Solveeit Logo
Login

Question

Question: In the magnetic meridian of a certain place the horizontal component of earth’s magnetic field is 0....

In the magnetic meridian of a certain place the horizontal component of earth’s magnetic field is 0.25 G and dip angle is 60°. The magnetic field of the earth at this location is

A

0.50 G

B

0.52 G

C

0.54 G

D

0.56 G

Answer

0.50 G

Explanation

Solution

: Here, , HE=0.25G\mathrm { H } _ { \mathrm { E } } = 0.25 \mathrm { G } and

∴\therefore The magnetic field of earth at the given location is

BE=HEcos⁡60∘=0.251/2=0.50GB _ { E } = \frac { H _ { E } } { \cos 60 ^ { \circ } } = \frac { 0.25 } { 1 / 2 } = 0.50 \mathrm { G }