Tag: electromagnetic induction

Questions Related to electromagnetic induction

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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. ${N}^{0}$

  2. $N$

  3. ${N}^{2}$

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

Show answer
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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $10s$

  2. $0.1s$

  3. $0.01s$

  4. $1s$

Show answer
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:

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $7.54\Omega $

  2. $0.754 \Omega $

  3. $75.4\Omega $

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

Show answer
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 :

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. Weber ampere$^{-1}$

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

  3. Joule ampere$^{-1}$

  4. ohm $s^{-1}$

Show answer
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.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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

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

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

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

Show answer
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-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
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)-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. number of turns

  2. radius of solenoid

  3. length of solenoid

  4. none of these

Show answer
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)-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $r$

  2. $r^2$

  3. $r^3$

  4. does not depend upon $r$

Show answer
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-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
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 :

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. same

  2. doubled

  3. trippled

  4. quadrupled

Show answer
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)