Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres, which lie at the origin. Let the two magnets lie along the $X$-axis and $Y$-axis, respectively. The magnitude of the magnetic field at a distance $R$ from the centre on the $Y$-axis is $\frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$. Assuming $R >> l$ (magnet length), the magnitude of $M$ is

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \; G$ experiences a torque of $0.016 \; Nm.$
$(a)$ What is the magnetic moment of the magnet?
$(b)$ What is the work done in moving it from its most stable to most unstable position?
$(c)$ The bar magnet is replaced by a solenoid of cross-sectional area $2 \times 10^{-4} \; m^{2}$ and $1000$ turns,but of the same magnetic moment. Determine the current flowing through the solenoid.

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

At a point on the equatorial line (right bisector) of a magnetic dipole,the magnetic:

Two identical short bar magnets,each having magnetic moment $M$,are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is