Question

A rectangular loop of N closely packed turns is positioned near a long straight wire as shown in Figure. What is the mutual inductance M for the loop–wire combination if $\mathbf{}\mathit{N}\mathbf{=}\mathbf{100}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathbf{}\mathit{b}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{l}\mathbf{=}\mathbf{30}\mathbf{}\mathit{c}\mathit{m}\mathbf{}$

Short answer

Mutual Inductance M for the loop – wire combination will be$1.32\times {10}^{-5}\mathrm{H}$

Step-by-step solution

Given

$$\begin{gathered} N=100 \\ a=1.0\mathrm{cm}=0.01\mathrm{m} \\ \mathrm{b}=8.0\mathrm{cm}=0.08\mathrm{m} \\ \mathrm{l}=30\mathrm{cm}=0.3\mathrm{m} \end{gathered}$$

Understanding the concept

We know the formula for the magnetic flux in terms of the magnetic field. We also have the formula for the mutual inductance in terms of the magnetic flux. So, we can calculate the required answer using the given information.

Formula:

$$\begin{gathered} \varphi =\int BdA \\ B=\frac{{\mu }_{0}I}{2\mathrm{\pi r}} \\ M=\frac{N\varphi }{I} \end{gathered}$$

Calculate mutual Inductance M for the loop – wire combination

We have,

$\varphi =\int BdA$

Magnetic flux through the loop can be written as,

$$\begin{gathered} {\varphi }_B={\int }_a^{a+b}{B}_{wire}\times ldr \\ {\varphi }_B={\int }_a^{a+b}\left(\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}\right)\times ldr \\ {\varphi }_B=\frac{{\mu }_{0}Il}{2\mathrm{\pi }}{\int }_a^{a+b}\left(\frac{1}{r}\right)dr \\ {\varphi }_B=\frac{{\mu }_{0}Il}{2\mathrm{\pi }}{\left[Inr\right]}_a^{a+b} \\ {\varphi }_B=\frac{{\mu }_{0}Il}{2\mathrm{\pi }}\left(ln\right(a+b)-lna) \\ {\varphi }_B=\frac{{\mu }_{0}Il}{2\mathrm{\pi }}\times \left(In\left(1+\frac{b}{a}\right)\right) \end{gathered}$$

We know that,

$$\begin{gathered} M=\frac{N\varphi }{I} \\ M=\frac{N}{I}\times \left(\frac{{\mu }_{0}Il}{2\mathrm{\pi }}\times \left(In\left(1+\frac{b}{a}\right)\right)\right) \\ M=\frac{N{\mu }_{0}I}{2\mathrm{\pi }}\times \left(In\left(1+\frac{0.08}{0.01}\right)\right) \\ M=1.32\times {10}^{-5}H \end{gathered}$$