The magnetic field of a proton is very similar to that of a circular current loo

Question

The magnetic field of a proton is very similar to that of a circular current loop 0.650 times 10^-15 m in radius carrying 1.05 x 104 A. An MRI machine is required to be able to manipulate these fields. To see why an MRI utilizes iron to increase the magnetic field produced by a coil, calculate the current needed in a 400 loop-per-meter circular coil 0.660 m in radius to produce 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

Here,

radius ,r = 0.650 *10^-15 m

current , I = 1.05 *10^4 A

part A)

let the current needed is I

as magnetic field of the coils is

B = u0 * N * I/L

1.2 = 4pi *10^-7 * 400 * I/0.660

solvning for I

I = 1575.6 A

the current needed is 1575.6 A

Part B)

at the centre of proton

magnetic field = u0 ( I)/(2R)

magnetic field = 4pi *10^-7 * 1.05 *10^4/(2 * 0.650 *10^-15 )

magnetic field = 1.015 *10^13 T

it is very large comprared to previous field

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