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 × 10-4 T θ =60° Be =? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c23773ace0ce7c0ea100c36159b936ff-.png" /> 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