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In the magnetic meridian of a certain pl...

In the magnetic meridian of a certain place the vertical component of the earth's magnetic field is 0.3464 G and the dip angle is `30^(@)` . The horizontal component of the earth 's magnetic field at this location is

A

0.4 G

B

0.6 G

C

0.7 G

D

0.8 G

Text Solution

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The correct Answer is:
B
`tan delta = (B_(v))/(B_(H)) implies B_(H)= (B_(v))/(tandelta)= (0.3464)/(1//sqrt(3))`
= `0.3464xx1.73 = 0.6` G
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