The horizontal component of the Earth's magnetic field is 3.6 × 10...

The horizontal component of the Earth's magnetic field is

3.6 \times {10}^{- 5}T

where the dip angle is

60 °

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

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

Options

A $2.8 \times {10}^{- 4}tesla$
B $2.1 \times {10}^{- 4}tesla$
C $7.2 \times {10}^{- 5}tesla$
D $3.6 \times {10}^{- 5}tesla$

Correct answer

C. $7.2 \times {10}^{- 5}tesla$

Step-by-step solution

Horizontal component of Earth’s field,

H = Bcos\theta,

since,

\theta = 60 °

3.6 \times {10}^{- 5} = B \times \dfrac{1}{2} ⇒ B = 7.2 \times {10}^{- 5}tesla

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