Question

A magnetic needle points down at \(30^\circ\) to the horizontal. Horizontal component of earth’s magnetic field = 0.3 G. Find the total magnetic field.

• A. \( \frac{\sqrt{3}}{2} \ \text{G} \)
• B. \( \sqrt{3}\ \text{G} \)
• C. \( \frac{20}{\sqrt{3}} \ \text{G} \)
• D. \( \frac{2}{\sqrt{3}} \ \text{G} \)

Hint:

Use \( B = \frac{B_H}{\cos \theta} \) to find total field from horizontal component.

Answer 1

The Correct Option is A

The horizontal component is: \[ B_H = B \cos \theta = 0.3,\quad \theta = 30^\circ \Rightarrow B = \frac{B_H}{\cos 30^\circ} = \frac{0.3}{\sqrt{3}/2} = \frac{0.6}{\sqrt{3}} = \frac{\sqrt{3}}{2}\ \text{G} \]