Tag: electromagnetic induction and alternating currents

Quizzes (10) Questions (183)

Questions Related to electromagnetic induction and alternating currents

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

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

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

Give the MKS units for the following quantities.
Magnetic Induction.

  1. Weber$/m^3$.
  2. Weber$/m^2$.
  3. Weber$/m^4$.
  4. Weber$/m^5$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

B. $Weber/m^2$

The term magnetic flux density refers to the fact that B is magnetic flux per unit surface. This relationship is based on Faraday's law of magnetic induction.
The SI unit measuring the strength of B is T(Tesla= $Weber/m^2$)

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

Two coaxial coils are very close to each other and their mutual inductance is $5mH$. If a current $50\sin{500t}$ is passed on one of the coils then the peak value of induced emf in the secondary coil will be:

  1. $5000V$
  2. $500V$
  3. $150V$
  4. $125V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By the principle of mutual inductance we know that flux induced in coil2 will depend on the current flowing in coil1

$\phi _1=Mi _1$
Here
M=Mutual inductance
$i _1$=current in 1st coil 
Now by Faraday's law induced emf in 2nd coil will be 
$\frac { d\Phi  }{ dt }=emf$
$emf=M\frac{di _1}{dt}$
Now put value of current and mutual inductance
$emf=(5\times 10^{-3}) \frac{d (50\sin500t)}{dt} $
$emf=(5\times 10^{-3})\times 50\times 500\times \cos500t$
$emf=125\cos500t$
So above is the equation of induced emf
So maximum value of induced emf will be
$emf _{max}=125 volts$

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

A solenoid would over a rectangular frame. If all the linear dimensions of the frame are increased by a factor $3$ and the number of turns per unit length remains the same, the inductance increased by a factor of :-

  1. $3$
  2. $9$
  3. $27$
  4. $63$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Firstly, the relation between inductance and length is calculated


$ emf=\dfrac{-Nd\phi }{dt}=-NA\dfrac{dB}{dt} $

$ B=\mu \dfrac{N}{l}I $

$ emf=-\dfrac{\mu {{N}^{2}}A}{l}\dfrac{dI}{dt} $

$ emf=-L\dfrac{dI}{dt} $

$ L=\dfrac{\mu {{N}^{2}}A}{l} $

So, L is inversely proportional to length of solenoid.

If all the linear dimensions of the frame are increased by a factor 3 and the number of turns per unit length remains the same

Then area $\begin{align}

$ A=3l\times 3b $

$ =9lb $

$ {{L}^{'}}=\dfrac{9{{\mu }_{0}}{{N}^{2}}A}{l} $

$ L'=9\,L $

So new inductance becomes 9 times.

 

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

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 

  1. $\dfrac{i _{1}}{i _{2}}=\dfrac{1}{4}$
  2. $\dfrac{i _{1}}{i _{2}}=4$
  3. $\dfrac{W _{2}}{W _{1}}=4$
  4. $\dfrac{V _{2}}{V _{1}}=\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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

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

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:

  1. 8 x $10^{-5}$ H
  2. 3 x $10^{-5}$ H
  3. 4 x $10^{-5}$ H
  4. 6 x $10^{-5}$ H
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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 $\left(\mu _o=4\pi\times 10^{-7} TmA^{-1}\right)$

  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. $2.4\pi \times 10^{-5}H$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. $30\pi\ V$
  2. $60\pi\ V$
  3. $15\pi\ V$
  4. $300\pi\ V$
  5. $3\pi\ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $