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 $^o$)?

The magnetic field of the Earth at the equator is approximately $4 \times 10^{-5} \, T$. The radius of the Earth is $6.4 \times 10^6 \, m$. Then the dipole moment of the Earth will be nearly of the order of:

The values of the apparent angles of dip in two planes at right angles to each other are $45^{\circ}$ and $30^{\circ}$ respectively. The true value of the angle of dip at the place is ............

Assertion: If a compass needle is kept at the magnetic north pole of the Earth,the compass needle may stay in any direction.
Reason: $A$ dip needle will stay vertical at the north pole of the Earth.

$A$ compass needle is free to rotate in a horizontal plane. Its magnetic moment is $60 \ Am^2$. When it is pointing geographical north,it experiences a torque of $1.2 \times 10^{-3} \ Nm$ due to the Earth's magnetic field. If the Earth's magnetic field in the horizontal direction is $40 \times 10^{-6} \ T$,what is the declination in degrees at that place?