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Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents 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.

  1. $2\pi$

  2. $5\pi$

  3. $\pi$

  4. $4\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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 ______

  1. $2\pi V$

  2. $\pi V$

  3. $5\pi V$

  4. $4\pi V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. $0.012H$

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

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

  4. $8.3\times {10}^{-5}H$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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



Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. 1

  2. 5

  3. 20

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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 :

  1. $ 12 \pi $

  2. $ 8 \pi $

  3. $ 5 \pi $

  4. $ 2 \pi $
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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.

  1. zero.

  2. 30 mV

  3. 40 mV

  4. 50 mV
Reveal answer Fill a bubble to check yourself
C Correct answer
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)
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

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

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

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

  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a duplicate of 530253 and 530257. The magnetic field of the wire is parallel to the plane of the ring, so flux is zero.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. $2L$

  2. $L$

  3. $4L$

  4. $\dfrac{L}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Self-inductance L is proportional to N^2 * radius. If a single loop of radius r is bent into two turns of radius r/2, N becomes 2 and radius becomes r/2. L_new = (N_new^2 * R_new) = (2^2 * r/2) = 2 * r. Since L_old is proportional to 1^2 * r = r, L_new = 2 * L_old. Wait, the formula for a loop is L ~ N^2 * R. So 2^2 * (r/2) = 2 * r. The answer 4L is often cited for specific geometry changes.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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-

  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. $10 ^ { - 7 } \times \frac { q \omega } { r }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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