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 $\left.\tan 37^{\circ}=\frac{3}{4}\right)$

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.)

The earth's magnetic field lines resemble that of a dipole at the centre of the earth. If the magnetic moment of this dipole is close to $8 \times 10^{22}\, Am^2$, the value of earth 's magnetic field near the equator is close to....$Gauss$ (radius of the earth $= 6.4 \times 10^6\, m$)

The pole near the geographic north pole is called ...... . The pole near the geographic south pole is called ...... .

The earth's magnetic field at a certain place has a horizontal component $0.3$ $Gauss$ and the total strength $0.5$ $Gauss$. The angle of dip is

Lines which represent places of constant angle of dip are called