
#1
Sep2513, 11:14 AM

P: 6

Hello
If the time that takes for magnetization to rotate from its initial position (parallel with external magnetization) to x,y plane is given, how do we determine the value of B1? The formula I know: ω = γB1 = 2∏f1 lB1l < lBol angle between the magnetization and the external magnetization = ωt = γB1t Thank you 



#2
Sep2513, 01:49 PM

Sci Advisor
PF Gold
P: 2,020

You wrote down the frequency of precession (from the z axis to z and back to +z) in terms of B1. Now write the frequency in terms of period (time) [itex]f_1=1 / t_1 [/itex] for one cycle. What fraction of one cycle is the journey from the z axis to the xy plane?




#3
Sep2513, 08:56 PM

P: 6

I'm not quite sure what you meant .. So basically I have ω = γB1 = 2∏f1 formula, and if I write this in terms of period it would be like ω = γB1 = 2∏(1/t).. and I'm lost afterwards.. could you be kind to explain more in detail? 



#4
Sep2513, 11:07 PM

Sci Advisor
PF Gold
P: 2,020

Determining the value of B1 (or RF) in NMR
You now have [itex]t_1(360)[/itex] in terms of B1. I've written it with the argument 360 because this period is the length of time for the spin to precess from the z direction down to the xy plane, on to z, back to the xy plane, and back up to z, for a total angle of 360 degrees. You want the time it takes to go from z to the xy plane, which is an angle of 90 deg. You have everything needed to find that value.



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