A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight wire carrying a current I = 2 A as shown below. The mutual inductance, in nH (rounded off to 2 decimal places), between conducting loop and the long wire is __________.

The correct answer is 138.63.

Step 1: Understand the Geometry and Formula
From the given image, we observe a long straight wire carrying a current $I=2\text{A}$ along the Z-axis. A square loop of side length $a=1\text{m}$ is placed in the same plane at a distance $d=1\text{m}$ from the wire.

The magnetic flux $\Phi$ passing through the loop due to the current $I$ in the long wire is related to the mutual inductance $M$ by the equation:

$\Phi =MI$

Step 2: Calculate the Magnetic Flux through the Loop
The magnetic field $B$ at a distance $r$ from an infinitely long straight wire carrying current $I$ is given by Ampere's Law:

$B\left(r\right)=\frac{{\mu }_{0}I}{2\pi r}$

To find the total magnetic flux $\Phi$ linked with the square loop, consider a thin elemental strip of width $dr$ and height $a$ at a distance $r$ from the long wire. The differential flux $d\Phi$ through this strip is:

$d\Phi =B\left(r\right)·adr=\frac{{\mu }_{0}Ia}{2\pi r}dr$

Integrating $r$ from the inner edge $r=d$ to the outer edge $r=d+a$:

$\Phi ={\int }_d^{d+a}\frac{{\mu }_{0}Ia}{2\pi r}dr=\frac{{\mu }_{0}Ia}{2\pi }\ln \left(\frac{d+a}{d}\right)$

Step 3: Derive the Mutual Inductance Formula
Using $M=\frac{\Phi }{I}$, we obtain:

$M=\frac{{\mu }_{0}a}{2\pi }\ln \left(\frac{d+a}{d}\right)$

Step 4: Substitute the Given Values
Given parameters from the diagram and text:

• Side of square loop, $a=1\text{m}$
• Distance from wire, $d=1\text{m}$
• Permeability of free space, ${\mu }_0=4\pi \times {10}^{-7}\text{H/m}$

Substituting these values into the mutual inductance equation:

$M=\frac{4\pi \times {10}^{-7}\times 1}{2\pi }\ln \left(\frac{1+1}{1}\right)$

$M=2\times {10}^{-7}\times \ln \left(2\right)$

Since $\ln \left(2\right)\approx 0.693147$:

$M=2\times {10}^{-7}\times 0.693147=1.386294\times {10}^{-7}\text{H}$

Converting Henries to nanohenries ($1\text{nH}={10}^{-9}\text{H}$):

$M=138.6294\text{nH}\approx 138.63\text{nH}$

Thus, the mutual inductance between the conducting loop and the long wire rounded off to 2 decimal places is 138.63 nH.