Tag: electromagnetic induction

Quizzes (6) Questions (197)

Questions Related to electromagnetic induction

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

A long solenoid  of diameter $0.1\ m$ has $2 \times {10^4}$ turns per metre.At the centre of the solenoid, a coil of $100$ turns and radius $0.01\ m$ is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to $0\ A$ from $4\ A$ in $0.05\ s$. If the resistance of the coil is $10 \ {\pi ^2}\Omega ,$ the total charge flowing through the coil during this time is.

  1. $32\ \pi \mu C$
  2. $16\ \mu C$
  3. $32\ \mu C$
  4. $16\ \pi \mu C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

Number of turns, $n=100$

Radius, $r=0.01\,m$

Resistance, $R=10\pi^2 \Omega$

As we know,

$\epsilon=-N\dfrac{d\phi}{dt}$

$=\dfrac{\epsilon}{R}=-\dfrac NR\dfrac{d\phi}{dt}$,   $\Delta I=-\dfrac NR\dfrac{d\phi}{dt}$

$\dfrac{\Delta}{\Delta t}=-\dfrac NR\dfrac{\Delta\phi}{\Delta t}\implies \Delta q=-[\dfrac NR(\dfrac{\Delta \phi}{\Delta t})]\Delta t$

$-$ve sign shoes that induced emf opposes the change in flux.

$\Delta q=\dfrac{\mu _0 ni\pi r^2}{R}$

$\Delta q=\dfrac{4\pi\times 10^{-7}\times 100\times 4\times \pi\times (0.01)^2}{10\pi^2}=32\mu C$
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

Two coils, a primary of $400$ turns and a secondary of $20$ turns are wound over an iron core of length $20\pi\ cm$ and cross-section of $2\ cm$ radius. If $\mu _{r}=800$, then the coefficient of mutual induction is approximately

  1. $1.6\times 10^{7}H$
  2. $1.6\times 10^{-2}H$
  3. $1.6\times 10^{3}H$
  4. $1.6\ H$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mutual inductance of a solenoid system is given by M = (mu_0 * mu_r * N1 * N2 * A) / l. Substituting N1 = 400, N2 = 20, length l = 20 pi cm = 0.2 pi m, cross-sectional area A = pi * r^2 = pi * (0.02)^2 m^2, and mu_r = 800 yields M = 1.6 x 10^-2 H.

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

A charge of ${10^{ - 6}}C$ is describing a circular path of radius $1$ cm making $5$ revolution per second . The magnetic induction field at the centre of the circle is 

  1. $\pi \times {10^{ - 10}}T$
  2. $\pi \times {10^{ - 9}}T$
  3. $\frac{\pi }{2} \times {10^{ - 10}}T$
  4. $\frac{\pi }{2} \times {10^{ - 9}}T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A moving charge constitutes a current I = q * f = (10^-6 C) * (5 rev/s) = 5 x 10^-6 A. The magnetic field at the centre of a circular current loop is B = (mu_0 * I) / (2 * r). Substituting r = 1 cm = 0.01 m gives B = (4 * pi * 10^-7 * 5 * 10^-6) / (2 * 0.01) = pi * 10^-10 T.

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

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right)  } $ then the maximum value of e.m.f.induced in coil B is 

  1. $\pi \quad volt$
  2. $\pi /2volt$
  3. $\pi /3volt$
  4. $\pi /4volt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The induced EMF is E = M * (di/dt). Given i = 5 * sin(10 * pi * t), di/dt = 5 * 10 * pi * cos(10 * pi * t) = 50 * pi * cos(10 * pi * t). The maximum EMF is E_max = M * (di/dt)_max = 2 * 10^-2 * 50 * pi = 1 * pi = pi V.

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

When the primary current in the spark-coil of a car changes from $4A$ to zero in $10\mu s$, an emf of $40000$V is induced in the secondary. The mutual inductance between the primary and the secondary winding of the spark-coil will be-

  1. 1 H

  2. 0.1 H

  3. 10 H

  4. zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the relation e = M * (di / dt), we can solve for mutual inductance M = e / (di / dt). Here e = 40000 V, di = 4 A, and dt = 10 microseconds = 10 x 10^-6 s. Thus M = 40000 / (4 / 10^-5) = 40000 / 400000 = 0.1 H.

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

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right)  } $ then the maximum value of e.m.f. induced in coil B is

  1. $\pi \quad volt$
  2. $\pi /2volt$
  3. $\pi /3volt$
  4. $\pi /4volt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 530241. The induced EMF is E = M * (di/dt). With i = 5 * sin(10 * pi * t), di/dt = 50 * pi * cos(10 * pi * t). E_max = 2 * 10^-2 * 50 * pi = pi V.

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

The electric field of an electromagnetic wave is given by, $E=(50N^{-1})\, \sin { \omega  } (t-x/c)$. Find the energy contained in a cylinder of cross section $10cm^2$ and length $50 cm$ along the x-axis.

  1. $5.5\times 10^{-12}J$
  2. $4.5\times 10^{-12}J$
  3. $5\times 10^{-13}J$
  4. $3.5\times 10^{-10}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The energy density of an EM wave is u = (1/2) * epsilon0 * E^2 + (1/2) * (B^2 / mu0). For an EM wave, the average energy density is u_avg = (1/2) * epsilon0 * E0^2. The total energy is U = u_avg * Volume. Volume = Area * length = 10 * 10^-4 m^2 * 0.5 m = 5 * 10^-4 m^3. E0 = 50 V/m. u_avg = 0.5 * 8.85 * 10^-12 * 50^2 = 1.1 * 10^-8 J/m^3. U = 1.1 * 10^-8 * 5 * 10^-4 = 5.5 * 10^-12 J.

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

An electron having kinetic energy T is moving in a circular orbit of radius R perpendicular to a uniform magnetic induction $\vec { \mathrm { B } }$  If kinetic energy is doubled and magnetic induction tripled, the radius will

  1. $\frac { 3 R } { 2 }$
  2. $ \frac{{\sqrt 2 }}{3}R$
  3. $\sqrt { \frac { 2 } { 9 } } R$
  4. $\sqrt { \frac { 4 } { 3 } } R$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know$,$

$R = \frac{{\sqrt {2mk} }}{{qB}}$
$R' = \frac{{\sqrt {2m2k} }}{{q\left( {3B} \right)}}$
$ = \frac{{\sqrt 2 }}{3}\frac{{\sqrt {2mk} }}{{qB}}$
$ = \frac{{\sqrt 2 }}{3}R$
Hence,
option $(B)$ is correct answer.