At a certain place the angle of $\operatorname{dip}$ is $30^{\circ}$ and the horizontal component of

Question

At a certain place the angle of $\operatorname{dip}$ is $30^{\circ}$ and the horizontal component of earth's magnetic field is 0.5 G . The earth's total magnetic field (in G ), at that certain place, is\

Professor PC Thomas Classes · Answer

(a) : Given, horizontal component of earth's magnetic field, $B_{H}=0.5 \mathrm{G}$, angle of dip, $	heta=30^{\circ}$\ We know that, $B_{H}=B \cos 	heta$\ $	herefore$ Total magnetic field,\ $B=\frac{B_{H}}{\cos 	heta}$ or $B=\frac{0.5}{\cos 30^{\circ}}=\frac{1}{\sqrt{3}} \mathrm{G}$\