$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{5/4} \ s$. One of the magnets is removed and if the other magnet oscillates in the same field,then the time period in seconds is:

$A$ dip circle lies initially in the magnetic meridian,it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

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

Two tangent galvanometers $A$ and $B$ have coils of radii $8 \text{ cm}$ and $16 \text{ cm}$ respectively and have a resistance of $8 \Omega$ each. They are connected in parallel with a cell of emf $4 \text{ V}$ and negligible internal resistance. The deflections produced in the tangent galvanometers $A$ and $B$ are $30^{\circ}$ and $60^{\circ}$, respectively. If $A$ has $2$ turns, then $B$ must have: (in $turns$)

$A$ short magnetic needle is pivoted in a uniform magnetic field of induction $1 \text{ T}$. Now,simultaneously another magnetic field of induction $\sqrt{3} \text{ T}$ is applied at right angles to the first field; the needle deflects through an angle $\theta$ whose value is (in $^{\circ}$)