The horizontal component of the earth's magnetic field at any place is `0.36 xx 10^(-4) Wb m^(-2)` If the angle of dip at that place is `60 ^@` then the value of the vertical component of earth's magnetic field will be ( in `Wb m^(-2))`

The value of horizontal component of earth's magnetic field at a place is -.35xx10^(-4)T . If the angle of dip is 60^(@) , the value of vertical component of earth's magnetic field is about:

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

At a place the earth's horizontal component of magnetic field is 0.36xx10^(-4) "Weber"//m^(2) . If the angle of dip at that place is 60^(@) , then the vertical component of earth's field at that place in "Weber"//m^(2) will be approxmately

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 horizontal component of earth's magnetic field at a place is 0*4xx10^-4T . If angle of dip 45^@ , what are the values of vertical component and total field strength of earth's field?

The horizontal component of the earth's magnetic field is 0.22 Gauss and total magnetic field is 0.4 Gauss. The angle of dip. Is

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