3 Marks Question

Question 13 Marks

What is Motional electromotive force $($emf$) ?$ Obtain motional emf $(\varepsilon)$= B I ${v}$ using proper example.

Answer

$\rightarrow $"The induced emf arising due to some motion is called Motional emf."

$\rightarrow $ As shown in figure, a rectangular conductor $\text{PQRS}$ is placed in a time independent, uniform magnetic field. If $\theta$ is the angle between $\vec{B}$ and area vector $\vec{A}$ of coil, then here $(\theta=0)$. Here, rod $PQ$ is free to do frictionless motion. Its effective length is $l$.
$\rightarrow $ On moving conductor $PQ$ with uniform velocity $\vec{v}$ as shown in figure, area enclosed by closed circuit $\text{PQRS}$ changes with time.
$\rightarrow $ Suppose, at any instant,
$RQ =x \ RS =l \text { then }$
Magnetic flux linked with closed loop $\text{PQRS}$ is
$\phi_{ B }= B I x(\because v=0 \rightarrow \cos \theta=1)$
$\rightarrow $ Distance $x$ changes with time, as a result, time rate of change of $\phi_{ B }$ induces emf.
$\therefore \varepsilon=-\frac{d \phi_{ B }}{d t}=-\frac{d}{d t} \text { (B } l x \text { ) [From eq. (1)] }$
$\therefore \varepsilon=- B l \frac{d x}{d t} \text { But } \frac{d x}{d t}=-0$
Where $v$ is speed of conductor $PQ$
$\therefore \varepsilon= B / v$
Equation $(2)$ is formula for motional emf.

$\rightarrow $ As shown in figure, a conducting rod having length $l$ is moving with velocity $\vec{v}$ in direction perpendicular to uniform magnetic field $\vec{B}$.
$\rightarrow $ Lorentz force $F =q v B$ acts on free electric charges $($free electrons$)$ in rod $PQ.$
$\rightarrow $ Free electric charges experience this force towards $Q$ end of conductor. Thus, work done by Lorentz force to shift electric charge from $P$ to $Q$ is
$W = F l=q B{v}l$
$\rightarrow $ By definition of emf,
$e m f =\text { Work done on unit electric charge }$
$\therefore \varepsilon =\frac{ W }{q}$
$ =\frac{q B {v} l}{q}$
$\therefore \varepsilon = B{v}l$
This is expression for motional emf.

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Question 23 Marks

For a system made of two very long coaxial solenoids, explain mutual inductance. Prove $M _{12}= M _{21}= M$ $($Reciprocity Theorem$).$

Answer

$\rightarrow$ As shown in figure, a system is prepared by adjusting two solenoids $S _1$ and $S _2$ of equal length $l$ in coaxial manner.
$\rightarrow$ For solenoid $S _1$,
radius $=r_1$, Total Number of turns $= N _1$, Number of turns per unit length $=n_1$.
$\rightarrow$ For solenoid $S _2$,
radius $r_2\left(r_2>r_1\right)$, Total Number of turns $= N _2$,
Number of turns per unit length $=n_2$.
$\rightarrow$ On passing electric current $I _2$ through solenoid $S _2$, magnetic flux linked per turn of $S _1$ is $\phi_1$, then total magnetic flux linked with $S _1$.
$N _1 \phi_1 \propto I _2$
$\therefore N _1 \phi_1= M _{12} I _2$
$\therefore M _{12}=\frac{ N _1 \phi_1}{ I _2}$
$\rightarrow$ In equation $(1), M _{12}$ is called Mutual inductance of solenoid $S_1$ with respect to solenoid $S_2$.
But, $N _1 \phi_1=\left(n_1 l\right)\left( A _1 B_2\right)$
$=\left(n_1 l\right)\left[\left(\pi r_1^2\right)\left(\mu_0 n_2 I _2\right)\right]$
$\therefore$ From equation $(1) (2),$
$M _{12}=\frac{\mu_0 n_1 n_2 \pi r_1^2 l I _2}{ I _2}=\mu_0 n_1 n_2 \pi r_1^2 l$
$\rightarrow$ In the same way, on passing electric current $I_1$ through $S _1$, magnetic flux linked per turn of solenoid $S _2$ is $\phi_2$ then total magnetic flux linked with $S _2$ will be $N _2 \phi_2$.
$\therefore N _2 \phi_2 \propto I _1$
$\therefore N _2 \phi_2= M _{21} I _1$
$\therefore M _{21}=\frac{ N _2 \phi_2}{ I _1}$
$\rightarrow$ In equation $(4), M _{21}$ is called Mutual inductance of solenoid $S_2$ with respect to solenoid $S_1$.
But, $N _2 \phi_2=\left(n_2 l\right)\left( A _1 B_1\right)=\left(n_2 l\right)\left(\pi r_1^2\right)\left(\mu_0 n_1 I _1\right) \ldots$
$\left(\because A _1< A _2\right)$
$\rightarrow$ From equation $(4) (5),$
$M _{21}=\mu_0 n_1 n_2 \pi r_1^2 l$
$\rightarrow$ From equation $(3)$ and $(6),$ it is clear that
$M _{12}= M _{21}= M$
Where $M$ is called Mutual inductance of entire system.
Equation $(7)$ is called Reciprocity Theorem.

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Question 33 Marks

Describe principle and construction of AC generator with figure and discuss its working.

Answer

• Principle of Working :
→When coil having area $\vec{A}$ rotates in a magnetic field $\vec{B}$, effective area $A \cos \theta$ of closed loop changes continuously and hence magnetic flux linked with loop changes continuously (where $\theta$ is angle between $\vec{A}$ and $\vec{B}$ ).

• Construction :
→Basic elements of AC generator are shown in figure.
→Magnetic field $\overrightarrow{ B }$ is formed by permanent magnetic poles N and S .
→As shown in figure, the generator consists of a coil mounted on a rotor shaft.
→The axis of rotation of this coil is perpendicular to the direction of the magnetic field. This coil is called armature.
→Coil connected with rotor shaft can be rotated with respect to its own axis keeping perpendicular to $\overrightarrow{ B }$, using external mechanical arrangement.
→Due to rotation of coil, magnetic flux linked with coil changes and hence induced emf is produced across two ends of coil which is joined with slip rings and brushes.
• Working :
→When coil is rotating with angular speed $\omega$, angle between $\vec{B}$ and area vector $\vec{A}$ at any instant is $\theta=\omega t . \quad\left(\because \omega=\frac{\theta}{t}\right)$ →Magnetic flux linked with coil at time $t$ is
$\phi_{ B }= BA \cos \theta= BA \cos \omega t$
→According to Faraday's law, induced emf in coil having N turns is
$\begin{aligned}
\varepsilon & =- N \frac{d \phi_{ B }}{d t} \\
\therefore \varepsilon & =- N BA \frac{d}{d t}(\cos \omega t) \\
\therefore \varepsilon & =+ N B \omega A \sin \omega t
\end{aligned}$
→When value of $\sin \omega t$ is $\pm 1$, induced emf becomes maximum equal to $\varepsilon_0$
$\therefore \varepsilon_0= NBA \omega$
→Equation (2) is expression for instantaneous induced emf and equation (3) for maximum induced emf.
from equation (2) & (3),
$\varepsilon=\varepsilon_0 \sin \omega t$
→Equation (4) is expression for induced emf obtained from AC generator which changes with time according to function $\sin \theta$ and also changes direction at fixed time interval. Thus it is called Alternating voltage and current. Equation (4) can be given as follows also,
$\varepsilon=\varepsilon_0 \sin 2 \pi v t$
Where $\omega=2 \pi \nu$ with $v$ as frequency of AC generator, which is equal to frequency of rotation of coil.
→Value of $\varepsilon$ changes periodically between $+\varepsilon_0$ and $-\varepsilon_0$.

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