Sunday, July 28, 2019
Physics MRI Essay Example | Topics and Well Written Essays - 1000 words
Physics MRI - Essay Example The Larmor (or resonant) frequency Ãâ°0 is the frequency at which the nuclide precesses about the magnetic field. The resonant frequency is equal to the magnetogyric ratio à ³ (specific to the nuclide) times the magnetic field B (Brandolini, 2004): the nuclide 13C at 75 MHz. From the equation above, the magnetogyric ratio à ³ is constant so that à ³ =Ãâ°0 /B = Ãâ°Ã¢â¬â¢0 /Bââ¬â¢, where Ãâ°Ã¢â¬â¢0 is the resonant frequency when the magnetic field Bââ¬â¢ = 1.5 T. Solving for Ãâ°Ã¢â¬â¢0 : Ãâ°Ã¢â¬â¢0 = (Ãâ°0 Bââ¬â¢)/B what is the mean B and B This is explained in the sentence directly above: the single prime corresponds to the resonant frequency when the magnetic field is 1.5 T. You are asking what is meant by Bââ¬â¢, but if you look at the sentence above, it was just defined: Bââ¬â¢ = 1.5T. It is the magnetic field at 1.5T. Bââ¬â¢Ã¢â¬â¢ is just a different value of the magnetic field (in this case 4 T) where we are trying to find the frequency w0ââ¬â¢Ã¢â¬â¢ that corresponds to it. From this equation, if you know the frequency Ãâ°0 and the magnetic field B, then the ratio of these is the gyromagnetic ratio. We know the frequency at 6.9T from the reference cited above. Therefore, to find the frequency at a different magnetic field, we just use the equation w0/B = à ³ = constant. So another set of corresponding values of w0 and B, call these new values w0ââ¬â¢Ã¢â¬â¢ and Bââ¬â¢Ã¢â¬â¢, will also have the same ratio: w0ââ¬â¢Ã¢â¬â¢/Bââ¬â¢Ã¢â¬â¢ =à ³ constant = w0/B. Since we now have w0ââ¬â¢Ã¢â¬â¢/Bââ¬â¢Ã¢â¬â¢ = w0/B, we can multiply both sides by Bââ¬â¢Ã¢â¬â¢ to get: w0ââ¬â¢Ã¢â¬â¢ = w0 * Bââ¬â¢Ã¢â¬â¢/B . Hopefully you can now see where that equation comes from. The reason I didnââ¬â¢t put the calculation down in this case, is because it is EXACTLY the same as the calculation before it, but with different values. You can just follow the equations that were used in the example above it, putting in the
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