The horizontal component of the Earth's magnetic field is $0.22 \ G$ and the total magnetic field is $0.4 \ G$. The angle of dip is:

At the place where the angle of dip is $60^{\circ}$,what is the relation between the vertical component $Z_{E}$ and the horizontal component $H_{E}$ of the 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 $8 \times 10^{22} \text{ Am}^2$,the value of earth's magnetic field near the equator is close to $.... \text{ Gauss}$ (radius of the earth $= 6.4 \times 10^6 \text{ m}$)

Answer the following questions:
$(a)$ The earth's magnetic field varies from point to point in space. Does it also change with time? If so,on what time scale does it change appreciably?
$(b)$ The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?
$(c)$ The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e.,the source of energy) to sustain these currents?
$(d)$ The earth may have even reversed the direction of its field several times during its history of $4$ to $5$ billion years. How can geologists know about the earth's field in such distant past?
$(e)$ The earth's field departs from its dipole shape substantially at large distances (greater than about $30,000\; km$). What agencies may be responsible for this distortion?
$(f)$ Interstellar space has an extremely weak magnetic field of the order of $10^{-12}\; T$. Can such a weak field be of any significant consequence? Explain.

$A$ small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^o$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

The vertical component of the earth's magnetic field is $6 \times 10^{-5} \text{ T}$ at a place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be (Given $\tan 37^{\circ} = \frac{3}{4}$)