Tag: electromagnetic induction

Quizzes (6) Questions (197)

Questions Related to electromagnetic induction

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 $

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

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

  1. $0.1\ H$
  2. $0.2\ H$
  3. $0.01\ H$
  4. $0.05\ H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two identical coils wound over each other, the mutual inductance M is equal to the self-inductance L of each coil, assuming perfect coupling (k=1). Given L = 0.1 H, M = 0.1 H.

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

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:

  1. $4.8\times10^{-2} V$
  2. $4.8\times10^{-3} V$
  3. $3.2\times10^{-3} V$
  4. $3.2\times10^{-2} V$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

  1. $0.0316$
  2. $0.063$
  3. $0.105$

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

The mutual inductance M is given by (mu_0 * N1 * N2 * A) / l. Using the solenoid parameters (n = 5 turns/cm = 500 turns/m, radius = 5 cm) and the inner coil (N = 100, radius = 1 cm), the calculation yields approximately 0.0316 mH.

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From DC, R = V/I = 100/1 = 100 ohm. From AC, Z = V/I = 100/0.5 = 200 ohm. Using Z^2 = R^2 + (2*pi*f*L)^2, we find 200^2 = 100^2 + (2*pi*50*L)^2, which solves to L approximately 0.55 H.

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

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 :- 

  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. $\dfrac { \mu _ { 0 } I } { 2 \pi ^ { 2 } R }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.