The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI
ID: 1536671 • Letter: T
Question
The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of 4.0 T, and the current in the solenoid is 3.0 102 A. What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.
______ turns/m
Explanation / Answer
magnetic field inside a solenoid B = uo*n*I
uo = magnetic permeability = 4*pi*10^-7
n = number of turns per m
I - crrent in the wire
4 = 4*pi*10^-7*n*3*10^2
turns/m = n = 10610 <<<==========ANSWER
direction of earth magnetic field is North
part(b)
magnetic force Fb = q*(v x B )
v = 1.38*10^5 i
B = 59.7*10^-6 j
q = 1.6*10^-19 C
Fb = 1.6*10^-19*( 1.38*10^5 i x 59.7*10^-6 j )
Fb = 1.6*10^-19*1.38*10^5*59.7*10^-6 k
Fb = 1.32*10^-18 k
magnetic forc = 1.32*10^-18 N
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(c)
Fgrav = G*mp*mE/R^2 = mp*g = 1.67*10^-27*9.8 = 1.64*10^-26 N
Felec = E*q = 1.5*10^2*1.6*10^-19 = 2.4*10^17 N
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