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

If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then:

Give the location of magnetic:
$(i)$ north pole
$(ii)$ south pole

If a magnet is suspended at an angle $30^o$ to the magnetic meridian,it makes an angle of $45^o$ with the horizontal. The real dip is

$A$ compass needle whose magnetic moment is $60 \, A \cdot m^2$ is pointing towards the geographical north at a certain place,where the horizontal component of the Earth's magnetic field is $40 \, \mu Wb/m^2$. It experiences a torque of $1.2 \times 10^{-3} \, N \cdot m$. What is the angle of declination at this place (in $^o$)?

$A$ long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth's magnetic field at the location is $0.33\;G,$ and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points,magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)