Question

The horizontal component of earth's magnetic field is $0.4\times {10}^{-4}T$ and the angle of dip is ${30}^{\circ }$. Calculate the value of
(i) vertical component
(ii) the total intensity of the earth's magnetic field.

Answer 1

i. Given $H=0.4\times {10}^{-4}Wb/m^2,\theta ={30}^o$
$\tan \theta =\frac{V}{H}\Rightarrow$ vertical component $V=H\tan \theta =0.4\times {10}^{-4}\times \tan {30}^o$
$=\frac{0.4\times {10}^{-4}}{\sqrt{3}}=0.23\times {10}^{-4}Wb/m^2$

ii.Total intensity of earth's megnetic field
$B_e=\sqrt{H^2+V^2}=\sqrt{{(0.4\times {10}^{-4})}^2+{\left(\frac{0.4\times {10}^{-4}}{\sqrt{3}}\right)}^2}=0.46\times {10}^{-4}Wb/m^2$