Question

In Figure, a rectangular loop of wire with length$\mathit{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{2}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{w}\mathit{i}\mathit{d}\mathit{t}\mathit{h}\mathbf{}\mathit{b}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{r}\mathit{e}\mathit{s}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathit{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathit{m}\mathit{\Omega }$is placed near an infinitely long wire carrying current i = 4.7 A. The loop is then moved away from the wire at constant speed v = 3.2 mm/s. When the center of the loop is at distance r = 1.5b, (a) what is the magnitude of the magnetic flux through the loop?(b) what is the current induced in the loop?

Short answer

1. Magnitude of the magnetic flux through the loop is$1.4x{10}^{-8}Wb$
2. The current induced in loop is$1\times {10}^{-5}A$.

Step-by-step solution

Step 1: Given

1. Length$a=2.2cm=0.022m$
2. Width$b=0.80cm=0.080m$
3. Resistance $R=0.40m\Omega$
4. Current i = 4.7 A
5. Speed v = 3.2 mm/s
6. Distance r = 1.5 b

Determining the concept

Use the magnetic flux formula to find the magnitude of magnetic flux. Substituting this value in Faraday’s law, find the emf induced in the coil. Substituting the value of emf and resistance in Ohm’s law, find the current induced in the loop.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follows:

$\varphi =\int \mathrm{Bd}A$

$i_{loop}=\left|\frac{\epsilon }{R}\right|$

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

Where,$\varphi$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance,𝜀 is emf.

(a) Determining the magnitude of the magnetic flux through the loop

Magnitude of magnetic flux through the loop:

Here, the magnetic flux is generated only due to the long straight wire.

The magnetic field due to the long straight wire is given as,

$B=\frac{{\mu }_{0}i}{2\tau \tau r}$

The magnitude of magnetic flux is given as,

$\varphi =\int \mathrm{Bd}A$

Consider a strip of height dr and the length a, then the area of the strip is,

dA = adr

The distance r varies from $r-\frac{b}{2}$to$r+\frac{b}{2}$

Thus, the magnitude of magnetic flux is,

$\left|\varphi \right|=\frac{{\mu }_{0}ia}{2\tau \tau }{\int }_{r-\frac{b}{2}}^{r+\frac{b}{2}}\frac{1}{r}dr$

$\left|\varphi \right|=\frac{{\mu }_{0}ia}{2\tau \tau }\ln \left(\frac{r+{\displaystyle \frac{b}{2}}}{r-\frac{b}{2}}\right)$

Since, r = 1.5b , therefore,

$r+\frac{b}{2}=1.5\left(0.80cm\right)+0.4cm=1.6cmandr-\frac{b}{2}=1.5(0.80cm)-0.4cm=0.8cm$

Thus,

$\frac{r+\frac{b}{2}}{r-\frac{b}{2}}=\frac{1.6\mathrm{cm}}{0.8\mathrm{cm}}=2$

Substituting the values,

$\left|\mathrm{\varphi }\right|=\frac{\left(4\mathrm{\tau \tau }\times {10}^{-7}\right)\left(4.7\mathrm{A}\right)\left(0.022\mathrm{m}\right)}{2\mathrm{\tau \tau }}\ln \left(2\right)\left|\mathrm{\varphi }\right|=1.4\times {10}^{-8}\mathrm{Wb}$

Hence, magnitude of the magnetic flux through the loop is$1.4\times {10}^{-8}Wb$

(b) Determining the current induced in loop

Current induced in loop :

The induced current in the loop is given by Ohm’s law,

$i_{loop}=\left|\frac{\epsilon }{R}\right|$

Where, R is the resistance of the loop and

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

$$\begin{gathered} \epsilon =\frac{{\mu }_{0}ia}{2\tau \tau }\left|\frac{d}{dt}\ln \left(\frac{r+{\displaystyle \frac{b}{2}}}{r-{\displaystyle \frac{b}{2}}}\right)\right| \\ \epsilon =\frac{{\mu }_{0}ia}{2\tau \tau }\left[\frac{1}{r+\frac{b}{2}}-\frac{1}{r-\frac{b}{2}}\right]\frac{dr}{dt} \end{gathered}$$

Since,$\frac{dr}{dt}=v$

Therefore,

$$\begin{gathered} \epsilon =\frac{{\mu }_{0}ia}{2\tau \tau }\left[-\frac{b}{r^2-{\left({\displaystyle \frac{b}{2}}\right)}^2}\right]v \\ \epsilon =\frac{{\mu }_{0}iabv}{2\tau \tau \left[r^2-{\left(\frac{b}{2}\right)}^2\right]} \\ {i}_{loop}=\left|\frac{\epsilon }{R}\right| \\ {i}_{loop}=\frac{{\mu }_{0}iabv}{2\tau \tau R\left[r^2-{\left(\frac{b}{2}\right)}^2\right]} \\ {i}_{loop}=\frac{\left(4\tau \tau \times {10}^{-7}{\displaystyle \frac{T.m}{A}}\right)\left(4.7A\right)\left(0.022m\right)\left(0.0080m\right)\left(3.2\times {10}^{-3}{\displaystyle \frac{m}{s}}\right)}{2\tau \tau \left(4\times {10}^{-4}\Omega \right)\left[2{\left(0.0080m\right)}^2\right]} \\ {i}_{loop}=1\times {10}^{-5}A \end{gathered}$$

Hence, the current induced in loop is$1\times {10}^{-5}A$

Therefore, calculate the magnitude of the magnetic flux using the formula for magnetic flux. Find the current induced in the coil by using Faraday’s law and Ohm’s law.