Tag: electromagnetic induction

Questions Related to electromagnetic induction

Which of the following does not have the same dimensions as the Henry?

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $\dfrac{\text {joule}}{(\text{ampere})^2}$

  2. $\dfrac{\text {tesla} - m^2}{(\text{ampere})^2}$

  3. $\text{ohm-second}$

  4. $\dfrac{1}{\text{Farad-second}}$

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Correct Option: D
Explanation:

Option $D$ is does not have the same dimensions as the Henry.

So, Option $D$ is correct.

The inductance is measured in 

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. ohm

  2. farad

  3. henery

  4. none of these

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Correct Option: C
Explanation:

The henry (symbolized H) is the Standard International ( SI ) unit of inductance and it is used to measure inductance.

Therefore, C is correct option.

Alternating current is flowing in inductance L and resistance R. The frequency of source is $\displaystyle\frac{\omega}{2\pi}$. Which of the following statement is correct.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. For low frequency the limiting value of impedance is L

  2. For high frequency the limiting value of impedance is $L\omega$

  3. For high frequency the limiting value of impedance is R

  4. For low frequency the limiting value of impedance is $L\omega$

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Correct Option: A
Explanation:

$\begin{array}{l} \, \, As\, frequency\, approaches\, zero\, or\, DC,\, the\, inducators\, reac\tan  ce\, would\, decrease\, tozero\, , \ acting\, like\, a\, short\, circuit.\, this\, means\, inductive\, reac\tan  ce\, is\, proportional\, to\, fequency \ \, \, \, \, \, \, \, \, \, \, \, \, \, so\, ,\, for\, low\, frequency\, the\, { { limimiting } }\, \, value\, of\, impedance\, is\, L,\, and\, \, alternating\,  \ current\, is\, flowing\, in\, inductance\, L\, and\, resistance\, R.\, \, The\, frequency\, of\, source\, is\, \frac { \omega  }{ { 2\pi  } } . \ so\, the\, correct\, option\, is\, A. \end{array}$

A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal to :



introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $-\in \;and\;r$

  2. $\in \;and\;-r$

  3. $-r\;and\;\in \;$

  4. $r\;and\;-\in$

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Correct Option: C
Explanation:

$E = V + Ir $
$\Rightarrow V=E-Ir$ 
$Comparing\;with\; y = mx + c$ 
$Slope = - r, intercept = E$




If circular coil with $N _{1}$ turns is changed in to a coil of $N _{2}$ turns. What will be the ratio of self inductances in both cases.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $\dfrac {N _{1}}{N _{2}}$

  2. $\dfrac {N _{2}}{N _{1}}$

  3. $\dfrac {N _{1}^{2}}{N _{2}^{2}}$

  4. $\sqrt {\dfrac {N _{1}}{N _{2}}}$

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Correct Option: A
Explanation:

If circular coil with N1 turns is changed in to a coil of N2 turns. 

$L=\dfrac{Nd \phi}{dt} $
$L _1=\dfrac{N _1d \phi}{dt} $
$L _2=\dfrac{N _2d \phi}{dt} $ 
 L is in Henries
        N is the Number of Turns
        Φ is the Magnetic Flux
        Ι  is in Amperes
$\dfrac{L _1}{L _2}=\dfrac{N _1}{N _2} $

If the number of turns per unit length of a coil of a solenoid is doubled the self-inductance of the solenoid will:

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. remain unchanged

  2. be halved

  3. be doubled

  4. become four times

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Correct Option: D

Henry, the SI unit of inductance can be written as :

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. weber ampere$^{-1}$

  2. volt second ampere$^{-1}$

  3. joule ampere$^{-1}$

  4. ohm s$^{-1}$

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Correct Option: B
Explanation:

The SI unit of inductance is Henry,
$\displaystyle L =-\dfrac{e}{\dfrac{di}{dt}}$
SI unit $= \displaystyle \dfrac{volt}{A} \times s$
$=volt \times second \times  ampere^{-1}$

Multiple Correct Answers Type
The SI unit of inductance, henry, can be written as

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. Weber/ampere

  2. Volt-second / ampere

  3. $Joule / (ampere)^2$

  4. Ohm-second

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Correct Option: A,B,C,D
Explanation:

$L=\cfrac{\phi}{i}$

$L=\cfrac{weber}{Ampere}$
$V=L\cfrac{Ldi}{dt}$
$L\to$ volt.second/ampere
$E=\cfrac{1}{2}Li^2$
$L\to $joule/(ampere$)^2$
$\omega L=X _L$
$L\to$ohm.second
Hence all are correct.

A lossless coaxial cable has a capacitance of $7\times { 10 }^{ -11 }$ F and an inductance of $0.39\mu H$. Calculate characteristic impedance of the cable.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 65

  2. 75

  3. 66

  4. 77

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Correct Option: B
Explanation:

Here,$C=7\times { 10 }^{ -11 }F$,
        $L=0.39\times { 10 }^{ -6 }H$

         ${ Z } _{ o }$ As the cable is lossless,
        $\therefore { Z } _{ o }\sqrt { \dfrac { L }{ C }  } =\sqrt { \dfrac { 0.39\times { 10 }^{ -6 } }{ 7\times { 10 }^{ -11 } }  } =75ohm$

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $\Omega$. Frequency & inductance are 50Hz & 0.7 H then wattless current is 

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 0.5 amp

  2. 0.7 amp

  3. 1.0 amp

  4. None

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Correct Option: A
Explanation:

A source= $220V$

The value of resistance= $220 \Omega$
Frequency= $50 Hz$
Inductance= $0.7H$
Find the wattless current= ?
Wattless component of current is $i=i _v\sin \theta$
                                                            $=\cfrac {Ev}{z}\sin \theta$
where, $z=$ impedance of $L-R$ circuit
                $=\sqrt {R^2+L^2W^2}$ so,
$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,
$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$
$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$
        $=\cfrac {220}{R^2+L^2W^2}LW$
        $=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$
        $=\cfrac {220 \times 220}{(220)^2+(220)^2}$
        $=0.5 A$ .