Q18P - Introduction to Electrodynamics

Question

A square loop, side a , resistance R , lies a distance from an infinite straight wire that carries current l (Fig. 7.29). Now someone cuts the wire, so l drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes a given point in the loop during the time this current flows? If you don't like the scissors model, turn the current down gradually:

$\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{)}\mathbf{I}\\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\end{array}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\begin{array}{c}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{1}\mathbf{/}\mathbf{a}\\ \mathbf{for}\mathbf{ }\mathbf{t}\mathbf{>}\mathbf{/}\mathbf{a}\end{array}$

                 

Step-by-Step Solution

Answer

The charge passing through the given point in the loop is $-\left(\frac{{\mu }_{0}al}{2\mathrm{\pi R}}\right)\ln \left(\frac{s+a}{s}\right)$ and direction of the induced current is counter clockwise.

Step 1: Write the given data from the question.

The side of the square loop is a.

The resistance of the loop is R.

The current in the infinite straight wire is l.

The distance between the square loop and straight wire is s.

Step 2: Determine the direction of the current and charges passes through the given point in the loop.

Let’s assume the small element dx on square loop and 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,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\pi x}}$ 

According to the Faraday’s law, the expression for the flux is given by,

$\varphi =\int B.dA$

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

$$\begin{gathered} \varphi ={\int }_s^{S+a}\left(\frac{{\mu }_{0}l}{2\mathrm{\pi x}}\right).adx \\ \varphi =\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right){\int }_s^{S+a}\frac{1}{x}dx \\ \varphi =\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right)\left[in(s+a)-\ln \left(s\right)\right] \\ \varphi =\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right) \end{gathered}$$ 

According to the Faraday’s law the generated emf is given by,

$e=-\frac{d\varphi }{dt}$ 

                        

Substitute  $\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)$ for $\varphi$ into above equation.

$$\begin{gathered} e=-\frac{d}{dt}\left[\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mu }_{0}la}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)\frac{dl}{dt} \end{gathered}$$

The current in term of charge is given by,

$l=\frac{dQ}{dt}$                                                                              …… (1) 

The current in terms of resistance and voltage is given by,

$l=\frac{\epsilon }{R}$                                                                                …… (2)

Equate equation (1) and (2).

  $\frac{dQ}{dt}=\frac{\epsilon }{R}$

 

Substitute $-\left(\frac{{\mu }_{0}a}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)\frac{dl}{dt}$ for $e$ into above equation.

$$\begin{gathered} \frac{dQ}{dt}=-\frac{1}{R}\times \left(\frac{{\mu }_{0}a}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)\frac{dl}{dt} \\ dQ=-\frac{1}{R}\times \left(\frac{{\mu }_{0}a}{2\mathrm{\pi }}\right)\ln \left(\frac{s+a}{s}\right)dl \\ Q=-\left(\frac{{\mu }_{0}a}{2\mathrm{\pi R}}\right)\ln \left(\frac{s+a}{s}\right) \end{gathered}$$

The field of the square loop is out of the page therefore the current direction must be out of page. Hence the induced current flows counter clockwise.

 

Hence the charge passing through the given point in the loop is $-\left(\frac{{\mu }_{0}al}{2\mathrm{\pi R}}\right)\ln \left(\frac{s+a}{s}\right)$ and direction of the induced current is counter clockwise.