CSIR-NET 2012 Dec

CSIR NET – DEC 2012

39) If ‘F’ is the magnetic field on the Earth’s surface, the magnetic field measured at a height equivalent to the Earth’s radius would be

1) 0.5 F

2) 0.25 F

3) 0.125 F

4) 0.0625 F

(Thanks to Suresh Sir, GSI)

(Thanks to Neeraja, AU)

Solution:

Magnetic field at the Earth surface is

$F_r=\frac{\mu_om}{4\pi r^3}\sqrt{1+3\sin^2\phi}---(1)$

r – radius of the Earth, ϕ – latitude

Magnetic field at the height equivalent to the mean radius is

Earth’s radius becomes '2r'

$F_{2r}=\frac{\mu_om}{4\pi (2r)^3}\sqrt{1+3\sin^2\phi}---(2)$

$\therefore \frac{F_{2r}}{F_r}=\frac{\frac{\mu_om}{4\pi 8r^3}\sqrt{1+3\sin^2\phi}}{\frac{\mu_om}{4\pi r^3}\sqrt{1+3\sin^2\phi}}$

$\frac{F_{2r}}{F_r}=\frac{1/8}{1}=\frac{1}{8}$

$F_{2r}=\frac{F_r}{8}$

$F_{2r}=0.125F_r$

GEOPHYSICS