Question

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22˚ with the horizontal. The horizontal component of the earth’ magnetic field at the place is known to be 0.35 G. Determine the strength of the earth’s magnetic field at the place.

Answer 1

Here,

δ = 22˚, 

H = 0.35 G, 

R = ? 

As, 

H = R cosδ 

⇒ R = \(\frac{H}{cosδ}\) 

= \(\frac{0.35}{cos22°}\)

= \(\frac{0.35}{0.9272}\)

= 0.38 G