With the help of a labelled diagram, explain how does an AC works. There are $N$ turns in coil of AC generator. Area of each turn is A and the coil is rotating with a uniform angular velocity. Establish an expression for generated induced electromotive force. What is the source of energy production by this method?
What do you understand by self-induction?Obtain an expression for self-induction of a solenoid.###Define self-induction. Obtain an expression for self-induction of a solenoid of N turns, length $l$ and area of cross-section A.
Define induction coefficient. Explain the phenomenon of mutual inductance with the help of an experiment. Obtain an expression for mutual inductance coefficient of two solenoids and prove that $M=\sqrt{L_1 L_2}$.
(i) How will you show that Lenz's law is a conclusion of energy conservation law? Explain with suitable examples.
(ii) Using the expression for Lorentz force applied on charge carriers of any conductor, obtain an expression for that induced emf which is induced in a conductor of length $l$ moving with a velocity $v$, perpendicular to any magnetic field $B$.
A rectangular wire loop of sides 8 cm and 2 cm having a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1 cm s ^{-1}$ in a direction normal to the (a) longer side (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer
Given: Length of loop $l=8 cm=8 \times 10^{-2} m$ Breadth of loop b $=2 cm=2 \times 10^{-2} m$ Magnetic field $B=0.3 T$ Velocity of the loop $v=1 cm s ^{-1}=10^{-2} ms^{-1}$ $\therefore \quad$ Area of loop $A =l \times b$ $\begin{array}{l}A=8 \times 2 \times 10^{-4} m^2 \\ A=16 \times 10^{-4} m^2\end{array}$ Induced electromotive force $\varepsilon=$ ? Time of stay of each e.m.f. in the loop $t=$ ? (a) When velocity is perpendicular to the longer side, $\begin{aligned} \varepsilon & = B l v \\ & =0.3 \times 8 \times 10^{-2} \times 10^{-2} \\ & =2.4 \times 10^{-4} V\end{aligned}$ Electromotive force will remain in the loop till the loop does not come out of the magnetic field which means till the time equal to the time taken in covering distance equal to length of the shorter size of the loop. $t=\frac{\text { Length of shorter size }}{v}$ $t=\frac{2 \times 10^{-2}}{10^{-2}}=2 s$ (b) When velocity is perpendicular to the shorter size, Breadth $b=2 \times 10^{-2} m$ $\varepsilon= B l v$ $\varepsilon= Bb v$$\because l=b$ $\begin{array}{l}=0.3 \times 2 \times 10^{-2} \times 10^{-2} \\ =0.6 \times 10^{-4} V\end{array}$ Time $(t)=\frac{\text { Length of longer size }}{v}$ $=\frac{8 \times 10^{-2}}{10^{-2}}=8 sec$.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s. What is the induced emf in the loop while the current is changing?
Answer
Given: Number of turns per em of length = 15 $\therefore$Number of turns per m of length = 1500 $n=1500 turns / m$ $A =2.0 cm^2$ $=2 \times 10^{-4} m^2$ $I _1=2.0 A, I _2=4.0 A$ $d I = I _2- I _1=4-2=2 A$ $d t=0.1 s$ $\begin{aligned} \therefore \quad \frac{d}{d t} & =\frac{2}{0.1}=20 As ^{-1} \\ \varepsilon & =\text { Induced electromotive force in } \\ \text { the loop } & \end{aligned}$ Magnetic field generated inside the solenoid $B =\mu_0 n I$ We know that $\phi_{ B }= BA =\mu_0 n IA$ $\therefore$ On using the relation, $\varepsilon=-\frac{d \phi_{ B }}{d t}$ $\varepsilon=-\frac{d}{d t} \phi_{ B }$ $=-\frac{d}{d t}\left(\mu_0 n IA \right)$ $\varepsilon=-\mu_0 n A \frac{d I }{d t}$ Therefore, $\quad|\varepsilon|=\left|-\mu_0 nA \frac{ dI }{ dt }\right|$ $\varepsilon=\mu_0 nA \frac{ dI }{ dt }$ On putting the values, $\varepsilon=4 \pi \times 10^{-7} \times 1500 \times 2 \times 10^{-4} \times 20$ $\begin{array}{l}=7.54 \times 10^{-6} V \\ =7.5 \mu V\end{array}$