Tag: electromagnetic induction

Quizzes (6) Questions (197)

Questions Related to electromagnetic induction

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)

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

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
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, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

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

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

$L$ becomes twice.

Answer-(B)

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

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. $1/L^2$
Reveal answer Fill a bubble to check yourself
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, $L\propto \dfrac{1}{l}$

Answer-(C)

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

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. $1/N^2$
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)

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

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. $4/3$
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
                                      
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)

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

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. $314 H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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