Tag: introduction to induction

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Questions Related to introduction to induction

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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. $\mu _0N^3\pi R^2l$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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 $r$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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)
Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. same

  2. doubled

  3. trippled

  4. quadrupled
Reveal answer Fill a bubble to check yourself
D Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. $A$
  2. $A^2$
  3. $A^3$
  4. $A^4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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)

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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. $N^4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)