Question

At a place, the earth's horizontal component of magnetic field is $0.38\times {10}^{-4}\text{T}$. If the angle of dip at that place is ${60}^{\circ }$, then the vertical component of earth's magnetic field at that place will be approximately -

• A. $0.12\times {10}^{-4}\text{T}$
• B. $0.24\times {10}^{-4}\text{T}$
• C. $0.40\times {10}^{-4}\text{T}$
• D. $0.65\times {10}^{-4}\text{T}$

Answer 1

The correct option is D $0.65\times {10}^{-4}\text{T}$

We know that,

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

$\Rightarrow \tan {60}^{\circ }=\frac{B_V}{0.38\times {10}^{-4}}$

$\Rightarrow B_V=0.65\times {10}^{-4}\text{T}$

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