Question
  1. Define mutual inductance and write its $S.I$. unit.
  2. A square loop of side $'a\ '$carrying a current $I_2$ is kept at distance $x$ from an infinitely long straight wire carrying a current $ I_1$ as shown in the figure. Obtain the expression for the resultant force acting on the loop.

Answer

  1. Mutual inductance is where the magnetic field generated by a coil of wire induces voltage in an adjacent coil of wire. $A$ transformer is a device constructed of two or more coils in close proximity to each other, with the express purpose of creating a condition of mutual inductance between the coils. It's $SI$ unit is: $\frac{\text{Wb}}{\text{A}}$
  2.  

According to the right $-$ hand screw rule, the magnetic field will be into the plane across the loop.
Force on length $AD$ :
$F =$ Bil
$\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi\text{x}} \ ($Towards Right$)$
Force on length $AD$ :
$F =$ Bil
$\text{F}_2=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi(\text{x}+\text{a})} \ ($Towards Left$)$
Force on $AB$ and $CD$ will be equal and opposite.
Hence, they'll cancel out.
Force on the loop:
$\text{F}_\text{Net}=\text{F}_1-\text{F}_2$
$=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi}\Big[\frac{1}{\text{x}}-\frac{1}{(\text{x}+\text{a})}\Big]$
$\text{F}_\text{Net}=\frac{\mu_0\text{I}_1\text{I}_2\text{a}}{2\pi}\Big[\frac{\text{x}+\text{a}-\text{x}}{(\text{x}+\text{a})\text{x}}\Big]=\frac{\mu_0\text{I}_1\text{I}_2\text{a}^2}{2\pi(\text{x}+\text{a})\text{x}}$
$\text{F}_\text{Net}=\frac{\mu_0\text{I}_1\text{I}_2\text{a}^2}{2\pi\text{x}(\text{x}+\text{a})}  \ ($Towards left$)$

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