The horizontal component of the earth's magnetic field at any place is 0.36× 10- 4 Wb m- 2 . If the angle of dip at that place is 60° then the value of the vertical component of earth's magnetic field will be (in Wb m- 2 )

Question

The horizontal component of the earth's magnetic field at any place is 0.36× 10- 4 Wb m- 2 . If the angle of dip at that place is 60° then the value of the vertical component of earth's magnetic field will be (in Wb m- 2 ) (A) 0.12× 10- 4 (B) 0.24× 10- 4 (C) 0.40× 10- 4 (D) 0.62× 10- 4

Tardigrade Admin · Accepted Answer

Given, BH=0.36× 10- 4Wbm- 2 Angle of dip tanθ =(BV/BH) tan60° =(BV/0.36 × 1 0- 4) BV=(0.36)(√3)× 10- 4 BV=0.62× 10- 4Wbm- 2

Q. The horizontal component of the earth's magnetic field at any place is $0.36 \times 1 0^{- 4} Wb m^{- 2}$ . If the angle of dip at that place is $6 0^{\circ}$ then the value of the vertical component of earth's magnetic field will be (in $Wb m^{- 2}$ )

$0.12 \times 1 0^{- 4}$

$0.24 \times 1 0^{- 4}$

$0.40 \times 1 0^{- 4}$

$0.62 \times 1 0^{- 4}$

Given,
$B_{H} = 0.36 \times 1 0^{- 4} Wb m^{- 2}$
Angle of dip
$t an θ = \frac{B _{V}}{B _{H}}$
$t an 6 0^{\circ} = \frac{B _{V}}{0.36 \times 1 0 ^{- 4}}$
$B_{V} = (0.36) (3) \times 1 0^{- 4}$
$B_{V} = 0.62 \times 1 0^{- 4} Wb m^{- 2}$

Q. The horizontal component of the earth's magnetic field at any place is 0.36×10−4Wbm−2 . If the angle of dip at that place is 60∘ then the value of the vertical component of earth's magnetic field will be (in Wbm−2 )

0.12×10−4

0.24×10−4

0.40×10−4

0.62×10−4