At a certain place, the angle of dip is 60° and the horizontal component of the earth's magnetic field (BH) is 0.8× 10- 4 T. The earth's overall magnetic field is

Question

At a certain place, the angle of dip is 60° and the horizontal component of the earth's magnetic field (BH) is 0.8× 10- 4 T. The earth's overall magnetic field is (A) 1.5× 10- 4 T (B) 1.6× 10- 3 T (C) 1.5× 10- 3 T (D) 1.6× 10- 4 T

Tardigrade Admin · Accepted Answer

Given, BH=0.8× 10- 4T , θ =60° , Be= ? <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-hk8agjpwqvmkqhgq.jpg" /> We know that, BH=Be cos θ 0.8× 10- 4=Be cos 60° Be=(0.8 × 10- 4/(1)2) =1.6× 10- 4 T

Q. At a certain place, the angle of dip is $6 0^{\circ}$ and the horizontal component of the earth's magnetic field $(B_{H})$ is $0.8 \times 1 0^{- 4} T .$ The earth's overall magnetic field is

$1.5 \times 1 0^{- 4} T$

$1.6 \times 1 0^{- 3} T$

$1.5 \times 1 0^{- 3} T$

$1.6 \times 1 0^{- 4} T$

Given,
$B_{H} = 0.8 \times 1 0^{- 4} T$ ,
$θ = 6 0^{\circ}$ ,
$B_{e} = ?$

We know that,
$B_{H} = B_{e} cos θ$
$0.8 \times 1 0^{- 4} = B_{e} cos 6 0^{\circ}$
$B_{e} = \frac{0.8 \times 1 0 ^{- 4}}{\frac{1}{2}}$
$= 1.6 \times 1 0^{- 4} T$

Q. At a certain place, the angle of dip is 60∘ and the horizontal component of the earth's magnetic field (BH​) is 0.8×10−4T. The earth's overall magnetic field is

1.5×10−4T

1.6×10−3T

1.5×10−3T

1.6×10−4T