Question

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known at to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer 1

Angle of dip, θ = 60°

H = 0.4 G = 0.4 × 10^–4 T

If Be is earth’s magnetic field, then

H = B_e cos θ \(\Rightarrow\) B_e = \(\cfrac{H}{cos\,\theta}\) = \(\cfrac{0.4\times 10^{-4}T}{cos\,60^\circ}\) = \(\cfrac{0.4\times 10^{-4}T}{0.5}\) = 0.8 × 10^-4 T = 0.8 G