A compass needle of magnetic moment $60{\mathrm{Am}}^2$ pointing geographical north at a certain place where the horizontal components of earth's magnetic field is $40\mathrm{\mu T}$, experiences a torque of $1.2\times {10}^{-3} \mathrm{Nm}$. What is the declination of the place?

The vertical component of the earth's magnetic field at a given place is $\sqrt{3}$ times its horizontal components. If the total intensity of the earth's magnetic field at the place is $0.4G$. Find the value of 

(i) angle of dip

(ii) the horizontal component of the earth's magnetic field.

The horizontal component of the earth's magnetic field's at a given place is $0.4\times {10}^{-4} \mathrm{Wb}/{\mathrm{m}}^2$ and angle of dip is $30°$. Calculate the value of 

(i) vertical component

The horizontal component of earth's magnetic field's at a given place is $0.4\times {10}^{-4} \mathrm{Wb}/{\mathrm{m}}^2$ and angle of dip is $30°$. Calculate the value of 

(ii) total intensity of earth's magnetic field.