Introduction to induction - class-XII

Self inductance, $L = \dfrac{\mu _0 N^2 A}{l}$ __(1)

both (a) and (b) 

$As , \varepsilon , = ,  L , \dfrac{dI}{dt},$
$L , = , \varepsilon \dfrac{dt}{dI} \Rightarrow , L , = , \dfrac{volt \times second}{ampere}$

 Henry (symbol H) is the SI derived unit of self-inductance. 

$\begin{array}{l} B=\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 2\left( { 2{ r _{ 1 } } } \right)  } }  \ =\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 2 } } } }  \ { B^{ ' } }=\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } }  \ \Rightarrow B=\frac { { { B^{ ' } } } }{ 2 }  \ \Rightarrow \frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 1 } } } } =\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 4{ r _{ 1 } } } }  \ \Rightarrow \frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 4{ r _{ 1 } } } } =\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 1 } } } }  \ \Rightarrow \frac { { { i _{ 1 } } } }{ { { i _{ 2 } } } } =1 \ Hence, \ option, , D, , is, , correct, , answer. \end{array}$

Self-inductance,
$\begin{array}{l}L \propto {N^2}\\therefore \dfrac{{{L _1}}}{{{L _2}}} = {\left( {\dfrac{{{N _1}}}{{{N _2}}}} \right)^2} = {(2)^2} = 4...........{ since,{N _2} = 2{N _1}} \\left[ {{L _2} = 4{L _1}} \right]\end{array}$

Option $D$ is does not have the same dimensions as the Henry.

The henry (symbolized H) is the Standard International ( SI ) unit of inductance and it is used to measure inductance.

$\begin{array}{l} , , As, frequency, approaches, zero, or, DC,, the, inducators, reac\tan  ce, would, decrease, tozero, , \ acting, like, a, short, circuit., this, means, inductive, reac\tan  ce, is, proportional, to, fequency \ , , , , , , , , , , , , , so, ,, for, low, frequency, the, { { limimiting } }, , value, of, impedance, is, L,, and, , alternating,  \ current, is, flowing, in, inductance, L, and, resistance, R., , The, frequency, of, source, is, \frac { \omega  }{ { 2\pi  } } . \ so, the, correct, option, is, A. \end{array}$

The self-inductance of a solenoid is proportional to the square of the number of turns per unit length (L is proportional to n^2). If n is doubled, L becomes 2^2 = 4 times the original value.

The SI unit of inductance is Henry,
$\displaystyle L =-\dfrac{e}{\dfrac{di}{dt}}$
SI unit $= \displaystyle \dfrac{volt}{A} \times s$
$=volt \times second \times  ampere^{-1}$

$L=\cfrac{\phi}{i}$

Here,$C=7\times { 10 }^{ -11 }F$,
        $L=0.39\times { 10 }^{ -6 }H$

A source= $220V$

In LR circuit,

$M=\mu _o\mu _r \cfrac {N _1N _2}{l}A$

$L1=0.01$

${\phi _{21}} = M.i,$

The SI unit of inductance is the Henry (H).

Self-inductance L is proportional to the square of the total number of turns (N^2). If N is doubled, L becomes 2^2 = 4 times the original.

$ L\dfrac{dI}{dt} = emf = \dfrac{d\phi}{dt}$

On removal of load from the circuit, the circuit suddenly becomes an open circuit.
Thus $ \dfrac{di}{dt} \rightarrow \infty $
For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the inductor is $ L\dfrac{di}{dt} $
Therefore, the circuit has high inductance.

Factual question, self inductance is proportional to square of the number of turns in coil.

Weber $=ML^2T^{-2}I^{-1}$
$=ML^2T^{-2}Q^{-1}T=ML^2T^{-1}Q^{-1}$   ($I=QT^{-1}$)
Henry H is  SI unit of inductance. 
$H=ML^2T^{-2}I^{-2}$     also $I=QT^{-1}$
so $H=ML^2T^{-2}Q^{-2}T^2=ML^2Q^{-2}$

Self inductance depend on the number of turns in a coils $L \propto N$.

$e=\cfrac { -\left( { \phi  } _{ 2 }-{ \phi  } _{ 1 } \right)  }{ t } =\cfrac { -\left( 0-NBA \right)  }{ t } =\cfrac { NBA }{ t } $
$ \therefore \quad t=\cfrac { NBA }{ e } =\cfrac { 50\times 2\times { 10 }^{ -2 }\times { 10 }^{ -2 } }{ 0.1 } \quad t=0.1s$

${ X } _{ L }=\omega L=2\pi fL$
$=2\times \cfrac { 22 }{ 7 } \times 60\times 0.2=75.4\Omega $

$V=L\dfrac{di}{dt}$

Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

$E=vBl$    where $v$=speed of conductor

Since the phase angle is $45^{\circ}$,

As per mechanical-electrical analog, displacement is analogous to charge and force analogous to voltage.

Inductor stores energy in the electromagnetic field.
Capacitor and condenser stores energy in the electrostatic field and resistance doesn't store energy in any field.

Given, $V=110v$ ,$P=10W$ ,$V _0=220 w$ ,$f=50Hz$