Suppose a magnetic monopole ${\mathbf{q}}_{\mathbf{m}}$ passes through a resistanceless loop of wire with self-inductance L . What current is induced in the loop?

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,

$\nabla \times E=-{\mu }_0{J}_m-\frac{\partial B}{\partial t}$

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

Integrating both sides of equation over the da surface,

$$\begin{gathered} \int (\nabla \times E).da=\int \left(-{\mu }_0{J}_m-\frac{\partial B}{\partial t}\right).da \\ \oint E.dl=\int -{\mu }_0{J}_m.da-\int \frac{\partial B}{\partial t}.da \\ \epsilon =-{\mu }_0{J}_m.da-\frac{d}{dt}\int B.da \\ \epsilon =-{\mu }_0{l}_{menc}-\frac{d\Phi }{dt} \end{gathered}$$

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

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

$\epsilon =-L\frac{dl}{dt}$

Equating both values,

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

Here,$∆Q_m$is the total magnetic charge passing through the surface,$∆\Phi$is the change in flux through the surface.

For theresistanceless loop of wire, and .

$$\begin{gathered} l=\frac{{\mu }_0}{L}\times q_m+\frac{1}{L}\times 0 \\ l=\frac{{\mu }_0{q}_m}{L} \end{gathered}$$

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