1. Obtain an expression for the mutual inductance between a long … - Vidyadip

Question

Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
Now assume that the straight wire carries a current of $50A$ and the loop is moved to the right with a constant velocity $, v = 10\ m/s.$ Calculate the induced emf in the loop at the instant when $x = 0.2\ m$. Take $a = 0.1\ m$ and assume that the loop has a large resistance.

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Answer

Take a small element dy in the loop at a distance $y$ from the long straight wire $ ($as shown in the given figure$)$.
Magnetic flux associated with element dy, $\text{d}\phi=\text{BdA}$
Where,
$dA =$ Area of element $dy = \text{ady}$
$B =$ Magnetic field at distance $y$
$=\frac{\mu_0\text{I}}{2\pi\text{y}}\text{s}$
$I = $ Current in the wire
$\mu_0=\text{Permeability of free space}=4\pi\times10^{-7}\text{Tm}\text{A}^{-1}$
$\therefore\ \text{d}\phi=\frac{\mu_0\text{Ia}}{2\pi}\frac{\text{dy}}{\text{y}}\text{s}$
$\phi=\frac{\mu_0\text{Ia}}{2\pi}\int\frac{\text{dy}}{\text{y}}$
$y$ tends from $x$ to $a + x.s$
$\therefore\ \phi=\frac{\mu_0\text{I}\text{a}^{\text{a}+\text{x}}}{2\pi}\int_{\text{x}}\frac{\text{dy}}{\text{y}}$
$=\frac{\mu_0\text{Ia}}{2\pi}\big[\log_{\text{e}}\text{ y}\big]_{\text{x}}^{\text{a}+\text{x}}$
$=\frac{\mu_0\text{Ia}}{2\pi}\log_{\text{e}}\Big(\frac{\text{a}+\text{x}}{\text{x}}\Big)$
For mutual inductance $M,$ the flux is given as:
$\phi=\text{mI}$
$\therefore\ \text{mI}=\frac{\mu_0\text{Ia}}{2\pi}\log_{\text{e}}\Big(\frac{\text{a}}{\text{x}}+1\Big)$
$\text{m}=\frac{\mu_0\text{a}}{2\pi}\log_{\text{e}}\Big(\frac{\text{a}}{\text{x}}+1\Big)$
Emf induced in the loop, $\text{e}=\text{B}'\text{av}\Big(\frac{\mu_0}{2\pi}\Big)\text{av}$ Given,
$I = 50A$
$x = 0.2m$
$a = 0.1m$
$v = 10m/s$
$\text{BQv}=\frac{\text{M}\text{v}^2}{\text{r}}$
$\therefore\ \text{B}2\pi\text{r}\lambda\text{r}^2=\frac{\text{Mv}}{\text{r}}$
$\text{v}=\frac{\text{B}2\pi\text{r}\lambda\text{r}^2}{\text{M}}$
For $\text{r}\leq\text{a}$ and $\text{a}<\text{R},$ we get
$(\text{i})=-\frac{2\text{B}_0\text{a}^2\lambda}{\text{MR}}\text{k}$
$e = 5 \times 10^{-5}V$

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