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In the magnetic meridian of a certain pl...

In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is `0*26G` and dip angle is `60^@`. What is the magnetic field of earth at this location?

A

0.50G

B

0.52G

C

0.54G

D

0.56G

Text Solution

Verified by Experts

The correct Answer is:
A
`"Here", H_(E)=0.25G and cos delta=(H_(E))/(B_(E))`
The magnetic field of earth at the given location is
`B_(E)=(H_(E))/(cos 60^(@))=(0.25)/(1//2)=0.50G`
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