$A$ bar magnet $A$ of magnetic moment $M_A$ is found to oscillate at a frequency twice that of magnet $B$ of magnetic moment $M_B$ when placed in a vibrating magnetometer. We may say that

$A$ short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane,the net magnetic induction at $P$ is (Horizontal component of earth's magnetic field = $B_H$)

$A$ very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained $20 \, cm$ away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point is $0.3 \, gauss$,the magnetic moment of the magnet is:

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

The tangent galvanometer,when connected in series with a standard resistance,can be used as:

The deflection produced in a tangent galvanometer,whose coil has a resistance of $9 \Omega$,is $30^{\circ}$. The potential difference across the coil is $4.5 \ V$. If the number of turns in the coil is $10$,the radius of the coil is (Given,$B_{H} = 3.14 \times 10^{-5} \ T$)