108)
A planet of 70,000 km radius exhibits a magnetic field of 4.2 Oe at its
equator. What is the rate of decrease of its magnetic field (in gammas/meter)
on its surface at the location of magnetic field $\arccos(\sqrt{\frac{3}{4}})$
(Special thanks to Y.Suresh Sir, GSI)
(Thanks to Chandrasekhar, ANU)
Given that radius of
the planet(r) = 70,000 km
Magnetic field at the equator = 4.2 Oe
The Earth’s total
magnetic field in terms of geographical latitude ($\phi$) is as below
$B_{F}=\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin
^{2}\phi +1}$
As per given data at the equator the total
magnetic field is 4.2 Oe means at a geographical latitude 00
$B_{F}=\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin
^{2}\phi +1}$
$B_{F(equator)}=4.2 $
$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin
^{2}\phi +1}=4.2$
$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin
^{2}(0) +1}=4.2$
$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{0
+1}=4.2$
$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}=4.2$
Now we have to
calculate the rate of decrease of magnetic field on its surface at the location
of magnetic latitude $\arccos(\sqrt{\frac{3}{4}})$
Mean the latitude
$\phi=\arccos(\sqrt{\frac{3}{4}})$
$\phi=\arccos(\frac{\sqrt{3}}{2})$
$\cos\phi
=\frac{\sqrt{3}}{2} $
$\cos\phi =\cos 30 $
$\phi=30$
The rate of decrease in
the magnetic field
$\frac{dB_{F}}{dr}=\frac{1}{dr}(\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin
^{2}\phi +1})$
$\frac{dB_{F}}{dr}=\frac{-3\mu_{0}}{4\pi}\frac{m}{r^{4}}\sqrt{3\sin
^{2}\phi +1}$
$\frac{dB_{F}}{dr}=\frac{-3}{r}(\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}})\sqrt{3\sin
^{2}(30) +1}$
$\frac{dB_{F}}{dr}=\frac{-3}{70,000}(4.2)\sqrt{1+\frac{3}{4}}$
$\frac{dB_{F}}{dr}=\frac{-3\times
4.2}{70,000}\sqrt{\frac{7}{4}}$
$\frac{dB_{F}}{dr}=0.0002376
Oe/km$
$\frac{dB_{F}}{dr}=0.0002376
\frac{10^{5} gamma}{10^{3}meter}$
$\frac{dB_{F}}{dr}=0.0002376
\times 10^{2} gamma/meter$
$\frac{dB_{F}}{dr}=0.02376
gamma/meter$
Gravity anomaly values of 1.2, 2.4 and 3.6 in mgals are located at distance coordinates 20, 24
ReplyDeleteand 28 km, respectively, along a gravity profile across a faulted basement of limited throw. The
depth to the basement (in km) is:
Please solve this. If it is a faulted basement, then at which point will we calculate the depth?
http://gpsurya.blogspot.com/2020/04/csir-net-2018-dec.html?m=1
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