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Introduction to induction - class-XII

Description: introduction to induction
Number of Questions: 49
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Tags: physics electromagnetic induction and alternating currents electromagnetic induction
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A solenoid coil is wound on a frame of rectangular cross section. If all the linear dimension of the frame are increased by a factor of two and the number of turns per unit length of the coil remains the same, the self-inductance increases by a factor of

  1. $4$

  2. $8$

  3. $12$

  4. $16$


Correct Option: B
Explanation:

Self inductance, $L = \dfrac{\mu 0 N^2 A}{l}$ _(1)


where N = total number of turns


$n = \dfrac{N}{l}$, where n = number of turns per unit length

$N = nl$ put it in (1)

$L = \dfrac{\mu _0 (n l )^2 A}{l} = \mu _0 n^2 l A$

In the given question n is same
Area will increase by $4$ times
length will increase by $2$ times

$L' = \mu _0 n^2 (2l) (4A) = 8 \mu _0 n^2 lA$

$L' = 8L$ inductance will increased by $8$ times.

The physical quantity which is measured in the unit of Wb $A^{-1}$ is

  1. self inductance

  2. mutual inductance

  3. magnetic flux

  4. both (a) and (b)


Correct Option: D
Explanation:

both (a) and (b) 

self inductance and mutual inductance has same unit both measure magnetic flux per unit area.

The unit of inductance is equivalent to

  1. $\dfrac {volt\times ampere}{second}$

  2. $\dfrac { ampere}{volt\times second}$

  3. $\dfrac {volt}{Ampere\times second}$

  4. $\dfrac {volt\times second}{ampere}$


Correct Option: D
Explanation:

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

What is the SI unit of self-inductance ?

  1. Henry

  2. Tesla

  3. Weber

  4. Gauss


Correct Option: A
Explanation:

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

Two concentric co-planar circular loops of radii $r _1$ and $r _2$ carry currents of respectively $i _1$ and $i _2$ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of loops is half that due to $i _1$ alone at the centre. If $r _2 = 2r _1$. the value of $i _2 / i _1$ is 

  1. $2$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{4}$

  4. $1$


Correct Option: D
Explanation:

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

When the number of turns in a solenoid is doubled without any change in the length of the solenoid, its self-inductance becomes :

  1. Half

  2. Double

  3. Four times

  4. Eight times


Correct Option: C
Explanation:

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

Which of the following does not have the same dimensions as the Henry?

  1. $\dfrac{\text {joule}}{(\text{ampere})^2}$

  2. $\dfrac{\text {tesla} - m^2}{(\text{ampere})^2}$

  3. $\text{ohm-second}$

  4. $\dfrac{1}{\text{Farad-second}}$


Correct Option: D
Explanation:

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

So, Option $D$ is correct.

The inductance is measured in 

  1. ohm

  2. farad

  3. henery

  4. none of these


Correct Option: C
Explanation:

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

Therefore, C is correct option.

Alternating current is flowing in inductance L and resistance R. The frequency of source is $\displaystyle\frac{\omega}{2\pi}$. Which of the following statement is correct.

  1. For low frequency the limiting value of impedance is L

  2. For high frequency the limiting value of impedance is $L\omega$

  3. For high frequency the limiting value of impedance is R

  4. For low frequency the limiting value of impedance is $L\omega$


Correct Option: A
Explanation:

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

A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal to :



  1. $-\in \;and\;r$

  2. $\in \;and\;-r$

  3. $-r\;and\;\in \;$

  4. $r\;and\;-\in$


Correct Option: C
Explanation:

$E = V + Ir $
$\Rightarrow V=E-Ir$ 
$Comparing\;with\; y = mx + c$ 
$Slope = - r, intercept = E$




If circular coil with $N _{1}$ turns is changed in to a coil of $N _{2}$ turns. What will be the ratio of self inductances in both cases.

  1. $\dfrac {N _{1}}{N _{2}}$

  2. $\dfrac {N _{2}}{N _{1}}$

  3. $\dfrac {N _{1}^{2}}{N _{2}^{2}}$

  4. $\sqrt {\dfrac {N _{1}}{N _{2}}}$


Correct Option: A
Explanation:

If circular coil with N1 turns is changed in to a coil of N2 turns. 

$L=\dfrac{Nd \phi}{dt} $
$L _1=\dfrac{N _1d \phi}{dt} $
$L _2=\dfrac{N _2d \phi}{dt} $ 
 L is in Henries
        N is the Number of Turns
        Φ is the Magnetic Flux
        Ι  is in Amperes
$\dfrac{L _1}{L _2}=\dfrac{N _1}{N _2} $

If the number of turns per unit length of a coil of a solenoid is doubled the self-inductance of the solenoid will:

  1. remain unchanged

  2. be halved

  3. be doubled

  4. become four times


Correct Option: D

Henry, the SI unit of inductance can be written as :

  1. weber ampere$^{-1}$

  2. volt second ampere$^{-1}$

  3. joule ampere$^{-1}$

  4. ohm s$^{-1}$


Correct Option: B
Explanation:

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

Multiple Correct Answers Type
The SI unit of inductance, henry, can be written as

  1. Weber/ampere

  2. Volt-second / ampere

  3. $Joule / (ampere)^2$

  4. Ohm-second


Correct Option: A,B,C,D
Explanation:

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

$L=\cfrac{weber}{Ampere}$
$V=L\cfrac{Ldi}{dt}$
$L\to$ volt.second/ampere
$E=\cfrac{1}{2}Li^2$
$L\to $joule/(ampere$)^2$
$\omega L=X _L$
$L\to$ohm.second
Hence all are correct.

A lossless coaxial cable has a capacitance of $7\times { 10 }^{ -11 }$ F and an inductance of $0.39\mu H$. Calculate characteristic impedance of the cable.

  1. 65

  2. 75

  3. 66

  4. 77


Correct Option: B
Explanation:

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

         ${ Z } _{ o }$ As the cable is lossless,
        $\therefore { Z } _{ o }\sqrt { \dfrac { L }{ C }  } =\sqrt { \dfrac { 0.39\times { 10 }^{ -6 } }{ 7\times { 10 }^{ -11 } }  } =75ohm$

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $\Omega$. Frequency & inductance are 50Hz & 0.7 H then wattless current is 

  1. 0.5 amp

  2. 0.7 amp

  3. 1.0 amp

  4. None


Correct Option: A
Explanation:

A source= $220V$

The value of resistance= $220 \Omega$
Frequency= $50 Hz$
Inductance= $0.7H$
Find the wattless current= ?
Wattless component of current is $i=i _v\sin \theta$
                                                            $=\cfrac {Ev}{z}\sin \theta$
where, $z=$ impedance of $L-R$ circuit
                $=\sqrt {R^2+L^2W^2}$ so,
$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,
$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$
$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$
        $=\cfrac {220}{R^2+L^2W^2}LW$
        $=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$
        $=\cfrac {220 \times 220}{(220)^2+(220)^2}$
        $=0.5 A$ .

The time constant of a circuit is 10 sec, When a resistance of $ 100 \Omega $ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

  1. $250 H$

  2. $50H$

  3. $150 H$

  4. $25 H$


Correct Option: A
Explanation:

In LR circuit,

The time constant $\tau=\dfrac{L}{R}$
$10=\dfrac{L}{R}$
$L=10R$. . . . . . .(1)
When Resistance $100\Omega $ is connect in series, than the time constant is
$\tau'=\dfrac{L}{R+100}=2s$
$L=2R+200$. . . . . . .(2)
Equating equation (1 ) and (2), we get
$2R+200=10R$
$8R=200$
$R=25\Omega$
From equation (1),
$L=10R=10\times 25$
$L=250H$

the number of turn of primary and secondary coil of the transformer is 5 and 10 respectively ad the mutual inductance is 25 H. if the number f turns of the primary and secondary is made 10 and 5 , then the mutual inductance of the coils will be

  1. 6.25 H

  2. 12.5 H

  3. 25 H

  4. 50 H


Correct Option: D
Explanation:

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

$M \propto N _1N _2$
Since $N _1N _2=10 \times 5= 5 \times 10=50$ in both cases.
Mutual inductance will remain same.

The coefficients of self induction of two inductance coils arc 0.0 1H and 0.03H respectively. When they are connected in series so as to support each other. then the resultant self inductance becomes 0.06 Henry. The value of coefficient of mutual induction will be-

  1. 0.02 H

  2. 0.05 H

  3. 0.01 H

  4. ZERO


Correct Option: C
Explanation:

$L1=0.01$

$L2=0.03$
$L eff =0.06$

$Leff = L1+L2+2M$
$0.06=0.01+0.03+2M$
$2M = 0.06-0.03$
$M=0.03/2$
$=0.015$

In an induction coil, the coefficient of mutual induction is 4 henry. If a current of 5 ampere in 1  the primary coil is cut off in $\frac { 1 }{ 1500 } $s, the e.m.f at the terminals of the secondary coil will be:

  1. 15 kV

  2. 60 kV

  3. 10 kV

  4. 30 kV


Correct Option: D
Explanation:

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

$ \Rightarrow v = \frac{{d{\phi _{21}}}}{{dt}} = M\frac{{di}}{{dt}}$
$ = 4 \times \frac{5}{{\left( {1/1500} \right)}}$
$ = 20 \times 1500$
$ = 30kv$
Hence,
option $(D)$ is correct answer.

The inductance is measured in 

  1. Ohm

  2. Farad

  3. henery

  4. None of these


Correct Option: C

if the length and area of cross section of an inductor remain same but the number of turns is doubled its self inductance will become:

  1. half

  2. four time

  3. double

  4. one- fourth


Correct Option: B

The SI unit of inductance, the henry, can be written as :

  1. weber / ampere

  2. volt second / ampere

  3. joule / ampere$^2$

  4. ohm second


Correct Option: A,B,C,D
Explanation:

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


a) $ [L] = weber / ampere $

b) $ [L] = volt.second / ampere $

c) $ Joule = volt \times ampere \times second $

$ \Rightarrow volt \times second = Joule / ampere$

$ \Rightarrow [L] = joule/ ampere^2$

d) $ V = IR \Rightarrow volt = ampere.ohm$

$\Rightarrow [L] = ohm.second$.

If a spark is produced on removing the load from an AC circuit then the element connected in the circuit is

  1. high resistance

  2. high capacitance

  3. high inductance

  4. high impedance


Correct Option: C
Explanation:

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.

If N is the number of turns in a coil, the value of self-inductance varies are

  1. N$^o$

  2. N

  3. N$^2$

  4. N$^{-2}$


Correct Option: C
Explanation:

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

Which of the following units denotes the dimensions ${M}{L}^{2}/{Q}^{2}$, where ${Q}$ denotes the electric charge?

  1. $Weber$ ($Wb$)

  2. $Wb$$/\mathrm{m}^{2}$

  3. $Henry$ ($H$)

  4. $\mathrm{H}/\mathrm{m}^{2}$


Correct Option: C
Explanation:

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 $H=ML^2T^{-2}Q^{-2}T^2=ML^2Q^{-2}$

If $N$ is the number of turns in a circular coil then the value of self inductance varies as

  1. ${N}^{0}$

  2. $N$

  3. ${N}^{2}$

  4. ${N}^{-2}$


Correct Option: B
Explanation:

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

A magnetic field of $2\times {10}^{-2}T$ acts at right angles to a coil of area $100{cm}^{2}$ with $50$ turns. The average emf induced in the coil is $0.1V$. When it is removed from the field in $t$ second, the value of $t$ is

  1. $10s$

  2. $0.1s$

  3. $0.01s$

  4. $1s$


Correct Option: B
Explanation:

$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$

The inductive reactance of a coil of $0.2H$ inductance at a frequency of $60Hz$ is:

  1. $7.54\Omega $

  2. $0.754 \Omega $

  3. $75.4\Omega $

  4. $7.54\times { 10 }^{ -3 }\Omega $


Correct Option: C
Explanation:

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

The SI unit of inductance, the henry, can be written as :

  1. Weber ampere$^{-1}$

  2. Volt - s ampere$^{-1}$

  3. Joule ampere$^{-1}$

  4. ohm $s^{-1}$


Correct Option: B
Explanation:

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

So the unit $L=\dfrac{Vdt}{di}$
so SI unit of L is $Volt-s\,ampere^{-1}$

A long solenoid with length $l$ and a radius $R$ consists of $N$ turns of wire,Neglecting the end effects, find the self-inductance.

  1. $\mu _0N^2\pi R^2/l$

  2. $\mu _0N\pi R^2/l$

  3. $\mu _0N^2\pi R^3l$

  4. $\mu _0N^3\pi R^2l$


Correct Option: A
Explanation:

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


Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Hence, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Answer-(A)

If cross section area and length of a long solenoid are increased 3 times then its self-inductance will be changed how many times-

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: A
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times A$

$\implies E=\dfrac{\mu _{o}N^2 A}{l}i=Li$
                                                                                    where $L$=self inductance
   
$\implies L=\dfrac{\mu _oN^2A}{l}$

Hence, $L\propto \dfrac{A}{l}$

Hence, on increasing both $A$ and $l$ three times,

 $L$ will remain the same.

Hence, answer-(A)

Self inductance of a long solenoid depends upon following(s)-

  1. number of turns

  2. radius of solenoid

  3. length of solenoid

  4. none of these


Correct Option: A,B,C
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L$ depends on $N$,  $R$ and $l$.

Answer-(A),(B),(C)

Self-Inductance of a Long Solenoid is proportional to(where $r$ is radius of solenoid)-

  1. $r$

  2. $r^2$

  3. $r^3$

  4. does not depend upon $r$


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto R^2$

Answer-(B)

Area of a long solenoid is doubled.So how many times we have to increase its length to keep its self inductance constant-

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times A$

$\implies E=\dfrac{\mu _{o}N^2A}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
$\implies L=\dfrac{\mu _o N^2A}{l}$

$\implies L\propto \dfrac{A}{l}$

The length of solenoid should be doubled also on doubling the area to keep $L$ constant.

Answer-(B)

If radius of long solenoid is doubled, then its self inductance will be :

  1. same

  2. doubled

  3. trippled

  4. quadrupled


Correct Option: D
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto R^2$

Hence, on doubling radius, $L$ becomes 4 times.

Answer-(D)

If length of a solenoid is increased then what change should be made on no. of turns to keep self inductance constant-

  1. increase

  2. remain same

  3. decrease

  4. none of these


Correct Option: A
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

For constant $L$,    $N^2\propto l$

Hence, on increasing length of coil, number of turns should be increased to keep $L$ constant.

Answer-(A)

Self inductance of long solenoid is directly proportional to-($A$ is area of cross section)

  1. $A$

  2. $A^2$

  3. $A^3$

  4. $A^4$


Correct Option: A
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times A=Li$

$\implies L\propto A$

Answer-(A)

If radius of long solenoid is reduced to half of original without changing other physical factor,then its self inductance will change-

  1. 1/3 times

  2. 1/2 times

  3. 1/5 times

  4. 1/4 times


Correct Option: D
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto R^2$

Hence, on reducing the radius to half, $L$ will becomes $\dfrac{1}{4}$ times.

Answer-(D)

Self inductance $L$ of long solenoid is being proportional to the number of turns $N$ as-

  1. $N$

  2. $N^2$

  3. $N^3$

  4. $N^4$


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto N^2$

Answer-(B)

If the number of turns and length of the long solenoid are doubled without changing the area, then its self-inductance $L$ will be:

  1. same

  2. 2 times

  3. 3 times

  4. 4 times


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

$\implies L\propto \dfrac{N^2}{l}$

Hence, on doubling both $N$ and $l$,

$L$ becomes twice.

Answer-(B)

Self inductance of a long solenoid is directly proportional to-
(Where $L$ is the length of solenoid)

  1. $L$

  2. $L^2$

  3. $1/L$

  4. $1/L^2$


Correct Option: C
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto \dfrac{1}{l}$

Answer-(C)

Self inductance of a long solenoid is directly proportional to 
($N$ is no. of turns in solenoid)

  1. $N$

  2. $N^2$

  3. $1/N$

  4. $1/N^2$


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

Hence, $L\propto N^2$

Answer-(B)

If area of a long solenoid is doubled,length is trippled and no. of turns are remained contant.Then its self-inductance will be changed how many times-

  1. $1/3$

  2. $2/3$

  3. $1/9$

  4. $4/3$


Correct Option: B
Explanation:
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times A$

$\implies E=\dfrac{\mu _{o}N^2A}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 A}{l}$

Hence, on doubling area and tripling the length of solenoid,

$L$ becomes $\dfrac{2}{3}$ times.


Answer-(B)

When the speed at which a conductor is moved through a magnetic field is increased, the induced voltage

  1. increases

  2. decreases

  3. remains constant

  4. reaches zero


Correct Option: A
Explanation:

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


Hence, on increasing speed of conductor , induced voltage increases..

Answer-(A)

Reactance of a coil is $157\Omega$. On connecting the coil across a source of frequency $ 100Hz$, the current lags behind e.m.f. by ${ 45 }^{ o }$. The inductance of the coil is _________.

  1. $0.25 H$

  2. $0.5 H$

  3. $4H$

  4. $314 H$


Correct Option: A
Explanation:

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

$\dfrac{X _L}{R}=tan\phi=tan45^{circ}=1$
$\implies X _L=R$
$\implies \omega L=R$
$\implies 2\pi f L=R$
$\implies L=\dfrac{R}{2\pi f}$
$=\dfrac{157}{2\pi\times 100}H$
$=0.25H$

The electrical analog of mass is

  1. Diode

  2. Capacitance

  3. Inductance

  4. Resistance


Correct Option: C
Explanation:

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

From newton's second law of motion,
$F = m \cfrac{d^2x}{dt^2}$

For a diode, voltage and current are exponentially related and is non-linear.
For capacitance,  $V = \cfrac{Q}{C}$
For inductance, $V = L\cfrac{dI}{dt} = L\cfrac{d^2 q}{dt^2}$
For resistance, $V = IR = R \cfrac{dq}{dt}$

By comparing the above equations, it can be concluded that electrical analog of mass is inductance. 

Which of following circuit element stores energy in electromagnetic field?

  1. inductor

  2. condenser

  3. variable resistor

  4. capacitor


Correct Option: A
Explanation:

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.

Find the necessary inductance. if 110 V, 10 W rating bulb is to be used with 220 V A.C source having frequency 50 Hz.

  1. L=8.90 H

  2. L=6.75 H

  3. L=7.25 H

  4. L=6.5 H


Correct Option: B
Explanation:

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

Current through series inductor,
Current through bulb=$\dfrac{P}{V}=\dfrac{10W}{110V}=0.09A$
Voltage across inductor,$V _{ind}=\sqrt{V _0^{2} -V^{2}}$=$\sqrt{220^{2}-110^2}=191V$
Reactance of inductor,$R=\dfrac{V _{ind}}{I}=\dfrac{191}{0.09}=2122.22$
Also,$R=2 \pi fL$ or $L$=$\dfrac{R}{2 \pi f}$=$\dfrac{2122.22}{2 \pi 50}$=$6.75H$

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