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 ${45}^{\circ }$, then the vertical component of earth's magnetic field at that place in $\mu \text{T}$ will be

Answer 1

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

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

$\therefore B_V=B_H\tan \delta$

$B_V=0.36\times {10}^{-4}\times \tan {45}^{\circ }$

$=0.36\times {10}^{-4}{\text{Wb/m}}^2$

$=36\mu \text{T}$

Accepted answers$:36,36.036.00$