104 / ( 2x 0.65x 10-15 )

Bo = 10.149 x 1012 T

That is why ferromagnetic materials are used as the core in electromagnets. Because they can increase the magnetic field strength to extremely high values.

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