Question

At a certain location in Africa, compass point ${12}^{\circ }$ west of geographic north. The north tip of magnetic needle of a dip circle placed in the plane of magnetic meridian points ${60}^{\circ }$ above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. The magnitude of the earth’s field (in gauss) at the location is

Answer 1

Here angle of dip $\delta ={60}^{\circ }$

we know that $B_H=B\cos \delta$

where, $B_H$ = horizontal component of earth's field

B = earth's total magnetic field intensity
$0.16\times {10}^{-4}T=B\cos {60}^{\circ }$

$B=\frac{0.16\times {10}^{-4}}{\frac{1}{2}}$

$=0.32\times {10}^{-4}T$
= 0.32 G

Thus, magnitude of earth's field is 0.32 G.

Note : the compass will point towards magnetic north at any place , here compass point ${12}^{\circ }$ west of geographic north which means that the magnetic north ( hence the magnetic meridian ) is at ${12}^{\circ }$ from geographical meridian at given point.
hence , the declination at that point is ${12}^{\circ }$ . However this information is not used in solving this problem.