The angle of dip at a particular place where the horizontal intensity is `30Am^(-1)` is found to be `38^(@)` . Calculate the total inte4nsity of the earth's magnetic field there.

The angle of dip at a place is 60^(@) . At this place the total intensity of earth's magnetic field is 0.64 units. The horizontal intensity of earth's magnetic field at this place is

The angle of dip at a place is 40.6^(@) and the intensity of the vertical component of the earth's magnetic field V=6xx10^(-5) Tesla. The total intensity of the earth's magnetic field (I) at this place is

The horizontal compound of the earth's magnetic field at a given place is 0.4xx10^(-4) Wb//m^(2) and angle of dip =30^(@) . What is the total intensity of the earth's magnetic field at that place?

The angle of dip at a place where horizontal and vertical components of earth magnetic field are same, is

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.

Our earth behaves as it has a powerful magnet within it. The value of magnetic field on the surface of earth is a few tenth of gauss (1G = 10^(-4) T) There are three elements of Earth’s magnetism (i) Angle of declination (ii) Angle of dip (iii) Horizontal component of Earth’s magnetic field. In the magnetic meridian of a certain place, the horizontal component of Earth’s magnetic field is 0.6 G and the dip angle is 53^@ . The value of vertical component of earth's magnetic field at this place is

The angle of dip at a certain place is 30^(@) . If the horizontal component of the earth's magnetic field is H, the intensity of the total magnetic field is