Finding the Magnetic Field (B_0) on a Graph [NMR]

In summary: I am unsure how to go about finding ##B_0## as the equation I am working with is in the form of amplitude vs. time. As the graph does not possesses any values of frequency, I am unable to use the above equation to obtain my ##B_0## for each of the data points.Assuming the equation you are referring to is the one shown below,$$f_0 = \big(\frac{\gamma}{2\pi} \big) B_0$$I believe the hint to finding the value for ##\frac{\gamma}{2\pi}## lies in the term (from the homework statement) "rare-earth magnets".
  • #1
Athenian
143
33
Homework Statement
Short Question Explanation:

I have a semi-logarithmic graph of free induction decay (FID) amplitude as a function of time. While the primary purpose of this graph was to find the value of the apparent spin-spin relaxation time (##{T_2}^*##) and its uncertainty for each corresponding data point, I am also tasked with finding the value of the magnetic field (##B_0##) and its uncertainty for the sample due to a pair of external rare-earth magnets.

Note that I do know the process for finding ##{T_2}^*## which will be outlined below as it may help with understanding how to obtain the value of ##B_0## (the problem I am having trouble with).

Additional details to the problem are shown below.
Relevant Equations
##f_0## (MHz) ##= 4.258 B_0## (kilogauss)
##\frac{1}{{T_2}^*} = \frac{1}{T_2} + \gamma \Delta B##
##2\Delta B_0 = \frac{2}{{T_2}^* \gamma}##
##M= M_0 e^{\frac{-t}{{T_2}^*}}##
Background Information (Not Strictly Necessary):

As a quick recap, the graph I am dealing with is a semi-logarithmic graph of free induction decay (FID) amplitude as a function of time. To acquire the value for ##{T_2}^*## (and its uncertainty) in the graph, I used the below equation to do so.

$$M= M_0 \; e^{\frac{-t}{{T_2}^*}}$$

However, the graph needs to be a semi-logarithmic one as opposed to an exponential one. Thus, after the "conversion", I found that:

$${T_2}^* = -\frac{\log{(e)}}{m}$$

Note that ##m## above signifies the slope (##m##) in the equation of a linear line on a graph.

For more details how I came up with the above equation, please refer to the below thread (the solution process is in the latter posts).

https://www.physicsforums.com/threads/calculating-numerical-uncertainties-in-an-equation.1001620/

Example Graph Data (Related to the Next Section):

t (##\mu##s)V_1(mV)
90​
156.3​
100​
131.3​
120​
75​
150​
45.3​
175​
34.4​
200​
17.2​

When plotted in a semi-log graph ...
##m=-0.008344 \pm 0.0004948## mV/##\mu##s
##b=2.931 \pm 0.07160## mV

Linear Equation for above ...
##\log{(V_1)} = y = mt+b##

My Attempt at Finding ##B_0##:

As shown in "Relevant Equations", I know that:
$$f_0 = 4.258 B_0$$

where ##f_0## is measured in MHz and ##B_0## measured in kilogauss.

However, my graph is in the form of amplitude vs. time. As the graph does not possesses any values of frequency, I am unable to use the above equation to obtain my ##B_0## for each of the data points.

I have seen an online PowerPoint (link shown below) that states I must translate my amplitude vs. time graph to one that is amplitude vs. frequency.

Link: https://chemistry.mit.edu/wp-content/uploads/2018/08/DCIF-IntroNMRpart1-theory-o07.pdf (Page 10)

If this is possible, I should be able to take my frequency value for each corresponding data point and find ##B_0## for each of the given data points via the equation ##f_0 = 4.258 B_0##. However, I have absolutely no idea how to approach the problem with this potential method.

In another online article (though I do not remember the link), it argues that as
##\frac{1}{{T_2}^*} = \frac{1}{T_2} + \frac{1}{T_1} + \gamma \Delta B_0## (this article adds ##\frac{1}{T_1}## for some reason), I would be able to obtain the below equation.
$$\Delta B_0 = \frac{1}{{T_2}^* \gamma}$$

While an intriguing discovery, knowing the value of ##\Delta B_0## doesn't exactly bring me any closer to finding the respective ##B_0## values for each of my data points (in the above table).

While perhaps an obvious hint, I was instructed to find the values of ##B_0## and its uncertanties for each corresponding data point on the graph. As the uncertainty values are required for this exercise, I know that the calculation process should involve the numerical data obtained on my linear (or potentially exponential) equation (e.g. from ##m##) as those are the only data set I have that possesses any uncertainty values.

Once again, the above question deals with NMR (Nuclear Magnetic Resonance). There are a lot of helpful information on the web, but none has led me to understand how to find for ##B_0##. In theory, though, the process of finding ##B_0## shouldn't be ridiculously difficult and ought to be relatively straightforward. That said, any assistance on the necessary steps I should take to find ##B_0## would be greatly appreciated.

If additional clarification or information is needed that could help streamline the assistance process, please do let me know! Thank you for reading through my question!
 
Physics news on Phys.org
  • #2
To find ##B_0## you need to measure it or measure ##f_0## and know the gyromagnetic ratio. You do not seem to understand the NMR process, nor the transverse and longitudinal relaxation times. Perhaps general study is required. This looks OK
 
  • Like
Likes Athenian
  • #3
@hutchphd, thank you for the helpful feedback and link. I read through the entirety of the link's content and it certainly helped me better understand the topic.

hutchphd said:
To find ##B_0## you need to measure it or measure ##f_0## and know the gyromagnetic ratio.

Assuming the equation you are referring to is the one shown below,
$$f_0 = \big(\frac{\gamma}{2\pi} \big) B_0$$
I believe the hint to finding the value for ##\frac{\gamma}{2\pi}## lies in the term (from the homework statement) "rare-earth magnets".

According to the text provided by my instructor (and I quote a short section of it):

"The precession rate (resonance frequency) of protons (hydrogen nuclei) depends linearly on the magnetic field according to the equation ##f_0## (MHz) ##= 4.258 B_0## (kilogauss). In the experiment, most of the field comes from a permanent (rare-earth magnet that has a field of about 3.6 kG".

As I was given all the experimental data (i.e. no hands-on experiment), this should theoretically be all the data I need to work with to find ##B_0##. As additional information (that may or may not be helpful), the table of data in the original post is data for the "Raw ##T_2## from free induction decay with Time Const. (0.01 ms) Gain (30%) Repetition Time (1s and 100%)".

That should be, in principle, all the information that is needed to acquire the value of ##B_0##. That said, I still don't know how to approach finding this value. If I could measure ##f_0##, this would solve my problem. However, it seems like this data was intentionally not given.

Regardless, I do know my values should all come close to 3.6 kG (with small differences in value due to chemical shifts).

If the above information is not enough to work with, please do let me know. Once again, thank you for your help!
 
  • #4
If I'm not mistaken, the free induction decay is a (decaying) sinusoidal waveform at the frequency that you need to compute ## B_o ##, and all you have in your data is the amplitude of this sinusoid during its decay. Typically, the frequency would be observed with an oscilloscope trace.
 
  • Like
Likes Athenian and hutchphd
  • #5
Yes absolutely and for the gyromagnetic ratio

Athenian said:
the equation f0 (MHz) =4.258B0 (kilogauss)
I really mean the ratio above ##4.258\frac {MHz} {kG}## as the ratio in question for the proton (there is a factor of ##2\pi## floating around in the definition)
 
  • Like
Likes Athenian and Charles Link
  • #6
Thank you both for the feedback. To get proper clarification, I decided to go ahead and ask my professor and see what answer I get. I'll get back here once I have any answers or questions.

Once again, thanks for all the help thus far!
 
  • Like
Likes hutchphd and Charles Link
  • #7
Quick Update:

Unsurprisingly, the numerical data of ##f_0## is needed to calculate for ##B_0##. However, the data was not provided to me until I let my professor know about it. Soon afterward, I was provided with the relevant ##f_0## data to help me proceed to find the value for ##B_0##.

Despite the "anti-climatic" solution, thank you both for all your help!
 
  • Like
Likes Charles Link

1. What is the purpose of finding the magnetic field (B_0) on a graph in NMR?

The magnetic field (B_0) is a crucial parameter in nuclear magnetic resonance (NMR) spectroscopy. It is used to determine the energy levels of the nuclei in a sample, which can provide information about the structure and composition of the sample.

2. How is the magnetic field (B_0) measured on a graph in NMR?

The magnetic field (B_0) is typically measured using a calibrated magnetic field probe, which is placed at the location of the sample in the NMR instrument. The probe measures the strength of the magnetic field and displays it on a graph.

3. What factors can affect the accuracy of the magnetic field (B_0) measurement on a graph in NMR?

Several factors can affect the accuracy of the magnetic field (B_0) measurement, including temperature fluctuations, magnetic field drift, and instrument calibration. It is important to carefully control these factors to ensure accurate measurements.

4. How does the magnetic field (B_0) affect the NMR spectrum?

The magnetic field (B_0) directly affects the energy levels of the nuclei in the sample, which in turn affects the NMR spectrum. A stronger magnetic field will result in a higher resolution and more distinct peaks in the spectrum.

5. Can the magnetic field (B_0) be adjusted in NMR experiments?

Yes, the magnetic field (B_0) can be adjusted in NMR experiments. This is often done to optimize the resolution and sensitivity of the NMR spectrum. However, it is important to carefully calibrate and control the magnetic field to ensure accurate and reproducible results.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
280
  • Introductory Physics Homework Help
Replies
2
Views
192
  • Introductory Physics Homework Help
Replies
1
Views
872
  • Introductory Physics Homework Help
Replies
3
Views
736
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
749
  • Advanced Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top