Kumar Mittal Solutions for Exercise 1: QUESTIONS

The horizontal component of earth’s magnetic field at a place is $\sqrt{3}$ times the vertical component. What is angle of dip at that place?

If the ratio of the horizontal component of earth’s magnetic field to the resultant magnetic field at a place is $\frac{1}{\sqrt{2}}$, what is the angle of dip at that place?

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at$60°$with the horizontal. The horizontal component of earth’s magnetic field is$0.3 \mathrm{x} {10}^{-4} \mathrm{Wb}/{\mathrm{m}}^2.$ Find the vertical component and total magnetic field of earth.

A magnetic needle free to rotate about the vertical direction (compass) points $3.5°$ west of the geographic north. Another magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian (dip needle) has its north tip pointing down at $18°$ with the horizontal. The horizontal component of earth’s magnetic field at the place is$0.40 \mathrm{G}$. Find the magnitude and direction of the earth’s magnetic field ${\overrightarrow{\mathrm{B}}}_{\mathrm{E}}$at the place $\left(\tan 18°=0.325\right)$

A short bar-magnet is placed with its north pole pointing north. The neutral point is $10 \mathrm{cm}$ away from the centre of the magnet. If ${\mathrm{B}}_{\mathrm{H}} = 0.4 \mathrm{G},$calculate the magnetic moment of the magnet. 

The magnetic moment of a small magnet is $4.8 \mathrm{A} {\mathrm{m}}^2$. It is kept in earth’s magnetic field in magnetic meridian with its north pole pointing towards geographical south. The neutral points are situated at a distance of $30 \mathrm{cm}$ from the centre of the magnet. Calculate the horizontal component of the earth’s magnetic field.

A small bar-magnet is placed in a horizontal plane with its axis in the magnetic meridian. Neutral points are obtained on its equatorial line at $12.5 \mathrm{cm}$ from the centre of the magnet. The earth’s magnetic field at the place is $0.38 \mathrm{G}$ and the angle of dip is zero, (a) Find the total magnetic field on the axis of the magnet at $12.5 \mathrm{cm}$ from the centre of the magnet, 

A small bar magnet is placed in a horizontal plane with its axis in the magnetic meridian. Neutral points are obtained on its equatorial line at $12.5 \mathrm{cm}$ from the centre of the magnet. The earth’s magnetic field at the place is $0.38 \mathrm{G}$ and the angle of dip is zero, (b) Locate the neutral points when the magnet is turned around by $180°.$