$A$ magnet,when placed perpendicular to a uniform magnetic field of strength $10^{-4} \ Wb/m^2$,experiences a maximum torque of $4 \times 10^{-5} \ N \cdot m$. What is its magnetic moment in $A \cdot m^2$?

$A$ cylindrical magnetic rod has length $5 \ cm$ and diameter $1 \ cm$. It has uniform magnetization $5.3 \times 10^3 \ A/m$. Its net magnetic dipole moment is nearly

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

Two points $A$ and $B$ are situated at a distance $x$ and $2x$ respectively from the nearer pole of a magnet $2 \ cm$ long. The ratio of the magnetic field at $A$ and $B$ is

$A$ cylindrical rod magnet has a length of $5 \, cm$ and a diameter of $1 \, cm$. It has a uniform magnetization of $5.30 \times 10^3 \, A/m$. What is its magnetic dipole moment?

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the $z$-axis from $z = a > 0$ to $z = R$,then along the circular arc of radius $R$ to the $x$-axis,then along the $x$-axis to $x = a$,and finally along the circular arc of radius $a$ back to the $z$-axis.