# A magnetic needle free to rotate in a vertical plane

Question:

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

Solution:

Horizontal component of earth’s magnetic field, BH = 0.35 G

Angle made by the needle with the horizontal plane $=$ Angle of dip $=\delta=22^{\circ}$

Earth’s magnetic field strength = B

We can relate and BHas:

$B_{H}=B \cos \theta$

$\therefore B=\frac{B_{H}}{\cos \delta}$

$=\frac{0.35}{\cos 22^{\circ}}=0.377 \mathrm{G}$

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.