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Electromagnetic Induction — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
  1. Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
  2. 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 = 0.2m. Take a = 0.1m and assume that the loop has a large resistance.

✓

Answer

  1. 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 = 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:

$\phi=\text{mI}$

$\therefore\ \text{mI}=\frac{\mu_0\text{Ia}}{2\pi}\log_{\text{e}}\Big(\frac{\text{a}}{\text{x}}+1\Big)$

$\text{m}=\frac{\mu_0\text{a}}{2\pi}\log_{\text{e}}\Big(\frac{\text{a}}{\text{x}}+1\Big)$

  1. 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

$\text{BQv}=\frac{\text{M}\text{v}^2}{\text{r}}$

$\therefore\ \text{B}2\pi\text{r}\lambda\text{r}^2=\frac{\text{Mv}}{\text{r}}$

$\text{v}=\frac{\text{B}2\pi\text{r}\lambda\text{r}^2}{\text{M}}$

$\text{For }\text{r}\leq\text{a}\text{ and }\text{a}<\text{R},\ \text{we get}:$

$(\text{i})=-\frac{2\text{B}_0\text{a}^2\lambda}{\text{MR}}\text{k}$

e = 5 × 10-5V

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