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 (Bt
) 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 $f_{2}$
$\frac{f_{1}}{f_{2}}=
\frac{\sqrt{1+3sin\phi_{1}^{2}}}{\sqrt{1+3sin\phi_{2}^{2}}}$
=
$\frac{\sqrt{1+3sin30^{2}}}{\sqrt{1+3sin60^{2}}}$
=$\frac{\sqrt{1+3\times\frac{1}{4}}}{\sqrt{1+3\times\frac{3}{4}}}$
=$\frac{\sqrt{1+\frac{3}{4}}}{\sqrt{1+\frac{9}{4}}}$
=$\frac{\sqrt{\frac{7}{4}}}{\sqrt{\frac{13}{4}}}$
=$\frac{\sqrt{7}}{\sqrt{13}}$
$\frac{f_{1}}{f_{2}}
= \frac{\sqrt{7}}{\sqrt{13}}$
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