$A$ thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

The magnetic field at the centre of a single-turn circular coil of a given length of wire carrying a current $I$ is $B$. If the same wire is bent into a circular coil of two turns and the same current $I$ is passed through it,the new magnetic field at the centre will be:

Magnetic field at the centre of a circular loop of area $A$ is $B$. Then the magnetic moment of the loop is ($\mu_0$ is the permeability of free space).

An electron revolving in a circular orbit of radius $r$ with velocity $v$ and frequency $f$ has an orbital magnetic moment $M$. If the frequency of revolution is doubled,then the new magnetic moment will be:

$A$ steady current $i$ flows in a small square loop of wire of side $L$ in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overrightarrow {{\mu _1}} $ and $\overrightarrow {{\mu _2}} $ respectively denote the magnetic moments due to the current loop before and after folding. Then

In a hydrogen atom,an electron revolves around the nucleus at a frequency of $6.6 \times 10^{15} \, \text{rev/sec}$ in an orbit of radius $0.528 \, \mathring{A}$. The magnetic moment in $\text{A} \cdot \text{m}^2$ is: