The magnetic needle of a tangent galvanometer is deflected at an angle $30^\circ$ due to a magnet. The horizontal component of Earth's magnetic field $0.34 \times 10^{-4} \, T$ is along the plane of the coil. The magnetic intensity is

The period of oscillation of a magnet is $2 \, s$. When it is remagnetised so that the pole strength is $4$ times its original value,its new period will be......$s$.

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ (about its center,perpendicular to its length) is cut into two equal pieces,perpendicular to its length. Let $T$ be the period of oscillation of the original magnet about an axis through its midpoint,perpendicular to its length,in a uniform magnetic field $B$. What would be the period $T'$ for each piece?

$A$ tangent galvanometer has a coil of $25$ turns and a radius of $15\, cm$. The horizontal component of the earth's magnetic field is $3 \times 10^{-5}\, T$. The current required to produce a deflection of $45^{\circ}$ in it is....$A$

The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?