$A$ current $i$ is flowing through a wire of length $L$. If it is made into a circular loop of one turn, then its magnetic moment is

In an atom,an electron of charge $(-e)$ performs uniform circular motion ($U$.$C$.$M$.) around a stationary positively charged nucleus with a period of revolution $T$. If $r$ is the radius of the orbit of the electron and $v$ is the orbital velocity,then the circulating current $I$ is proportional to:

Two wires of the same length are used to form a circle and a square. They carry the same current. The ratio of the magnetic moment of the circle to that of the square is:

The dipole moment of a circular loop carrying a current $I$ is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant,the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is

The figure shows a square current-carrying loop $ABCD$ of side $10\,cm$ and current $i = 10\,A$. The magnetic moment $\vec{M}$ of the loop is:

The magnetic field at the centre of a current carrying circular coil of area $A$ is $B$. The magnetic moment $(M)$ of the coil is $(\mu_0 = \text{permeability of free space})$