Tag: electromagnetic induction

Questions Related to electromagnetic induction

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)

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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. same

  2. 2 times

  3. 3 times

  4. 4 times

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

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

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)

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $L$

  2. $L^2$

  3. $1/L$

  4. $1/L^2$

Show answer
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)

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $N$

  2. $N^2$

  3. $1/N$

  4. $1/N^2$

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)

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-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $1/3$

  2. $2/3$

  3. $1/9$

  4. $4/3$

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
                                      
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

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. increases

  2. decreases

  3. remains constant

  4. reaches zero

Show answer
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 _________.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $0.25 H$

  2. $0.5 H$

  3. $4H$

  4. $314 H$

Show answer
Correct Option: A
Explanation:

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

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