Question

The horizontal component of earth’s magnetic field at a place is \(0.4 \times 10^{-4}\,T\). If the angle of dip is \(45^\circ\), the value of total intensity is:

• A. \(0.5 \times 10^{-4}\,T\)
• B. \(0.4 \times 10^{-4}\,T\)
• C. \(0.5 \times 10^{-6}\,T\)
• D. \(0.4 \times 10^{-6}\,T\)

Hint:

Use \( B = \frac{B_H}{\cos\theta} \).

Answer 1

The Correct Option is A

To determine the total intensity of Earth's magnetic field at a place given the horizontal component and the angle of dip, we need to apply the relationship between these quantities.

The total intensity of Earth's magnetic field \( B \) is related to the horizontal component \( B_H \) and the angle of dip \( \delta \) by the formula:

\(B = \frac{B_H}{\cos \delta}\)

Here, the horizontal component \( B_H \) is given as \( 0.4 \times 10^{-4}\,T \) and the angle of dip \( \delta \) is \( 45^\circ \).

Recall that \(\cos 45^\circ = \frac{1}{\sqrt{2}}\)

Substituting the given values into the formula:

\(B = \frac{0.4 \times 10^{-4}}{\cos 45^\circ} = \frac{0.4 \times 10^{-4}}{\frac{1}{\sqrt{2}}} = 0.4 \times 10^{-4} \times \sqrt{2}\)

Simplifying further, knowing that \(\sqrt{2} \approx 1.414\):

\(B = 0.4 \times 10^{-4} \times 1.414 = 0.5656 \times 10^{-4}\,T\)

Rounding to a reasonable number of significant figures gives \( 0.5 \times 10^{-4}\,T \), which matches with option 1.

Thus, the total intensity of Earth's magnetic field at the place is \( \mathbf{0.5 \times 10^{-4}\,T} \).

The correct answer is:

• \(0.5 \times 10^{-4}\,T\)