Q7.23P - Introduction to Electrodynamics

Question

A square loop of wire, of side $\mathbf{a}$, lies midway between two long wires,$\mathbf{3}\mathbf{a}$ apart, and in the same plane. (Actually, the long wires are sides of a large rectangular loop, but the short ends are so far away that they can be neglected.) A clockwise current  $\mathbf{I}$in the square loop is gradually increasing: $\raisebox{1ex}{$\mathbf{d}\mathbf{l}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{d}\mathbf{t}$}\right.\mathbf{=}\mathbf{k}$ (a constant). Find the emf induced in the big loop. Which way will the induced current flow?

Step-by-Step Solution

Answer

The induced emf is $-\frac{{\mathrm{\mu }}_0\mathrm{ka}}{\mathrm{\pi }}\ln \left(2\right)$ and direction of the current in big loop is counter clockwise.

Step 1: Write the given data from the question.

The side of the square loop is $\mathrm{a}$

The distance between the long wires is $3\mathrm{a}$

The current is the square loop is $\mathrm{l}$

The increasing rate of the current $\raisebox{1ex}{$\mathrm{dl}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.=\mathrm{k}$,

Step 2: Determine the emf induced in the big loop.

Consider the diagram of the system as,        

The expression for the magnetic field of one wire is given by,

 $\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi s}}$

The magnetic flus due to one wire is given by,

 ${\mathrm{\varphi }}_1-{\int }_{\mathrm{\theta }}^{2\mathrm{a}} \mathrm{B}\times \mathrm{da}$

Substitute $\frac{{\mathrm{\mu }}_0\mathrm{f}}{2\mathrm{\pi s}}$ for $\mathrm{B}$ and $\mathrm{ads}$ for $\mathrm{da}$ into the above equation.

${\mathrm{\varphi }}_1={\int }_{\mathrm{a}}^{2\mathrm{a}} \frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi s}}\times \mathrm{ads}$

${\mathrm{\varphi }}_1=\frac{{\mathrm{\mu }}_0\mathrm{la}}{2\mathrm{\pi }}{\int }_{\mathrm{a}}^{2\mathrm{a}} \frac{1}{\mathrm{s}}\mathrm{ds}$

${\mathrm{\varphi }}_1=\frac{{\mathrm{\mu }}_0\mathrm{la}}{2\mathrm{\pi }}(\ln \mathrm{s}{)}_{\mathrm{a}}^{2\mathrm{a}}$

${\mathrm{\varphi }}_1=\frac{{\mathrm{\mu }}_0\mathrm{la}}{2\mathrm{\pi }}(\ln 2)$

The magnetic flue due to two wire is given by,

$\mathrm{\varphi }=2{\mathrm{\varphi }}_1$

$\mathrm{\varphi }=2\frac{{\mathrm{\mu }}_0\mathrm{la}}{2\mathrm{\pi }}\ln \left(2\right)$

$\mathrm{\varphi }=\frac{{\mathrm{\mu }}_0\mathrm{la}}{\mathrm{\pi }}\ln \left(2\right)$

The induced emf is given by,

 $\mathrm{\epsilon }=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}$

Substitute ${\mathrm{\mu }}_{\mathrm{\pi }}{\mathrm{\mu }}_0\ln \left(2\right)$ for $\mathrm{\varphi }$ into above equation

  $\mathrm{\epsilon }=-\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{{\mathrm{\mu }}_0\mathrm{la}}{\mathrm{\pi }}\ln \left(2\right)\right)$

$\mathrm{\delta }=-\frac{{\mathrm{\mu }}_0/\mathrm{a}}{\mathrm{\pi }}\ln \left(2\right)\frac{\mathrm{di}}{\mathrm{dt}}$

Substitute $\mathrm{k}$ for $\raisebox{1ex}{$\mathrm{dl}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.$ into the above equation.

$\mathrm{\epsilon }=-\frac{{\mathrm{\mu }}_0\mathrm{a}}{\mathrm{\pi }}\ln \left(2\right)\mathrm{k}$

$\mathrm{\epsilon }=-\frac{{\mathrm{\mu }}_0\mathrm{ka}}{\mathrm{\pi }}\ln \left(2\right)$

Therefore, the induced emf is $-\frac{{\mathrm{a}}_0\mathrm{ka}}{\mathrm{\pi }}\ln \left(2\right)$.

The net flux in inward and increasing. According to Len’s law, the emf should oppose the increment in the flux. Therefore, the magnetic flux should be outward, and the current flows in a counter clockwise direction in the big loop.

Hence the induced emf is $-\frac{{\mathrm{a}}_0\mathrm{ka}}{\mathrm{\pi }}\ln \left(2\right)$ and direction of the current in big loop is counter clockwise.