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Question 13 Marks

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

Answer

→ "The induced emf arising due to some motion is called Motional emf."

→As shown in figure, a rectangular conductor 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 $l$.
→On moving conductor PQ with uniform velocity $\vec{v}$ as shown in figure, area enclosed by closed circuit PQRS changes with time.
→Suppose, at any instant,
$RQ =x \& RS =l \text { then }$
Magnetic flux linked with closed loop PQRS is
$\phi_{ B }= B I x(\because v=0 \rightarrow \cos \theta=1)$
→Distance $x$ changes with time, as a result, time rate of change of $\phi_{ B }$ induces emf.
$\begin{array}{l}
\therefore \varepsilon=-\frac{d \phi_{ B }}{d t}=-\frac{d}{d t} \text { (B } l x \text { ) [From eq. (1)] } \\
\therefore \varepsilon=- B l \frac{d x}{d t} \quad \text { But } \frac{d x}{d t}=-0
\end{array}$
Where $v$ is speed of conductor PQ
$\therefore \varepsilon= B / v$
Equation (2) is formula for motional emf.

→As shown in figure, a conducting rod having length $l$ is moving with velocity $\vec{v}$ in direction perpendicular to uniform magnetic field $\vec{B}$.
→Lorentz force $F =q v B$ acts on free electric charges (free electrons) in rod PQ.
→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
$W = F l=q B{v}l$
→By definition of emf,
$\begin{aligned}
e m f & =\text { Work done on unit electric charge } \\
\therefore \varepsilon & =\frac{ W }{q} \\
& =\frac{q B {v} l}{q} \\
\therefore \varepsilon & = B{v}l
\end{aligned}$
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

→As shown in figure, a system is prepared by adjusting two solenoids $S _1$ and $S _2$ of equal length $l$ in coaxial manner.
→For solenoid $S _1$,
radius $=r_1$, Total Number of turns $= N _1$, Number of turns per unit length $=n_1$.
→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 $=n_2$.
→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 $S _1$.
$\begin{aligned}
& N _1 \phi_1 \propto I _2 \\
\therefore \quad & N _1 \phi_1= M _{12} I _2 \\
\therefore \quad & M _{12}=\frac{ N _1 \phi_1}{ I _2}
\end{aligned}$
→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),
$M _{12}=\frac{\mu_0 n_1 n_2 \pi r_1^2 l I _2}{ I _2}=\mu_0 n_1 n_2 \pi r_1^2 l$
→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 $N _2 \phi_2$.
$\begin{array}{l}
\therefore N _2 \phi_2 \propto I _1 \\
\therefore N _2 \phi_2= M _{21} I _1 \\
\therefore M _{21}=\frac{ N _2 \phi_2}{ I _1}
\end{array}$
→In equation (4), $M _{21}$ is called Mutual inductance of solenoid $S_2$ with respect to solenoid $S_1$.
But, $N _2 \phi_2=\left(n_2 l\right)\left( A _1 B_1\right)=\left(n_2 l\right)\left(\pi r_1^2\right)\left(\mu_0 n_1 I _1\right) \ldots$
$\left(\because A _1< A _2\right)$
→From equation (4) & (5),
$M _{21}=\mu_0 n_1 n_2 \pi r_1^2 l$
→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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