At a place, the earth's horizontal component of magnetic field is 0.36 × 10-4 Wb / m 2. If the angle of dip at that place is 60°, then the vertical component of earth's field at that place in Wb / m 2 will be approximately

Question

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

Tardigrade Admin · Accepted Answer

From the relation, BH=B cos φ and BV=B sin φ (BV/BH)= tan φ or BV=BH tan φ =0.36 × 10-4 × tan 60° =0.623 × 10-4 Wb / m 2

Q. At a place, the earth's horizontal component of magnetic field is $0.36 \times 1 0^{- 4} Wb / m^{2}$ . If the angle of dip at that place is $6 0^{\circ}$ , then the vertical component of earth's field at that place in $Wb / m^{2}$ will be approximately

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

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

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

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

From the relation, $B_{H} = B cos ϕ$ and $B_{V} = B sin ϕ$
$\frac{B _{V}}{B _{H}} = tan ϕ$ or $B_{V} = B_{H} tan ϕ$
$= 0.36 \times 1 0^{- 4} \times tan 6 0^{\circ}$
$= 0.623 \times 1 0^{- 4} Wb / m^{2}$

Q. At a place, the earth's horizontal component of magnetic field is 0.36×10−4Wb/m2. If the angle of dip at that place is 60∘, then the vertical component of earth's field at that place in Wb/m2 will be approximately

0.12×10−4

0.24×10−4

0.40×10−4

0.62×10−4