The horizontal component of the Earth's magnetic field is 3.6× 10- 5 T where the dip angle is 60° . The magnitude of the Earth's magnetic field is

Question

The horizontal component of the Earth's magnetic field is 3.6× 10- 5 T where the dip angle is 60° . The magnitude of the Earth's magnetic field is (A) 2.8× 10- 4tesla (B) 2.1× 10- 4tesla (C) 7.2× 10- 5tesla (D) 3.6× 10- 5tesla

Tardigrade Admin · Accepted Answer

Horizontal component of Earthâs field, H = B cos θ , since, θ =60° <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-rriver6jkckt.png" /> 3.6 × 10- 5 = B × (1/2) ⇒ B = 7.2 × 10- 5 tesla

Q. The horizontal component of the Earth's magnetic field is $3.6 \times 1 0^{- 5} T$ where the dip angle is $6 0^{\circ}$ . The magnitude of the Earth's magnetic field is

$2.8 \times 1 0^{- 4} t es l a$

$2.1 \times 1 0^{- 4} t es l a$

$7.2 \times 1 0^{- 5} t es l a$

$3.6 \times 1 0^{- 5} t es l a$

Horizontal component of Earth’s field,
$H = B cos θ,$ since, $θ = 6 0^{\circ}$

$3.6 \times 1 0^{- 5} = B \times \frac{1}{2} \Rightarrow B = 7.2 \times 1 0^{- 5}$ $t es l a$

Q. The horizontal component of the Earth's magnetic field is 3.6×10−5T where the dip angle is 60∘ . The magnitude of the Earth's magnetic field is

2.8×10−4tesla

2.1×10−4tesla

7.2×10−5tesla

3.6×10−5tesla