The magnetic field due to the earth has a horizontal component of `26 muT` at a place where the dip is `60^@`. Find the vertical component and the magnitude of the field.

In the magnetic meridian of a certain place, the horizontal component of earth's magnetic fied is 0.26 G and the dip angle is 60^@ . Find Vertical component of earth's magnetic field b. the net magnetic field at this place

The horizontal component of earth’s magnetic field at a place is sqrt""3 times the vertical component. The angle of dip at that place is

The horizontal component of earth's magnetic field at a place is sqrt(3) Time the vertical component. What is the angle of dip at this place?

In the magnetic meridian of a certain place, the horizontal component of earth's magnetic fied is 0.26 G and the dip angle is 60^@ . Find a. Vertical component of earth's magnetic field b. the net magnetic field at this place

Horizontal component of earth's magnetic field at a place is sqrt3 times the vertical component. What is the value of angle of dip at this place?

If the horizontal component of earth's magnetic field at a place where the angle of dip is 60^@ is 0*4xx10^-4T , calculate the vertical component and the resultant magnetic field at that place.

The horizontal component of the earth's magnetic field is 3.6 xx 10^(-5) T where the dip is 60^@ . Find the magnitude of the earth's magnetic field.