Tag: physics

Questions Related to physics

Two coils have a mutual inductance of $0.005\ H$. The current changes in the first coil according to equation $I=I _0sin\omega t$, where $I _0=10A$ and $\omega=100\pi rad/s$. The maximum value of emf (in volt) in the second coil is.

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $2\pi$

  2. $5\pi$

  3. $\pi$

  4. $4\pi$

Show answer
Correct Option: B
Explanation:

$EMF=\frac {MdI}{dt}$
$=(0\cdot 005)I _0 w cos wt$
Maximum EMF$=(0\cdot 005)\times 10\times 100\pi$
$=5\pi$

The mutual inductance of the system of two coils is $5mH$. The current in the first coil varies according to the equation $I={ I } { o }\sin { wt } $ where ${ I } _{ o }=10A$ and $W=100\pi \, rad/s$. The value of maximum induced emf in the second coil is ______

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $2\pi V$

  2. $\pi V$

  3. $5\pi V$

  4. $4\pi V$

Show answer
Correct Option: C
Explanation:

$Emf=M\cdot \cfrac { di }{ dt } =5\times { 10 }^{ -3 }\times { I } _{ o }\omega \cos { \omega t } \ { \left( Emf \right)  } _{ max }=5\times { 10 }^{ -3 }\times { I } _{ o }\omega =5\times { 10 }^{ -3 }\times 10\times 100\pi =5\pi V$

A short solenoid of length $4cm$, radius $2cm$ and $100$ turns is placed inside and on the axis of a long solenoid of length $80cm$ and $1500$ turns. A current of $3A$ flows through the short solenoid. The mutual inductance of two solenoids is

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $0.012H$

  2. $5.3\times {10}^{-5}H$

  3. $5.91\times {10}^{-3}H$

  4. $8.3\times {10}^{-5}H$

Show answer
Correct Option: C
Explanation:

As $M = \cfrac{\mu _0N _1N _2A}{l}$


where,
$A =$ common cross-sectional area
$l =$ length of small coil
$N _1 =$ No. of turns of small coil
$N _1 =$ No. of turns of long coil

$M = \cfrac{4\pi \times 10^{-7} \times 100 \times 1500 \times \pi \times (\cfrac{2}{100})^2}{(\cfrac{4}{100})} = 59157.6 \times 10^{-7} = 5.91 \times 10^{-3} H$



When the current in a coil changes from 8 ampere to 2 ampere in $3 \times 10^{-2}$ second, the e.m.f. induced in the coil is 2 volt. The self inductance of the coil (in millinery) is

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. 1

  2. 5

  3. 20

  4. 10

Show answer
Correct Option: A
Explanation:

$E.M.F. = L \dfrac{di}{dt}$


$2 = L \times \dfrac{8-2}{3 \times 10^{-2}}$

L = 1 millinery

Here (A) is correct answer

Two coils have mutual inductance $0.005 H$. The current changes in the form coil according to equation, $ I = I _0 \sin \omega t . $ Where $ I _0 = 10 A. $ and $ \omega = 100 \pi $ rads/s. The maximum value of emf in the second coil is :

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $ 12 \pi $

  2. $ 8 \pi $

  3. $ 5 \pi $

  4. $ 2 \pi $

Show answer
Correct Option: C
Explanation:
Mutual inductance between two coils
M = 0.005 H 
Peak current $ l _0 = 10 A $
Angular frequency $ \omega = 100 \pi $ rad/s
Current $ l = l _0 \sin \omega t $
$ \dfrac {d}{dt} = \dfrac {d}{dt} ( l \sin \omega t ) $
$ = l _0 \cos \omega t . \omega $
$ = 10 \times 1 \times 100 \pi $
$ = 1000 \pi $
Hence, induced emf is given by 
$ E = M \times \dfrac {dl}{dt} $
$ = 0.005 \times 1000 \times \pi = 5 \pi V $

A coil of area 500 $cm^2$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude $4 \times 10^{-5}$ $weber/m^2$. If it is rotated by 180 about an axis passing through one of its diameter in 0.1 sec, find the average induced emf.

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. zero.

  2. 30 mV

  3. 40 mV

  4. 50 mV

Show answer
Correct Option: C
Explanation:
0iven that :-  $N=1000, B=4\times 10^{-5}weber/m^2, A=500cm^2=0.05m^2$

Initial flux linked with the coil, $\phi _1=1000\times 4\times 10^{-5}\times 0.05$

$\implies \phi _1=2\times 10^{-3}weber$

After rotation of $180^{o}$, B remains same but normal vector gets reversed, hence $\phi _2=-\phi _1$

Average EMF=$E=\dfrac{-\Delta \phi}{t}$

$\implies E=-\dfrac{\phi _2-\phi _1}{t}$

$\implies E=\dfrac{\phi _1-\phi _2}{t}$

$\implies E=\dfrac{2\phi _1}{t}$

$\implies E=\dfrac{4\times 10^{-3}}{0.1}V$

$\implies E=40mV$

Answer-(C)

A long straight wire is placed along the axis of a circular ring of radius $R$. The mutual inductance of this system is

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $\dfrac{\mu _{0}R}{2}$

  2. $\dfrac{\mu _{0}\pi R}{2}$

  3. $\dfrac{\mu _{0}}{2}$

  4. $0$

Show answer
Correct Option: D

A coil of $Cu$ wire (radius $-r$, self-inductance-$L$) is bent in two concentric turns each having radius $\dfrac{r}{2}$. The self-inductance is now

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $2L$

  2. $L$

  3. $4L$

  4. $\dfrac{L}{2}$

Show answer
Correct Option: C

Mutual inductance of a system of two thin coaxial conducting loops of radius 0.1 m, and then center separated by distance 10 m is (Take $\mu^{2} = 10$)

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $2 \times 10^{-20}$ H

  2. $2 \times 10^{-13}$ H

  3. $2 \times 10^{-18}$ H

  4. $2 \times 10^{-15}$ H

Show answer
Correct Option: A

A ring of radius $r$ is uniformly charged with charge $q.$ If the ring is rotated about it's axis with angular frequency $\omega$, then the magnetic induction at its centre will be-

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $10 ^ { - 7 } \times \frac { \omega } { q r }$

  2. $10 ^ { - 7 } \times \frac { 9 } { \omega r }$

  3. $10 ^ { - 7 } \times \frac { r } { q \omega }$

  4. $10 ^ { - 7 } \times \frac { q \omega } { r }$

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} T=\dfrac { { 2\pi  } }{ w }  \ i=\dfrac { { qw } }{ { 2\pi  } }  \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi  } }  \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r }  \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$