$A$ bar magnet of magnetic moment $5 \,Am^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5} \,T$. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \,N$, then the magnetic length of the magnet is: (in $\,m$)

If a magnet of pole strength $m$ is divided into four parts such that the length and width of each part is half that of the initial one,then the pole strength of each part will be

$A$ wire of magnetic moment $M$ is moulded according to the figure. What is the new magnetic moment?

$A$ magnetised wire of magnetic moment $3.14 \, A-m^2$ is bent in the form of a semi-circle. The new magnetic moment will be.....$A-m^2$

If two identical bar magnets, each of length $l$, pole strength $m$ and magnetic moment $M$ are placed perpendicular to each other with their unlike poles in contact, the magnetic moment of the combination is

Each atom of an iron bar $(5 \, cm \times 1 \, cm \times 1 \, cm)$ has a magnetic moment of $1.8 \times 10^{-23} \, A \cdot m^2$. Given that the density of iron is $7.78 \times 10^3 \, kg/m^3$,the atomic weight is $56 \, g/mol$,and Avogadro's number is $6.02 \times 10^{23} \, mol^{-1}$,calculate the magnetic moment of the bar in the state of magnetic saturation in $A \cdot m^2$.