Magnetic Moment Dipole for Different Particles - Assignment Example

MERS7001 Assignment Name: Course: Instructor: Institution: City & State: Date: MERS7001 Assignment Question 1: The magnetic moment is the amount of tendency of a magnet to adjust with its magnetic field. Since the magnetic field and the magnetic moment of a magnet have both direction and magnitude, they can therefore be termed as vectors (Furlani, 2001). The magnetic moment has its direction pointing from north to South Pole; hence the magnet produces the magnetic field which is proportional to the magnetic moment. To understand the magnetic moment, one needs to understand the atomic behaviour and its construction. The atomic composition is centrally located in the nucleus of the atom with orbiting electrons around the nucleus. An atom has characteristics that are determined by the chemical assembly of the atom, which includes the atomic number (Z) that is composed of protons present in the nucleus and the atomic mass number (atomic weight) which is composed of the sum of the number of protons and neutrons in the nucleus. The active nucleus has a magnetic moment therefore both the atomic and atomic mass numbers play a significant role in producing MR. Using the electromagnetic laws, the nucleus spins about its axis and as a result, a magnetic field is developed around the nucleus (Tipler et al., 2002). The nucleus spin is at right angles with the magnetic field produced by the nucleus. Since magnetic moment has both magnitude and direction; the vector can be used to define the direction and magnitude of the magnetic moment. Figure 1: The axis for the magnetic dipole moment Figure 2: Rotation of the magnetization vector along the z axis Table 1: magnetic moment dipole for different particles Particle Magnetic moment in SI units (10-27 J/T) Spin quantum number Proton +14.11 1/2 Neutron -9.66 1/2 Electron -9284.76 1/2 Magnetic moment can be presented as a vector by the use of size and its magnitude. If the spin value is a half integral, that is, when the atomic mass number is odd, then the magnetic moment can give rise to an MR signal. For example, for 1H composed of a positive proton and no neutrons, the atomic number is an odd number. 7Li has an atomic mass number of seven (7) hence it has 3 protons and 4 neutrons and therefore it can produce the MR signal. The spin for 1H is ½ and that of 7Li is 3/2. This gives an integral of 1, 2, 3..., which explains the occurrence of the atomic number as odd when the mass number is even. Magnetic moment can be present in the following cases: a bar magnet, an electric current loop, a molecule, an electron and a planet. Magnetic dipole moment is more elaborate and this explains the magnetic moment of a system. The magnetic dipole moment uses a series of multiple expansion of the magnetic moment. The magnetic dipole moment has a symmetrical alignment with the magnetic field that is produced by the object and it is said to decrease with an inverse distance from the object. Applying the magnetic field, the net vectors of the magnetic moment will be in line with the low energy or will be anti-parallel with high energy thus producing an angular momentum which produces a spin around the magnetic field. As a result, the magnitude of the magnetic moment increases. If a magnet is present in the system, the magnetic moment will be spinning up and down. Therefore, it is of significance to note that there will be a random spin of the vectors present in the active MR nucleus giving rise to random directions of the nucleus which in turn defines the size of the magnetic moment. When there is a nucleus that is aligned in different directions, there is a possibility that the net magnetic moment will tend to shift towards zero magnitude (Arlart et al., 2002). Using the gyromagnetic ratio, the magnetic moment of a system can be calculated using the following equation:  Where: f is the frequency measured in hertz, γ is the gyromagnetic ratio and  is the external magnetic field. Figure 3: Magnetic spin for a magnetic dipole moment Question 2: For each atom, the gyromagenetic ratio is constant in a given magnetic field. For hydrogen atom, the gyromagenetic ratio γ is  The Planck’s constant h is  The static magnetic field  is 1.5T. The energy difference is measured in units of Joules J or electron volts eV where  Joules. Using the data given above, the energy difference is calculated as:   Joules If the magnetic field was doubled, it would be 1.5  2 = 3T and so, the energy difference changes to: E = Joule From the above results, it is clear that the energy difference between parallel and anti-parallel states is directly proportional to the magnetic field B0. Figure 4: The energy of the splitting states of parallel and anti-parallel Nucleus Spin 1H ½ 4H 2 13C ½ 31P ½ Table 2: table of the nucleus and their spin The SI units that are used to measure energy are Joules J or Newton meter N m or m2 kg/s2 Given the frequency of an x-ray as 2 x 1019 Hz, the energy of the photon can be calculated as follows: Using Planck’s constant: Where:  is the energy of the photon to be calculated  is the Planck’s constant = 6.63 J/s  is the oscillatory frequency  HZ of the X-proportional to E Therefore:  Hz → Comparing the energy of the x-ray photon used to penetrate, there is much greater energy than the energy of the MRI active nucleus. This is said to be a harmful ionization level for the tissues. Since CT is a special type of X-rays for imaging, they require energy for transmitting these images. The energy is used to help the radiologist to study the internal parts of the body. Also, the energies present in the CT have an advantage of increasing the image resolution for detailed analysis. MRI and CT serve the same purpose but MRI utilizes radio waves and magnetic field to generate images. The energy required for the MRI is used to ensure that the radio waves and the magnetic field strength are enough to produce the required images (Morin, 2008). Question 3: Table 3: table of Gyroelectromagnetic ratio for different nuclei Nucleus Gyroelectromagnetic Ratio, (106 rad/s/T) Ff0 at 1 Tesla (MHz) 1H 268 42.58 4H 82 13.06 13C 67 10.66 31P 108 17.19 The resonant frequencies of nuclides at 1.5T and 4T: 1H, 13C and 31P Using the gyromagnetic ratio,  Where:  is the angular precession frequency and  is the magnetic field. For 1H, γ is 42.58 MHz/T, which is a constant for H for all magnetic fields that is for 1.5T and 4T For 13C, the γ is 10.704 MHz/T which is a constant in magnetic field at 1.5T and 4T For 31P, γ is 17.234 MHz/T for magnetic fields that is at 1.5T and 4T From the relation,  Where  Therefore, for 1H at 1.5T, And at 4T, For 13C at 1.5T, And at 4T, For 31P at 1.5T, And at 4T, Question 4: After the 900 rotating frame pulses is applied, total magnetization present in the system will gyrate away from the z axis towards the x and y axis. The net magnetisation present will assume a path that will gyrate around the magnetic field. For the laboratory frame, the rotations will be moving while the frame will be constant. As a result, the receiver coiler for the MRI will appear to be fixed and non-moving (Cullity & Graham, 2008). This therefore explains the continuous motion of the MRI which produces the magnetization to rotate. This explanation becomes complex when trying to explain the MRI behaviour. Using the rotating frame of reference, one can explain the concept of the magnetization which follows the rotating frame pulse. At a given frequency in the transverse plane, the magnetic field rotates around the z axis thus producing a coordinate system for the rotating frame pulse. In the x and y axes, the rotating frame will tend to appear stationary and fixed due to the fact that the rotating frame pulse is at Larmour frequency (Gupta, 2011). In the rotating frame of reference, the x and y axis give the transverse planes that are two dimensional and they are used to explain the motion of gyrating rotating frame. Therefore, for this reason, the gyrations that are faster than the rotating frame will rotate in a clockwise direction while the ones at a lower gyration will rotate in an anti-clockwise direction. Table 4: table of different frequency positioning in the Cartesian plane along the X and Y axis Y-position taking X= 0 Frequency X position taking Y=0 0 15.33 15.61 1 15.32 15.52 2 15.33 15.42 3 15.34 15.32 4 15.35 15.34 5 15.36 15.39 Figure 5: Longitudinal relaxation of the rotating frame Figure 6: 900 effects when RF spins Fig 6 and fig 5 give a 90o RF pulse, thus explaining how the spins will move in spiral way to the transverse plane. References Arlart, I. P. et al., 2002. Magnetic Resonance Angiography. 2nd ed. Amsterdam: Springer. Tipler, P. A. et al., 2002. Modern Physics. 4th ed. New York: Macmillan. Furlani, E. P., 2001. Permanent Magnet and Electromechanical Devices: Materials, Analysis, and Applications. London: Academic Press. Cullity, B. D. & Graham, C. D., 2008. Introduction to Magnetic Materials. 2nd ed. New York: Wiley-IEEE Press. Gupta, H., 2011. Encyclopedia of Solid Earth Geophysics. Amsterdam: Springer. Morin, D., 2008. Introduction to Classical Mechanics: With Problems and Solutions. New York: Cambridge University Press. Read More

The magnetic dipole moment has a symmetrical alignment with the magnetic field that is produced by the object and it is said to decrease with an inverse distance from the object. Applying the magnetic field, the net vectors of the magnetic moment will be in line with the low energy or will be anti-parallel with high energy thus producing an angular momentum which produces a spin around the magnetic field. As a result, the magnitude of the magnetic moment increases. If a magnet is present in the system, the magnetic moment will be spinning up and down.

Therefore, it is of significance to note that there will be a random spin of the vectors present in the active MR nucleus giving rise to random directions of the nucleus which in turn defines the size of the magnetic moment. When there is a nucleus that is aligned in different directions, there is a possibility that the net magnetic moment will tend to shift towards zero magnitude (Arlart et al., 2002). Using the gyromagnetic ratio, the magnetic moment of a system can be calculated using the following equation:  Where: f is the frequency measured in hertz, γ is the gyromagnetic ratio and  is the external magnetic field.

Figure 3: Magnetic spin for a magnetic dipole moment Question 2: For each atom, the gyromagenetic ratio is constant in a given magnetic field. For hydrogen atom, the gyromagenetic ratio γ is  The Planck’s constant h is  The static magnetic field  is 1.5T. The energy difference is measured in units of Joules J or electron volts eV where  Joules. Using the data given above, the energy difference is calculated as:   Joules If the magnetic field was doubled, it would be 1.5  2 = 3T and so, the energy difference changes to: E = Joule From the above results, it is clear that the energy difference between parallel and anti-parallel states is directly proportional to the magnetic field B0.

Figure 4: The energy of the splitting states of parallel and anti-parallel Nucleus Spin 1H ½ 4H 2 13C ½ 31P ½ Table 2: table of the nucleus and their spin The SI units that are used to measure energy are Joules J or Newton meter N m or m2 kg/s2 Given the frequency of an x-ray as 2 x 1019 Hz, the energy of the photon can be calculated as follows: Using Planck’s constant: Where:  is the energy of the photon to be calculated  is the Planck’s constant = 6.63 J/s  is the oscillatory frequency  HZ of the X-proportional to E Therefore:  Hz → Comparing the energy of the x-ray photon used to penetrate, there is much greater energy than the energy of the MRI active nucleus.

This is said to be a harmful ionization level for the tissues. Since CT is a special type of X-rays for imaging, they require energy for transmitting these images. The energy is used to help the radiologist to study the internal parts of the body. Also, the energies present in the CT have an advantage of increasing the image resolution for detailed analysis. MRI and CT serve the same purpose but MRI utilizes radio waves and magnetic field to generate images. The energy required for the MRI is used to ensure that the radio waves and the magnetic field strength are enough to produce the required images (Morin, 2008).

Question 3: Table 3: table of Gyroelectromagnetic ratio for different nuclei Nucleus Gyroelectromagnetic Ratio, (106 rad/s/T) Ff0 at 1 Tesla (MHz) 1H 268 42.58 4H 82 13.06 13C 67 10.66 31P 108 17.19 The resonant frequencies of nuclides at 1.5T and 4T: 1H, 13C and 31P Using the gyromagnetic ratio,  Where:  is the angular precession frequency and  is the magnetic field. For 1H, γ is 42.58 MHz/T, which is a constant for H for all magnetic fields that is for 1.5T and 4T For 13C, the γ is 10.

704 MHz/T which is a constant in magnetic field at 1.5T and 4T For 31P, γ is 17.234 MHz/T for magnetic fields that is at 1.5T and 4T From the relation,  Where  Therefore, for 1H at 1.5T, And at 4T, For 13C at 1.5T, And at 4T, For 31P at 1.5T, And at 4T, Question 4: After the 900 rotating frame pulses is applied, total magnetization present in the system will gyrate away from the z axis towards the x and y axis.

Read More