Question

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at \(22^{\circ} \) 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.

Answer 1

Given, Horizontal component of earth’s Magnetic field, \( B_H =0.35~G\)
Angle made by the needle with the horizontal plane= angle of dip, \( \delta =22^{\circ} \)
Now, if \(B\) is the earth’s magnetic field at that point, then

\( B_H = B~\cos \delta \)

\( B = \dfrac{B_H}{\cos \delta} \)

\( B = \dfrac{0.35}{\cos ~22^{\circ} } =0.35~\sec~22^{\circ} \)

\( B = 0.35 \times 1.078 \approx 0.377~G\)

Final Answer: \( 0.377~G\)