Calculation of true thickness of the bed

 

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)

Solution:

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

$\frac{dB_{F}}{dr}=\frac{1}{dr}(\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin ^{2}\phi +1})$

 

$\frac{dB_{F}}{dr}=\frac{-3\mu_{0}}{4\pi}\frac{m}{r^{4}}\sqrt{3\sin ^{2}\phi +1}$

 

$\frac{dB_{F}}{dr}=\frac{-3}{r}(\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}})\sqrt{3\sin ^{2}(30) +1}$

 

$\frac{dB_{F}}{dr}=\frac{-3}{70,000}(4.2)\sqrt{1+\frac{3}{4}}$

 

$\frac{dB_{F}}{dr}=\frac{-3\times 4.2}{70,000}\sqrt{\frac{7}{4}}$

 

$\frac{dB_{F}}{dr}=0.0002376 Oe/km$

 

$\frac{dB_{F}}{dr}=0.0002376 \frac{10^{5} gamma}{10^{3}meter}$

 

$\frac{dB_{F}}{dr}=0.0002376 \times 10^{2} gamma/meter$

 

$\frac{dB_{F}}{dr}=0.02376 gamma/meter$