The time period of a magnet is $2 \, s$. The horizontal component of the Earth's magnetic field is $H$. When an external magnetic field $F$ is applied in the same direction,the new time period becomes $1 \, s$. What is the ratio $H/F$?

Magnetic moments of two bar magnets may be compared with the help of

If the number of turns and the radius of the cross-section of the coil of a tangent galvanometer are both doubled,then the reduction factor $K$ will become:

$A$ 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 is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

The radius of the coil of a Tangent galvanometer,which has $10$ turns,is $0.1\,m$. The current required to produce a deflection of $60^{\circ}$ $(B_H = 4 \times 10^{-5}\,T)$ is.....$A$