91Ó°ÊÓ

This law is used to determine the electromotive force (emf) generated due to the interaction between a magnetic field and an electric conductor.

Based on this law, the amount of emf induced in a conductor directly relies upon the changes in the magnetic flux

The magnetic monopole is,$q_m.$

The self-inductance of the resistanceless loop of wire is, L .

The generalised equation of Faraday’s law for the resistanceless loop of wire is given by,

$∇×E=-{\mathrm{μ}}_0{J}_m-\frac{∂B}{∂t}$

Here,represents the electric field,${\mathrm{μ}}_0$is the permeability of free space,$J_m$is the current of magnetic charge and$\frac{∂B}{∂t}$is the change in magnetic field.

Integrating both sides of equation over the da surface,

$$\begin{gathered} ∫(∇×E).da=∫\left(-{\mathrm{μ}}_0{J}_m-\frac{∂B}{∂t}\right).da \\ âˆ\circledR E.dl=∫-{\mathrm{μ}}_0{J}_m.da-∫\frac{∂B}{∂t}.da \\ \mathrm{Î\mu }=-{\mathrm{μ}}_0{J}_m.da-\frac{d}{dt}∫B.da \\ \mathrm{Î\mu }=-{\mathrm{μ}}_0{l}_{menc}-\frac{d\mathrm{Φ}}{dt} \end{gathered}$$

Here,the induced emf,$l_{m_{enc}}$is induced electro-magnetic current and$\frac{d\mathrm{Φ}}{dt}$is the change in the magnetic flux.

Also, the induced emf in the wire loop is given by,

$\mathrm{Î\mu }=-L\frac{dl}{dt}$

Equating both values,

$$\begin{gathered} -L\frac{dl}{dt}=-{\mathrm{μ}}_0{l}_{m_{enc}}-\frac{d\mathrm{Φ}}{dt} \\ \frac{dl}{dt}=\frac{-{\mathrm{μ}}_0}{L}{l}_{m_{enc}}+\frac{1}{L}\frac{d\mathrm{Φ}}{dt} \\ l=\frac{{\mathrm{μ}}_0}{L}∆Q_m+\frac{1}{L}∆\mathrm{Φ} \end{gathered}$$

Here,$∆Q_m$is the total magnetic charge passing through the surface,$∆\mathrm{Φ}$is the change in flux through the surface.

For theresistanceless loop of wire, and .

$$\begin{gathered} l=\frac{{\mathrm{μ}}_0}{L}×q_m+\frac{1}{L}×0 \\ l=\frac{{\mathrm{μ}}_0{q}_m}{L} \end{gathered}$$

Hence, the induced current in the loop is$l=\frac{{\mathrm{μ}}_0{q}_m}{L}.$