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 $50\ Hz$ $AC$ current of crest value $1\ A$ flows, through the primary of transformer. If the mutual inductance between the primary and secondary be $0.5\ H$, the crest voltage induced  in the secondary is

  1. 75 V

  2. 150 V

  3. 100 V

  4. 300V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The induced crest voltage in the secondary coil is given by E_s = M * (di/dt)_max. The rate of change of current is di/dt = omega * I_0 = (2 * pi * f) * I_0. Substituting M = 0.5 H, f = 50 Hz, and I_0 = 1 A gives E_s = 0.5 * (2 * pi * 50 * 1) = 50 * pi approx 150.7 V, which rounds to 150 V.

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

Which of the following statement is correct?

  1. when the magnetic flux linked with conducting loop is zero then emf induced is always zero

  2. when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero

  3. transformer works on mutual induction

  4. all of these
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

 Statement is.

A) When the magnetic flux linked with conducting loop is zero then emf induced is always zero.
     $emf=\dfrac{d\phi}{dt}$
  If $\phi=0$, $emf=\dfrac{d0}{dt}=0$
B) when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero.
    $emf=\dfrac{d\phi}{dt}=0$
    $d\phi=0$
   $\phi=constant$ magnetic flux is constant.
This is the wrong statement
C) The transformer works on mutual induction.
The correct statement is (A) and (C).


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

An electron originates at a point $A$ lying on the axis of a straight solenoid and moves with velocity $v$ at an angle $\alpha$ to the axis. The magnetic induction of the field is equal to $BA$ screen is oriented at right angles to the axis and is located at a distance $1$ from the point $a$. Find the distance from the axis to the point on the screen into which the electron strikes.

  1. $d = 5r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 2\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.
  2. $d = 2r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.
  3. $d = 3r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 3\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.
  4. $d = 4r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electron moves in a helical path in a uniform magnetic field. The radius of the helix is r = (mv*sin(alpha))/(eB) and the pitch angle/period determines the displacement. The distance from the axis is calculated using the geometry of the circular projection of the helical motion.

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

Two conducting circular loops of radii $R _{1}$ and $R _{2}$ are placed in the same plane with their centres coinciding. If $R _{1} \gg R _{2}$, the mutual inductance $M$ between them will be directly proportional to

  1. $R _{1}/R _{2}$
  2. $R _{2}/R _{1}$
  3. $R _{1}^{2}/R _{2}$
  4. $R _{2}^{2}/R _{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For two concentric loops where R1 >> R2, the magnetic field produced by the larger loop (R1) at its center is B = (mu0 * I) / (2 * R1). The flux through the smaller loop is Phi = B * Area2 = (mu0 * I * pi * R2^2) / (2 * R1). Since M = Phi / I, M = (mu0 * pi * R2^2) / (2 * R1), which is proportional to R2^2 / R1.

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

The mutual inductance $M _{12}$ of coil 1 with respect to coil 2

  1. increases when they are bought nearer.

  2. depends on the current passing through the coils.

  3. increases when one of them is rotated about an axis.
  4. is not same as $M _{21}$ of coil 2 with respect to coil 1.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mutual inductance depends on the geometry, orientation, and separation of the coils. Bringing them closer increases the magnetic flux linkage between them, thereby increasing the mutual inductance.

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.