⇒Magnetic Field due to Electric Current

Correct Option is (B)

As the length of the wire remains the same, we can write

$\begin{array}{rl}& N_{1}2\pi R=N_2\times 2\pi \frac{R}{3}\\ & N_1=\frac{N_2}{3}\\ & N_2=3N_1\end{array}$

$\begin{array}{ll}& \text{ Magnetic moment of coil }\mu =\text{ NIA }\\ \therefore & {\mu }_1=N_{1}I{A}_1=N_{1}I\pi R^2\\ \therefore & {\mu }_2=N_{2}I{A}_2=3\frac{N_{1}I\pi R^2}{9}\\ \therefore & \frac{{\mu }_2}{{\mu }_1}=3\frac{N_{1}I\pi R^2}{9}\times \frac{1}{N_{1}I\pi R^2}=\frac{1}{3}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& \mathrm{m}=\mathrm{nI}A\\ & \mathrm{m}=\mathrm{nI}\left(\pi {\mathrm{r}}^2\right)\end{array}$

Correct Option is (C)

The magnetic moment $\mu$ associated with the angular momentum $L$ of a revolving electron is given by :

$\mu =\frac{-e}{2m}L$

The negative sign indicates the direction of the magnetic moment is opposite to the direction of angular momentum. However, for the sake of magnitude, we'll focus on :

$\mu =\frac{eL}{2m}$

So, the correct answer is Option C : $\frac{eL}{2m}$.

Correct Option is (A)

Currently no explanation available

Correct Option is (B)

$\mathrm{M}=\frac{\mathrm{enh}}{4\pi {\mathrm{m}}_{\mathrm{e}}}$