Q. The earth’s magnetic field at a certain place has a horizontal component $0.3\ gauss$ and the total strength $0.5\ gauss$. The angle of dip is
A) $\tan^{- 1}\frac{3}{4}$
B) $\sin^{- 1}\frac{3}{4}$
C) $\tan^{- 1}\frac{4}{3}$
D) $\sin^{- 1}\frac{3}{5}$
Explanation
(c) $B^{2} = B_{V}^{2} + B_{H}^{2} \Rightarrow B_{V} = \sqrt{B^{2} - B_{H}^{2}} = \sqrt{(0.5)^{2} - (0.3)^{2}} = 0.4$
Now $\tan\phi = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3} = \frac{4}{3} \Rightarrow \phi = \tan^{- 1}\left( \frac{4}{3} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.