Question

At a location, vertical component of earth’s magnetic field is $1.6\times {10}^{-5}\text{T}$. If dip angle at that location is ${37}^{\circ }$, then find the magnitude of magnetic field at that point.

Answer 1

Given,
$\text{Vertical component}B=1.6\times {10}^{-5}\text{T}$

$\text{Angle of dip}\theta ={37}^{\circ }$

$B_v=B\sin \theta$

$\Rightarrow B=\frac{1.6\times {10}^{-5}}{\sin {37}^{\circ }}=\frac{1.6}{\frac{3}{5}}\times {10}^{-5}$

$\Rightarrow B=\frac{8}{3}\times {10}^{-5}=2.67\times {10}^{-5}\text{T}$

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