101) What will be the approximate Precessional frequency of the signal, if the intensity of magnetic field is 30,000nT and gyromagnetic ratio value is $2.67513\times10^{8}S^{-1}T^{-1}$ ?
(Thanks to Chandrasekhar and Rajkumar, AKNU)
Solution:
Given
that
Gyromagnetic ratio
$\gamma_{p}=2.67513\times10^{8}S^{-1}T^{-1}$
Intensity of magnetic
field $B_{t}$=30,000nT
The
Preccesional frequency of the signal can be defined as
$f=\frac{\gamma_{p}}{2\pi}B_{t}$-----------------------(1)
By
substituting the given values in the above equation we get
$f=\frac{\gamma_{p}}{2\pi}B_{t}$
$f=(\frac{2.67513\times10^{8}}{2\times3.14})(30,000nT)$
$f=\frac{2.67513\times10^{8}\times3\times10^{4}nT}{2\times3.14}$
$f=\frac{2.67513\times10^{8}\times3\times10^{4}\times10^{-9}T}{2\times3.14}$
[ since $1nT= 10^{-9}T$]
$f=\frac{2.67513\times3\times10^{3}T}{2\times3.14}(S^{-1}\times
T^{-1} \times T)$
$f=1277.928
Hz$
Reference : Fundamental of Geophysics, William Lowrie.
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