At a certain research station, the vertical component of the Earth's magnetic field is mea… — JAMB UTME

At a certain research station, the vertical component of the Earth's magnetic field is measured as $2.4 \times 10^{-5}\text{ T}$ . If the angle of dip at this station is $37^\circ$ (given $\sin 37^\circ = 0.60$ and $\cos 37^\circ = 0.80$ ), calculate the horizontal component of the Earth's magnetic field in $\mu\text{T}$ .

Answer: 32 μT

Answer

The horizontal component of the Earth's magnetic field is

32\ \mu\text{T}

The horizontal component

B_h

and vertical component

B_v

of the Earth's magnetic field are related by

\tan \theta = \frac{B_v}{B_h}

, where

\theta

is the angle of dip. Given

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

and

\tan 37^\circ = 0.75

, rearranging gives

B_h = \frac{24}{0.75} = 32\ \mu\text{T}

Step-by-Step Solution

Express the relationship between the vertical component (

B_v

), horizontal component (

B_h

), and angle of dip (

\theta

\tan \theta = \frac{B_v}{B_h}

By definition of the angle of dip in the magnetic meridian, the tangent of the dip angle equals the ratio of the vertical component to the horizontal component.

Calculate

\tan 37^\circ

using the provided trigonometric values.

\tan 37^\circ = \frac{0.60}{0.80} = 0.75

Tangent of an angle is the ratio of sine to cosine of that angle.

Convert

B_v

from teslas to microteslas.

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

Since

1\ \mu\text{T} = 10^{-6}\text{ T}

2.4 \times 10^{-5}\text{ T} = 24 \times 10^{-6}\text{ T} = 24\ \mu\text{T}

Rearrange the equation to solve for

B_h

and substitute the values.

B_h = \frac{B_v}{\tan \theta} = \frac{24\ \mu\text{T}}{0.75} = 32\ \mu\text{T}

Dividing

24

0.75

yields

32

Key Concept

Resolution of Earth's magnetic field into horizontal and vertical components

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic