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. $\left(cos22^{\circ }=0.9272\right)$

• 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 $R=\frac{H}{cos\delta }=\frac{0.35}{cos22^{\circ }}=\frac{0.35}{0.9272}=0.38G$
Thus, the value of the earth's magnetic field at that place is $0.38G$ .