Q. The value of the horizontal component of the earth’s magnetic field and angle of dip are $1.8 \times 10^{- 5}weber/m^{2}$ and $30{^\circ}$ respectively at some place. The total intensity of earth’s magnetic field at that place will be
A) $2.08 \times 10^{- 5}weber/m^{2}$
B) $3.67 \times 10^{- 5}weber/m^{2}$
C) $3.18 \times 10^{- 5}weber/m^{2}$
D) $5.0 \times 10^{- 5}weber/m^{2}$
Explanation
(a) Horizontal component $B_{H} = B\cos\phi$
Total intensity of earth’s magnetic field $B = \frac{B_{H}}{\cos\phi}$
$= \frac{1.8 \times 10^{- 5}}{\cos{30{^\circ}}} = \frac{1.8 \times 10^{- 5}}{\sqrt{3}/2} = 2.08 \times 10^{- 5}Wb/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.