Question

In an NMR experiment, the RF source oscillates at 38 MHz and the magnetic resonance of the hydrogen atoms in the sample being investigated occurs when the external field has magnitude 0.75 T. Assume that the internal and external magnetic fields are both along a z axis, with the external field in the positive direction, and take the proton magnetic moment component μz to be 2.30 × 10-26 J/T. What is the z component of the internal magnetic field?

Answer 1

Answer: Bint = 0.204T

Explanation:

From the question, given that;

the RF source oscillates at 38MHz.

The external field has a magnitude of 0.75T.

From the relationship between external and internal magnetic field,

the external magnetic field is along the positive z-axis, while the internal field is along the positive z-axis.

From the expression;

Bnet = Bext - Bint ...........(1)

Proton magnetic moment along z - direction gives

U2 = 2.3×10^-26 J/T

U2 = rhi = r(h/2π)I........(2)

Energy change when an external magnetic field is applied becomes

∆E = h∆r =

∆E = rh/2π Bnet

where Bnet = Bext - Bint = h∆r/2u

Bint = Bext - Bnet

Bint = Bext - h∆r/2u =

= 0.75 - ((6.61×10^-34)×(38×10^6))/(2×(2.3×10^-26))

Bint = 0.204T