The vertical component of the earth's magnetic field is $6 \times 10^{-5} T$ at any place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be $\left(\right.$ Given $\left.\tan 37^{\circ}=\frac{3}{4}\right)$

The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$

Let $V $ and $H $ be the vertical and horizontal components of earth's magnetic field at any point on earth. Near the north pole

At a point on the surface of the earth, the value of the horizontal component of earth's magnetic field is equal to value of vertical magnetic component of earths magnetic field the angle of dip is

Intensity of magnetic field due to earth at a point inside a hollow steel box is

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} Wb / m ^{2}$. If the angle of dip at that place is $60^{\circ}$, then the value of vertical component of the earth's magnetic field will be ........ $\times 10^{-4}\;W b / m^{2}$