Question

The vertical component of the earth’s magnetic field is 6 × 10^–5 T at any place where the angle of dip is 37°. The earth’s resultant magnetic field at that place will be (Given tan 37° = ¾)

• A. 8x10^-5 T
• B. 6x10^-5 T
• C. 5x10^-4 T
• D. 1x10^-4 T

Answer 1

The Correct Option is D

To solve this question, we need to determine the earth's resultant magnetic field at a place where the vertical component is given, and the angle of dip is provided.

Concept: The earth’s magnetic field at any place can be resolved into two components:

• Vertical component \(B_V\) 
• Horizontal component \(B_H\)

The angle of dip (\( \delta \)) is related to these components by the following relations:

• \( B_V = B \cdot \sin(\delta) \)
• \( B_H = B \cdot \cos(\delta) \)

Given:

• The vertical component \(B_V = 6 \times 10^{-5} \, \text{T}\)
• Angle of dip \( \delta = 37^\circ \)
• \( \tan(37^\circ) = \frac{3}{4} \)

Let's calculate the earth's resultant magnetic field:

First, use the relation for the vertical component:

\(B_V = B \cdot \sin(37^\circ)\)

But we know:

\(\sin(37^\circ) = \frac{\tan(37^\circ)}{\sqrt{1 + \tan^2(37^\circ)}}\)

Calculating \( \sin(37^\circ) \):

\( \tan(37^\circ) = \frac{3}{4} \) therefore:

\(\sin(37^\circ) = \frac{\frac{3}{4}}{\sqrt{1 + \left(\frac{3}{4}\right)^2}} = \frac{\frac{3}{4}}{\sqrt{1+\frac{9}{16}}} = \frac{\frac{3}{4}}{\sqrt{\frac{25}{16}}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}\)

Now substitute the values:

\( B_V = B \cdot \frac{3}{5} \)

\( 6 \times 10^{-5} = B \cdot \frac{3}{5} \)

Solving for \( B \):

\(B = \frac{6 \times 10^{-5} \times 5}{3} = \frac{30 \times 10^{-5}}{3} = 1 \times 10^{-4} \, \text{T}\)

Therefore, the earth's resultant magnetic field at that place is 1 × 10^-4 T.

Conclusion: The correct answer is:

1 × 10^-4 T