Earth's Magnetic Field Simulation
Example 5.9
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is \(0.26G\) and the dip angle is \(60^\circ\). What is the magnetic field of the earth at this location?
Given: \(H_E = 0.26 \, \text{G}\), \(\theta = 60^\circ\)
\(\cos \theta = \frac{H_E}{B_E}\)
\(B_E = \frac{H_E}{\cos \theta} = \frac{0.26}{\cos 60^\circ} = \frac{0.26}{0.5} = 0.52 \, \text{G}\)
\(\cos \theta = \frac{H_E}{B_E}\)
\(B_E = \frac{H_E}{\cos \theta} = \frac{0.26}{\cos 60^\circ} = \frac{0.26}{0.5} = 0.52 \, \text{G}\)
Interactive Simulation
Adjust the dip angle to see how it affects the Earth's magnetic field components:
60°
Dip Angle (θ)
60°
Horizontal Component (HE)
0.26 G
Total Field (BE)
0.52 G
Vertical Component (VE)
0.45 G
