Question

On suspending a magnet at 30° with the magnetic meridian, it makes an angle of 45° with the horizontal. What will be the actual angle of dip?

Hint:

The angle of dip can be determined using the angle of suspension and the angle with the magnetic meridian. The formula involves the tangent and cosine of the given angles.

Answer 1

The actual angle of dip \( \delta \) can be determined using the relationship between the angle of suspension \( \theta \), the magnetic declination (the angle with the magnetic meridian), and the angle of dip. The formula to calculate the actual angle of dip is given by:
\[ \tan(\delta) = \tan(\theta) \times \cos(\alpha) \] where:
- \( \delta \) is the actual angle of dip,
- \( \theta = 45^\circ \) is the angle between the magnet and the horizontal,
- \( \alpha = 30^\circ \) is the angle the magnet makes with the magnetic meridian.
Substituting the values into the equation:
\[ \tan(\delta) = \tan(45^\circ) \times \cos(30^\circ) \] We know that \( \tan(45^\circ) = 1 \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \). Therefore:
\[ \tan(\delta) = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \] Now, to find \( \delta \), we take the inverse tangent of both sides:
\[ \delta = \tan^{-1}\left( \frac{\sqrt{3}}{2} \right) \] Using a calculator:
\[ \delta \approx 60^\circ \] Thus, the actual angle of dip is approximately:
\[ \boxed{60^\circ} \]