Tag: electromagnetic induction and alternating currents

Questions Related to electromagnetic induction and alternating currents

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)

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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. increase

  2. remain same

  3. decrease

  4. none of these

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= $E=NBA$

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

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    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)

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $A$

  2. $A^2$

  3. $A^3$

  4. $A^4$

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= $E=NBA$

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

$\implies L\propto A$

Answer-(A)

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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 1/3 times

  2. 1/2 times

  3. 1/5 times

  4. 1/4 times

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 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-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $N$

  2. $N^2$

  3. $N^3$

  4. $N^4$

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= $E=NBA$

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

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

Hence, $L\propto N^2$

Answer-(B)