Question

At a given place on earth’s surface the horizontal component of earth’s magnetic field is 2 x 10-5 T and resultant magnetic field is 4 x 10-5 T. The angle of dip at this place is

• A. 30°
• B. 60°
• C. $90 ^ { \circ }$
• D. $45 ^ { \circ }$

Answer 1

Answer: (B)

60°

Answer 2

: Since $\mathrm { B } _ { \mathrm { H } } = \mathrm { B } \cos \delta$

Hence,

$\therefore \cos \delta = \frac { B _ { H } } { B } = \frac { 2 \times 10 ^ { - 5 } } { 4 \times 10 ^ { - 5 } } = \frac { 1 } { 2 } = \cos 60 ^ { \circ }$

$\Rightarrow \delta = 60 ^ { \circ }$