The vertical component of the earth's magnetic field is $6 imes 10^{-5} \mathrm{~T}$ at any place

Question

The vertical component of the earth's magnetic field is $6 	imes 10^{-5} \mathrm{~T}$ at any place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be (Given $	an 37^{\circ}=\frac{3}{4}$ )\

Professor PC Thomas Classes · Answer

(d) : Given, vertical component of the Earth's magnetic field, $B_{V}=6 	imes 10^{-5} \mathrm{~T}, \delta=37^{\circ}$\ $	an \delta=\frac{B_{V}}{B_{H}} \Rightarrow 	an 37^{\circ}=\frac{6 	imes 10^{-5}}{B_{H}}$\ $B_{H}=\frac{6 	imes 10^{-5}}{	an 37^{\circ}}=\frac{6 	imes 10^{-5}}{3 / 4}$\ $\Rightarrow B_{H}=6 	imes 10^{-5} 	imes \frac{4}{3}=8 	imes 10^{-5} \mathrm{~T}$\ $	herefore B=\sqrt{B_{V}^{2}+B_{H}^{2}}$\ $=\sqrt{\left(6 	imes 10^{-5}ight)^{2}+\left(8 	imes 10^{-5}ight)^{2}}$\ $=10^{-5} \sqrt{36+64}=10^{-4} \mathrm{~T}$\