The torque on a bar magnet due to the earth's magnetic field is maximum when the axis of the magnet is
Perpendicular to the field of the earth
Parallel to the vertical component of the earth's field
At an angle of $33°$ with respect to the $N-S $ direction
Along the North-South $(N-S)$ direction
A short magnetic needle is pivoted in a uniform magnetic field of strength $1 \,T.$ When another magnetic field of strength $\sqrt 3 \,T$ is applied to the needle in a perpendicular direction, the needle deflects through an angle $\theta$, where $\theta$ is....$^o$
The angle of dip at a place is $40.6°$ and the intensity of the vertical component of the earth's magnetic field $V = 6 \times {10^{ - 5}}$ $Tesla$. The total intensity of the earth's magnetic field $ (I) $ at this place is
At which place, earth's magnetism become horizontal
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
Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $3\,s$ and $4\,s$ respectively. If their moments of inertia are in the ratio of $3: 2$ then the ratio of their magnetic moments will e.