Calculation of true thickness of the bed

CSIR_NET_JUNE_2019

92) What is the ratio of the Larmor Precession frequencies measurable on the earth’s surface at the magnetic latitude’s 30⁰ and 60⁰

 

A) $\frac{\sqrt{2}}{\sqrt{3}}$

B) $\frac{\sqrt{3}}{\sqrt{5}}$

C) $\frac{\sqrt{5}}{\sqrt{7}}$

D) $\frac{\sqrt{7}}{\sqrt{13}}$

(Thanks to Chandrasekhar, ANU)

 

Solution:

The strength of the magnetic field (B_t ) is directly proportional to the frequency of the signal(f) 

 

$B_{t}= \frac{2\pi f}{r_{p}}$ 

 

(or)

$f = \frac{r_{p}B_{t}}{2\pi}$

Where f is the Larmor Precession frequency

$(B_{t_{1}} ) Magnetic Field Strenght= \frac{\mu_{0}m\sqrt{1+3sin\phi_{1}^{2}}}{4\pi r^{3}}$

$f_{1}=(\frac{r_{p}}{2\pi})\frac{\mu_{0}m\sqrt{1+3sin\phi_{1}^{2}}}{4\pi r^{3}}$

simillarly 

 

$f_{2}=(\frac{r_{p}}{2\pi})\frac{\mu_{0}m\sqrt{1+3sin\phi_{2}^{2}}}{4\pi r^{3}}$

The ratio between $f_{1}$ and $f_{2}$ 

 

$\frac{f_{1}}{f_{2}}= \frac{\sqrt{1+3sin\phi_{1}^{2}}}{\sqrt{1+3sin\phi_{2}^{2}}}$

= $\frac{\sqrt{1+3sin30^{2}}}{\sqrt{1+3sin60^{2}}}$

=$\frac{\sqrt{1+3\times\frac{1}{4}}}{\sqrt{1+3\times\frac{3}{4}}}$

=$\frac{\sqrt{1+\frac{3}{4}}}{\sqrt{1+\frac{9}{4}}}$

=$\frac{\sqrt{\frac{7}{4}}}{\sqrt{\frac{13}{4}}}$

=$\frac{\sqrt{7}}{\sqrt{13}}$

$\frac{f_{1}}{f_{2}} = \frac{\sqrt{7}}{\sqrt{13}}$