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

CSIR-NET 2015

87. Two rock units A and B of identical physical and chemical 
properties are inductively magnetized in the Earth's magnetic field 
at the magnetic latitudes 30° and 60° respectively. If ( IA and θA) 
and ( IB and θB) are intensities and dips of their induced 
magnetism, respectively, then

A)     IB= 2 IA              ;   tan θB=2 tan θA

B)     2 IB= √3 IA         ;   2 tan θB =√3 tan θA

C) √7 IB =  √5 IA         ;   2 tan θB = 3 tan θA

D) √7 IB =  √13 IA       ;   tan θB =3 tan θA

Solution:

1) If at any place, the angle of dip ‘i’ and magnetic latitude ‘Φ’ is

tani = 2 tanΦ -----------------(1)

magnetic latittudes at A = 30⁰ and B = 60⁰

substitute above values in eq (1)

tani_A= 2 tan 30⁰

tani_A = 2* 0.577

tanθ_A = 1.154

substitute above value eq (1)

tani_B = 2 tan60⁰

tani_B = 2* 1.732

tani_B = 3.464

tanθ_B = 3.464

therefore, the relation between the tanθ_A and tanθ_B is

tanθ_A / tanθ_B = 1.154 / 3.464

1.154* tanθ_B = 3.464* tanθ_A

therefore,

tanθ_B = 3 tanθ_A

2) The total intensity of Earth’s magnetic field

I = I₀ ( √(1+3 sin² Φ) ) ---------------(2)

Magnetic latitudes at A = 30⁰ and B= 60⁰

I_A = I₀ (√(1+3 sin² 30⁰) )

I_A = I₀ (√(1+(3*(1/2)²)))

I_A = I₀ (√(1+(3/4)) )

I_A = I₀ √(7/4)

I_A = (I₀* √7) / 2

at B= 60⁰

I_B = I₀ (√(1+3 sin² 60⁰) )

I_B = I₀ (√(1+3 * ((√3)/2)² ))

I_B = I₀ (√(1+3* (3/4) ))

I_B = I₀ (√(1+(9/4)) )

I_B = I₀ (√(13/4) )

I_B = I₀ (√(13) /2 )

I_B = (I₀ *√(13)) /2

Therefore, the relation between the intensities of Earth field is

I_A /I_B = ((I₀* √7) / 2) / ((I₀ *√(13)) /2)

I_A /I_B = √(7/13)

√13 I_A = √7 * I_B