91Ó°ÊÓ

The side of the square loop is a.

The current in the long straight wire is l .

(a)

Let assume the distance between the square loop and straight wire is s and take a small element dx on square loop.

Consider the distance between the small element and straight wire is x .

The area of the small strip of the square loop is given by,

$dA=adx$

The magnetic field in the lone wire is given by,

localid="1658557154740" $\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{Ï€³\ae }}$

According to the Faraday’s law, the flus of any closed of open surface area is calculated by the integral of normal component of magnetic field over the area.

localid="1658557165144" $\mathit{f}\mathbf{=}\mathbf{∫}\mathit{B}\mathbf{.}\mathit{d}\mathit{A}$

Substitute$\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}x}$for B and adx for dA into above equation.

localid="1657616767520" $$\begin{gathered} \mathrm{ϕ}={∫}_S^{S+A}\left(\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}x}\right).adx \\ \mathrm{ϕ}=\left(\frac{{\mathrm{μ}}_{0}la}{2x}\right){∫}_S^{S+A}\frac{1}{x}dx \\ \mathrm{ϕ}=\left(\frac{{\mathrm{μ}}_{0}la}{2x}\right)\left[\mathrm{In}\left(\mathrm{s}+\mathrm{a}\right)-\mathrm{In}\left(\mathrm{s}\right)\right] \\ \mathrm{ϕ}=\left(\frac{{\mathrm{μ}}_{0}la}{2x}\right)In\left(\frac{s+a}{s}\right) \end{gathered}$$

Hence the flux through the loop is$\left(\frac{{\mathrm{μ}}_{0}la}{2x}\right)In\left(\frac{s+a}{s}\right)$.

(b)

According to the Faraday’s law the generated emf when the loop is pulled away from the straight wire is given by,

$e=-\frac{d\mathrm{Ï•}}{dt}$

Substitute $\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)In\left(\frac{s+a}{s}\right)$ for into above equation.

role="math" localid="1657616507725" $$\begin{gathered} e=-\frac{d}{dt}\left[\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\ln \left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\frac{d}{dt}\left[In\left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\left[\frac{1}{{\displaystyle \frac{s+a}{s}}}\left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\left(\frac{s+a}{s}\right)\left(\frac{s{\displaystyle \frac{ds}{dt}}-\left(s+a\right){\displaystyle \frac{ds}{dt}}}{s^2}\right) \end{gathered}$$

The rate of change displacement is known as velocity.

Substitute v for$\frac{ds}{dt}$into the generated emf’s equation.

$$\begin{gathered} e=-\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\left(\frac{s+a}{s}\right)\left(\frac{sv-\left(s+a\right)v}{s^2}\right) \\ e=-\left(\frac{{\mathrm{μ}}_{0}la}{2\mathrm{π}}\right)\left(\frac{1}{s+a}\right)\left(-\frac{av}{s}\right) \\ e=\frac{{\mathrm{μ}}_{0}l{a}^{2}v}{2\mathrm{π}s\left(s+a\right)} \end{gathered}$$

Hence the expression for the induced emf is $\frac{{\mathrm{μ}}_{0}l{a}^{2}v}{2\mathrm{π}s\left(s+a\right)}$.

(c)

When the loop is pulled right, the flux remains the same therefore the emf will not be generated.

Hence the induced emf I zero when the loop is pulled to the right.