Q8P - Introduction to Electrodynamics

Question

A square loop of wire (side a) lies on a table, a distance s from a very long straight wire, which carries a current I, as shown in Fig. 7.18.

(a) Find the flux of B through the loop. 

(b) If someone now pulls the loop directly away from the wire, at speed , V what emf is generated? In what direction (clockwise or counter clockwise) does the current flow? 

(c) What if the loop is pulled to the right at speed V ?

Step-by-Step Solution

Answer

(a) The flux through the loop is $\left(\frac{{\mu }_{0}la}{2\pi }\right) \ln \left(\frac{s+a}{s}\right)$.

(b) The expression for the induced emf is $\frac{{\mu }_{0}l{a}^{2}v}{2\pi s\left(s+a\right)}$ .

(c) The induced emf I zero when the loop is pulled to the right.

Step 1: Write the given data from the question.

The side of the square loop is a.

The current in the long straight wire is l .

Step 2: Determine the equation to calculate the flux through the loop.

(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,

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

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.

$\mathit{f}\mathbf{=}\mathbf{\int }\mathit{B}\mathbf{.}\mathit{d}\mathit{A}$

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

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

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

 

Step 3: Calculate the emf generated and its direction.

(b)

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

$e=-\frac{d\varphi }{dt}$

 

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

$$\begin{gathered} e=-\frac{d}{dt}\left[\left(\frac{{\mu }_{0}la}{2\pi }\right)\ln \left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mu }_{0}la}{2\pi }\right)\frac{d}{dt}\left[In\left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mu }_{0}la}{2\pi }\right)\left[\frac{1}{{\displaystyle \frac{s+a}{s}}}\left(\frac{s+a}{s}\right)\right] \\ e=-\left(\frac{{\mu }_{0}la}{2\pi }\right)\left(\frac{s+a}{s}\right)\left(\frac{s{\displaystyle \frac{ds}{dt}} -\left(s+a\right){\displaystyle \frac{ds}{dt}}}{s^2}\right) \end{gathered}$$

The rate of change displacement is known as velocity.

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

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

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

Step 4: When the loop is pulled right at the speed of .

(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.