Question

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

Answer 1

We know that the angle of dip can be expressed as,

$\tan \delta =\frac{B_V}{B_H}$

$\therefore B_V=B_H\tan \delta$

$B_V=0.36\times {10}^{-4}\times \tan {60}^{\circ }$

$B_V=0.62\times {10}^{-4}{\text{Wb/m}}^2$

Hence, option $\left(D\right)$ is the correct answer.