The horizontal component of Earth's magnetic field at a place is $3.5 \times 10^{-5} \,T$. $A$ very long straight conductor carrying a current of $\sqrt{2} \,A$ in the direction from South-East to North-West is placed. The force per unit length experienced by the conductor is $..............$ $\times 10^{-6} \,N/m$.

At a place,the earth's horizontal component of magnetic field is $0.36 \times 10^{-4} \ Wb/m^2$. If the angle of dip at that place is $60^o$,then the vertical component of the earth's magnetic field at that place in $Wb/m^2$ will be approximately:

The lines joining the places of the same horizontal intensity are known as

The angle of dip at a place on the earth gives:

$A$ long vertical wire carries a steady current of $5.0 \, A$. $A$ sensitive magnetic compass is placed in a plane perpendicular to the wire and $10.0 \, cm$ south of it. It registers a deflection of $60^{\circ}$ north of east. The magnitude of the horizontal component of the earth's magnetic field is (permeability of free space is $4 \pi \times 10^{-7} \, N/A^2$):

At a certain place,the horizontal component ${B_0}$ and the vertical component ${V_0}$ of the Earth's magnetic field are equal in magnitude. The total intensity at the place will be