GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET

51. Assuming the Earth's magnetic field to be purely dipolar, the ratio of total magnetic field intensity at 45 degrees magnetic latitude to that at the equator will be

A) 2

B) √(5/2)

C) 1/√2

D) √5/2 

Solution:

Magnetic field intensity B is

$B= \frac{\mu_0 m}{4\pi r^3}\sqrt{1+3sin^2\phi}$ -------(1)

Magnetic field intensity at 45 degrees is

$B_{45^0}= \frac{\mu_0 m}{4\pi r^3}\sqrt{1+3sin^245^0}$ ---(2)

$sin45^0 =\frac{1}{\sqrt{2}}$ ---(3)

substitute eq(3) value in eq(2)

$B_{45^0}= \frac{\mu_0 m}{4\pi r^3}\sqrt{1+(3\times\frac{1}{2})}$

$B_{45^0}= \frac{\mu_0 m}{4\pi r^3}\sqrt{\frac{5}{2}}$ ---(4)

Magnetic field intensity at 0 degress (equator)

$B_{0^0}= \frac{\mu_0 m}{4\pi r^3}\sqrt{1+3sin^20^0}$ ---(5)

$sin0^0=0$ -----(6)

substitute eq(6) value in eq(5)

$B_{0^0}= \frac{\mu_0 m}{4\pi r^3}$ ----------(7)

The ratio of total magnetic field intensity at 45 degrees magnetic latitude to that at the equator will be

eq (4) divided by eq (7)

Therefore,

$\frac{B_{45^0}}{B_{0^0}} = \frac{ \frac{\mu_0 m}{4\pi r^3}\sqrt{\frac{5}{2}} }{ \frac{\mu_0 m}{4\pi r^3}}$

$\frac{B_{45^0}}{B_{0^0}} =\sqrt{\frac{5}{2}}$