A magnetic needle suspended horizontally by an unspun silk fibre, oscillates in the horizontal plane because of the restoring force originating mainly from
The torsion of the silk fibre
The force of gravity
The horizontal component of earth's magnetic field
All the above factors
A bar magnet of length $10 \,cm$ and having the pole strength equal to $10^{-3}$ weber is kept in a magnetic field having magnetic induction $ (B)$ equal to $4\pi \times {10^{ - 3}}$ Tesla. It makes an angle of $30°$ with the direction of magnetic induction. The value of the torque acting on the magnet is
Due to a small magnet intensity at a distance $x$ in the end on position is $9$ $Gauss$. What will be the intensity at a distance $\frac{x}{2}$ on broad side on position..... $Gauss$
Points $A$ and $B$ are situated perpendicular to the axis of a small bar magnet at large distances $x$ and $3 x$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to
A magnetic needle is kept in a non-uniform magnetic field. It experiences
Two bar magnets with magnetic moments $ 2 M $ and $M$ are fastened together at right angles to each other at their centres to form a crossed system, which can rotate freely about a vertical axis through the centre. The crossed system sets in earth’s magnetic field with magnet having magnetic moment $2M $ making and angle $ \theta $ with the magnetic meridian such that