Question

At a place, the earth's horizontal component of magnetic field is $0.36 \times 10^{-4} Wb / m ^{2}$. If the angle of dip at that place is $60^{\circ}$, then the vertical component of earth's field at that place in $Wb / m ^{2}$ will be approximately

• A. $0.12 \times 10^{-4}$
• B. $0.24 \times 10^{-4}$
• C. $0.40 \times 10^{-4}$
• D. $0.62 \times 10^{-4}$

Answer 1

Answer: (D)

From the relation, $B_{H}=B \cos \phi$ and $B_{V}=B \sin \phi$
$\frac{B_{V}}{B_{H}}=\tan \phi$ or $B_{V}=B_{H} \tan \phi$
$=0.36 \times 10^{-4} \times \tan 60^{\circ}$
$=0.623 \times 10^{-4} Wb / m ^{2}$