Calculation of true thickness of the bed

Csir_Net_Dec_2017

112) what is the value of the Earth’s Magnetic field (in Oe) at a place where it’s vertical and horizontal components are equal if the magnetic field at the equator is 0.30 Oe.

 (Special thanks to Y.Suresh sir, GSI)

(Thanks to Chandrasekhar, ANU)

 

Solution:

 

Given that Magnetic field at the equator = 0.30 Oe

Find the Magnetic field at a place where its $B_{V} =B_{H}$

 

We know that Earth’s total magnetic field

$B_{F}=\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\cos^{2}\theta+1}$-------------------(1)

Where m= magnetic moment

            R= radius

            $\theta$ = co-latitude

 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 0.30 Oe means at a geographical latitude 0⁰

i.e

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

 

$B_{F(equatotr}=0.30$

 

$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin ^{2}\phi +1}=0.30$

 

$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\sin ^{2}(0) +1}=0.30$

 

$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{0 +1}=0.30$

 

$\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}=0.30$

 

$m=\frac{0.30 \times4\pi r^{3}}{\mu_{0}}$--------------------------(2)

 

  Now as per given assume that $B_{Z} =B_{H}$

 We know that  the radial and the tangential components are as below

 

$B_{r}=B_{Z}=\frac{\mu_{0}}{4\pi}\frac{2m\cos\theta}{r^{3}}$

And

$B_{\theta}=B_{H}=\frac{\mu_{0}}{4\pi}\frac{m\sin\theta}{r^{3}}$

 Therefore

$B_{Z} =B_{H}$

 

$\frac{\mu_{0}}{4\pi}\frac{2m\cos\theta}{r^{3}}=\frac{\mu_{0}}{4\pi}\frac{m\sin\theta}{r^{3}}$

 

$2\cos\theta=\sin\theta$

 

$\frac{\cos\theta}{\sin\theta}=\frac{1}{2}$

 

$\cot\theta=0.5$

 

$\theta=\cot^{-1}(0.5)$

 

$\theta=63.43^{0}$

 Now value of the magnetic field at $B_{V} =B_{H}$ is

$B_{F}=\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\cos^{2}\theta+1}$

By substituting the m and $\theta$ values in the above equation we got

$B_{F}=\frac{\mu_{0}}{4\pi}\frac{m}{r^{3}}\sqrt{3\cos^{2}\theta+1}$

 

$B_{F}=(\frac{\mu_{0}}{4\pi r^{3}}){(\frac{0.30 \times4\pi r^{3}}{\mu_{0}})}{}\sqrt{3\cos^{2}(63.43^{0})+1}$

 

$B_{F}=0.30\sqrt{3\cos^{2}(63.43^{0})+1}$

 

$B_{F}=(0.30)(\sqrt{1.6})$

 

$B_{F}=0.379 Oe$