Question

The horizontal component of earth's magnetic field at any place is $0.3\times 10^{-4}Wb/m^2$ . If the angle of dip at that place is $60^{\circ }$ , then what will be the value of vertical component of earth's magnetic field ?(in $Wb/m^2$ )

• A. $0.12\times 10^{-4}$
• B. $0.017\times 10^{-4}$
• C. $0.40\times 10^{-4}$
• D. $0.0519\times 10^{-4}$

Answer 1

Answer: (D)

Given; $B_{H=}0.3\times 10^{-4}T$
We know; $tan\delta =\frac{B_v}{B_H}$
Where $\delta$ is the angle of dip, $B_v$ is the vertical component of earth's magnetic field and $B_H$ is the horizontal component of earth's magnetic field.
$\Rightarrow B_v=B_{H}tan\delta =0.03\times 10^{-4}\times tan60^{\circ }=0.03\times 10^{-4}\times \sqrt{3}=0.0519\times 10^{-4}T$