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In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being investigated occurs when the external field B→exthas magnitude 0.78 T . Assume that B→intand B→ext are in the same direction and take the proton magnetic moment component μ→zto be1.4×10-26J/T . What is the magnitude of B→int?

Short Answer

Expert verified

The magnitude of internal magnetic fiels Bint→is 19 mT .

Step by step solution

01

The given data:

  1. Oscillation frequency of the RF source, f=34×106Hz
  2. Magnitude of the external magnetic field, Bext=0.78T
  3. The internal and external magnetic fields are in the same direction.
  4. The component of the magnetic moment of proton,μz=9.27×10-24J/T
02

Understanding the concept of magnetic resonance:

Magnetic resonance imaging (MRI) is a medical imaging procedure that creates detailed pictures of your body's organs and tissues by using a magnetic field and computer-generated radio waves.

Using the condition of total magnetic field and the derived equation of magnetic field from energy equations of Stern-Gerlach experiment and Planck's relation, the value of the internal magnetic field by substituting the given values.

Formulas:

The energy equation from the Stern-Gerlach experiment is,

∆E=2μBB ….. (1)

Here, the magnetic momentum of the proton is

The energy due to the Planck-Einstein relation is,

∆E=hf ….. (2)

Here, the Plank’s constant is h=6.63×10-34J.s.

03

Calculation of the magnitude of the internal magnetic field:

At first from the given data, we can say that here the total magnetic field is given by:

B=Bint+Bext ….. (3)

Now, comparing equations (1) and (2), the value of the magnetic is as follows.

hf=2μBBB=hf2μB

Now, by substituting the given data and the above equation of total magnetic field in equation (3), the value of the internal field is given by,

B=hf2μµþ-Bext=6.63×10-34J.s34×106Hz29.27×10-24J/T-0.78T=19×10-3T=19mT

Hence, the value of the magnetic field is 19 mT.

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