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 $\text{0} \text{.35 G}$ . Determine the magnitude of the earth's magnetic field at the place. $\left(cos 22 ^\circ = 0.9272\right)$

• A. $0.38G$
• B. $0.35G$
• C. $0.30G$
• D. $0.40G$

Answer 1

Answer: (A)

Given, angle of dip $\delta=22^\circ $
Horizontal component of the earth's magnetic field $H=0.35G$
Let the magnitude of the earth's magnetic field at the place is $R$ .
Using the formula, $H=Rcos\delta$
or $R=\frac{H}{cos \delta}=\frac{0 . 35}{cos 22 ^\circ }=\frac{0 . 35}{0 . 9272}=0.38G$
Thus, the value of the earth's magnetic field at that place is $0.38G$ .