Tag: electromagnetic induction

Questions Related to electromagnetic induction

Two different coils have self inductance $L _{1}=8\ mH, L _{2}=2\ mH$. The current in the second coil is also increased at the same constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are $i _{1}, V _{1}$ and $W _{1}$ respectively. Corresponding values for the second coil at the same instant are $i _{2}, V _{2}$ and $W _{2}$ respectively. Then 

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $\dfrac{i _{1}}{i _{2}}=\dfrac{1}{4}$

  2. $\dfrac{i _{1}}{i _{2}}=4$

  3. $\dfrac{W _{2}}{W _{1}}=4$

  4. $\dfrac{V _{2}}{V _{1}}=\dfrac{1}{4}$

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Correct Option: A,C,D
Explanation:

We know

$e=L\dfrac{di}{dt}$
$e\propto L$

So,

$\dfrac{e _1}{e _2}=\dfrac{L _1}{L _2}=\dfrac{8}{2}=\dfrac{4}{1}$

Since $P=el=Constant$

Therefore,

$\dfrac{di _1}{dt}=\dfrac{di _2}{dt}$

$P _1=P _2=P$

$e _1i _1=e _2i _2$

$\therefore \dfrac{i _1}{i _2}=\dfrac{e _2}{e _1}=\dfrac{1}{4}$

Thus, ratio of current is 1:4

Two coils A and B having turns 300 and 600 respectively are placed near each other, on passing a current of 3.0 ampere in A, the flux linked with A is 1.2 x $10^{-4}$  weber and with B it is 9.0 x $10^{-5}$ weber. The mutual induction of the system is:

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1.  8 x $10^{-5}$ H

  2.  3 x $10^{-5}$ H

  3.  4 x $10^{-5}$ H

  4.  6 x $10^{-5}$ H

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Correct Option: A
Explanation:
In passing current $(I=3A)$ through coil $A$, the flux linked $(\phi _1)$ is given by:
$\phi _1=\mu _0 N _1^2I \left (\dfrac {A}{L}\right)\quad [N _1=300,\ given\\ A=Area\ of\ coil.\ L=Length]$
and material inductance $(M)$ between $A$ & $B$ given by:
$M=\mu _0 N _1 N _2\left (\dfrac {A}{L}\right)$ (Assuming both coils have same $A$ & $L$)
so from above,
$M=\dfrac {N _2}{N _1}\dfrac {\phi _1}{I}=\dfrac {600}{300}\times \dfrac {1.2\times 10^{-4}}{3}$
$\Rightarrow \ M=8\times 10^{-5}H$ (Ans)

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A=10cm^2$ and length $=20cm$. If one of the solenoids has $300$ turns and the other $400$ turns, their mutual inductance is $\left(\mu _o=4\pi\times 10^{-7} TmA^{-1}\right)$

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $4.8\pi \times 10^{-4}H$

  2. $4.8\pi \times 10^{-5}H$

  3. $2.4\pi \times 10^{-4}H$

  4. $2.4\pi \times 10^{-5}H$

Show answer
Correct Option: D
Explanation:

$\mu _{12}=\mu _on _1n _2lA$
$4\pi \times 10^{-7}\times 300\times 400 \times 10\times 10^{-4}\times 20\times 10^{-2}$
$=2.4\times 10^{-5}$

A $50\ Hz$ ac current of peak value $2$ A flows through one of the pair of coils. If the mutual inductance between the pair of coils is $150\ mH$. then the peak value of voltage induced in the second coil is

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $30\pi\ V$

  2. $60\pi\ V$

  3. $15\pi\ V$

  4. $300\pi\ V$

  5. $3\pi\ V$

Show answer
Correct Option: A
Explanation:

Let the current flows through coil 1 is, ${{I} _{1}}={{I} _{0}}\sin \omega t$ where, ${{I} _{0}}$is peak value of current.

Magnetic flux linked with coil 2 is ${{\phi } _{2}}=M   {{I} _{1}}=M{{I} _{0}}\sin \omega t$

Emf in coil 2 is

${{\varepsilon } _{2}}=\dfrac{d{{\phi } _{2}}}{dt}=\dfrac{d(M{{I} _{0}}\sin \omega t)}{dt}=M \omega {{I} _{0}}\cos \omega t$

So, peak value of voltage induced in coil 2 is $=M{{I} _{o}}\omega ....(1)$

Given that,

$ \nu =50\,Hz $

$ {{I} _{0}}=2\,A $

$ L=150\,mH $

$ \omega =2\pi \nu =2\pi \times 50 $

Put all values in equation (1)

$ current=150\times {{10}^{-3}}\times 2\times 100\pi  $

$ I=30\pi \,V $

The self-inductances of two identical coils are $0.1\ H$. They are wound over each other. Mutual inductance will be-

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $0.1\ H$

  2. $0.2\ H$

  3. $0.01\ H$

  4. $0.05\ H$

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Correct Option: A

The length of solenoid is 0.3 m and the number of turns is 2000. The area of cross-section of the solenoid is $1.2\times 10^{-3} m^2$. Another coil of turns 200 is wrapped over the solenoid. a current of 2 A is passed through the solenoid and its direction is changed in 0.25 sec. Then the induced emf in the coil:

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $4.8\times10^{-2} V$

  2. $4.8\times10^{-3} V$

  3. $3.2\times10^{-3} V$

  4. $3.2\times10^{-2} V$

Show answer
Correct Option: A
Explanation:
$N _1=2000$
$N _2=300$
$A=1.2\times 10^{-3}m^2$
$l=0.3\ m$
$t=0.25\ s$
$I=0.2\ A$
we know,

Mutual inductance $(M)=\dfrac{\mu _0N _1N _2A}{l}$

$\Rightarrow M=\dfrac{\mu _0N _1N _2A}{l}$

also,
induced emf $(\varepsilon)=M.\dfrac{dI}{dt}$

and $\dfrac{dI}{dt}=\dfrac{2-(-2)}{0.25}=\dfrac{4}{0.25}=16$

so $\varepsilon =M\dfrac{dI}{dt}=\dfrac{\mu _0N _1N _2A}{l}.\dfrac{dI}{dt}$

$\varepsilon=\dfrac{4\pi \times 10^{-7}\times 2000\times 300\times 1.2\times 10^{-3}\times 16}{0.3}$

$\boxed{\varepsilon=0.048\ V=4.8\times 10^{-2}\ V}$ hence $(A)$ option is correct

A coil of radius $1\ cm$ and of turns $100$ is placed in the middle of a long solenoid of radius $5\ cm$ and having $5\ turns/ cm$. The mutual induction in millihenry is

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $0.0316$

  2. $0.063$

  3. $0.105$

  4. Zero

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Correct Option: A

When 100 volts d.c. is applied across solenoid a current of 1.0 amp flows in it. When 100 volts a.c. is applied across the same coil, the current drops to 0.5 amp. If the frequency of the a.c. source is 50 Hz the impedance and inductance of the solenoid are

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. 200 ohm and 0.55 Henry

  2. 100 ohm and 0.86 Henry

  3. 200 ohm and 1.0 Henry

  4. 100 ohm and 0.93 Henry

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Correct Option: A

The inductance of a solenoid 0.5 m long of cross-sectional area $420 cm^{2}$ and with $500$ turns is 

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

  1. $12.5\ mH$

  2. $1.25\ mH$

  3. $15.0\ mH$

  4. $0.12\ mH$

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Correct Option: B

A current I flows in an infinity long wire with cross section in the from of a semicircular ring of radius R the magnitude of the magnetic induction along its axis is :- 

mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

  2. $\dfrac { \mu _ { 0 } \mathbf { I } } { 4 \pi R }$

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

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

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Correct Option: C
Explanation:

Let semicircular ring of radius $R$ as shown in figure 

An elementary length $dl$ cut for finding magnetic field,
So, 
$dl = Rd\theta $
So, elementary current along $dl$ is $di = \frac{i}{{\pi R}} \times Rd\theta  = \frac{{id\theta }}{\pi }$.
Now, you can see that elementary part of length is perpendicular upon $dB$.
so,
$dB = \frac{{{\mu _0}di}}{{2\pi R}}$

$ = \frac{{{\mu _0}di}}{{2{\pi ^2}R}}$


Now, magnetic filed along axis $B = \int\limits _0^\pi  {db.\sin \theta d\theta }  = 2dB$
Now, put $dB$ in here,
So, $B = \frac{{{\mu _o}i}}{{{\pi ^2}R}}$
Hence, 
Option $C$ is the correct answer.