Q. The horizontal component of earth's magnetic field at any place is $0.3\times 10^{- 4}Wb/m^{2}$ . If the angle of dip at that place is $60^\circ $ , then what will be the value of vertical component of earth's magnetic field ?(in $Wb/m^{2}$ )
NTA AbhyasNTA Abhyas 2020
Solution:
Given; $B_{H =}0.3\times 10^{- 4}T$
We know; $tan\delta=\frac{B_{v}}{B_{H}}$
Where $\delta$ is the angle of dip, $B_{v}$ is the vertical component of earth's magnetic field and $B_{H}$ is the horizontal component of earth's magnetic field.
$\Rightarrow B_{v}=B_{H}tan\delta=0.03\times 10^{- 4}\times tan60^\circ =0.03\times 10^{- 4}\times \sqrt{3}=0.0519\times 10^{- 4}T$
We know; $tan\delta=\frac{B_{v}}{B_{H}}$
Where $\delta$ is the angle of dip, $B_{v}$ is the vertical component of earth's magnetic field and $B_{H}$ is the horizontal component of earth's magnetic field.
$\Rightarrow B_{v}=B_{H}tan\delta=0.03\times 10^{- 4}\times tan60^\circ =0.03\times 10^{- 4}\times \sqrt{3}=0.0519\times 10^{- 4}T$