Tag: electromagnetic induction and alternating currents

Quizzes (10) Questions (183)

Questions Related to electromagnetic induction and alternating currents

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

In a given transformer for a given applied voltage, losses which remain constant irrespective of load changes are 

  1. friction and windage losses

  2. copper losses

  3. hysteresis and eddy current losses

  4. none of the above

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

In a given transformer for a given applied voltage, losses which remain constant irrespective of load changes are hysteresis and eddy current losses The losses that can occur in a material are: Iron losses: Iron loss is caused by the alternating flux in the core and consists of hysteresis and eddy current losses. of coercivity on the curve. (The reversed magnetizing force has flipped enough of the domains so that the net flux within the material is zero.)

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

Eddy currents are produced in a metallic conductor when

  1. The magnetic flux linked with it changes

  2. It is placed in a changing magnetic field.

  3. It is placed in a magnetic field.

  4. Both A and B
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Eddy currents are produced when the magnetic flux passing through the metal object continuously changes. This may happen due to many reasons:
1) The object is placed in a region with changing magnetic field.
2) The object continuously moves in and out of the magnetic field region (may be uniform or non uniform).

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

A magnet is dropped down an infinitely long vertical copper tube

  1. The magnet moves with continuously increasing velocity and ultimately acquires a constant terminal velocity

  2. The magnet moves with continuously decreasing velocity and ultimately comes to rest

  3. The magnet moves with continuously increasing velocity but constant acceleration

  4. The magnet moves with continuously increasing velocity and acceleration
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the magnet falls through the copper tube, changing magnetic flux induces eddy currents in the tube. According to Lenz's law, these eddy currents oppose the motion of the magnet, creating an upward magnetic force. Eventually, this magnetic drag balances gravity, and the magnet falls at a constant terminal velocity.

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics
Which is the correct formula for calculating the power lost due to eddy currents per unit mass for a thin sheet or wire?? Where $P$ is the power lost per unit mass $(W/kg)$, $B _p$ is the peak magnetic field $(T)$, $d$ is the thickness of the sheet or diameter of the wire $(m)$, $f$ is the frequency $(Hz)$, $k$ is a constant equal to 1 for a thin sheet and 2 for a thin wire
  1. $P= \dfrac{{\pi}^2 {B _p}^2 d^2 f^2}{6k\rho D}$
  2. $P= \dfrac{{\pi}^2 {B _p}^2 d^2 f}{k\rho D}$
  3. $P= \dfrac{{\pi}^2 {B _p}^2 d^2 f^2}{6k\rho D^3}$
  4. $P= \dfrac{{\pi}^2 {B _p}^2 d^2 f}{6k\rho D^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A)$P=\dfrac { { \pi  }^{ 2 }{ B } _{ p  }^{ 2 }{ d }^{ 2 }{ f }^{ 2 } }{ 6k\rho D } \ m{ L }^{ 2 }{ T }^{ -3 }=\dfrac { \left( M{ T }^{ -2 }{ A }^{ -1 } \right) ^{ 2 }\left( { L }^{ 2 } \right) \left( { T }^{ -2 } \right)  }{ M{ L }^{ -3 } } \ m{ L }^{ 2 }{ T }^{ -3 }=M{ L }^{ 2 }{ T }^{ -3 }\ m{ L }^{ 2 }{ T }^{ -3 }\neq m{ L }^{ 2 }{ T }^{ -2 }$ where A = amphere, T = Time, L = Length ,M = mass.

similarly C and D not matches so option A is correct .

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

Eddy currents are used in

  1. electrolysis

  2. making a galvanometer dead beat

  3. electroplating

  4. to increase the sensitivity of galvanometer
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In general the coil of galvanometer oscillates about it's equilibrium due to rotational inertia which consumes some time. To avoid this coils is bound over  a metallic frame or plate oscillates in a magnetic field the eddy currents generated in the frame or plate oppose the motion and bring the frame to rest as the oscillations die out quickly. This is known as making galvanometer dead beat. 

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

The working of magnetic braking of trains is based on

  1. Steady current

  2. Eddy current

  3. Alternating current

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

 It works the same as a disk eddy current brake, by inducing closed loops of eddy current in the conductive rail, which generate counter magnetic fields which oppose the motion of the train.

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

Read the following statements and answer whether the given statement is true or false.

Eddy current involves loss of energy in the form of heat.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Eddy currents (also called Foucault currents) are loops of electrical current induced within conductors by a changing magnetic field in the conductor due to Faraday's law of induction. 

Eddy current involves loss of energy in the form of heat. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. They can be induced within nearby stationary conductors by a time-varying magnetic field created by an AC electromagnet or transformer. 
The statement is true.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A.c across L-R,L-C and L-C-R series circuits. In an LR circuit, $R=10\Omega$ and $L=2H$, If an alternating voltage of $120V$ and $60Hz$ is connected in this circuit, then the value of current flowing in it will be _____ A (nearly)

  1. $0.32$
  2. $0.16$
  3. $0.48$
  4. $0.8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Impedance Z = sqrt(R^2 + (2*pi*f*L)^2). R = 10, L = 2, f = 60. XL = 2 * 3.14 * 60 * 2 = 753.6. Z = sqrt(100 + 567913) approx 754. Current I = V/Z = 120 / 754 approx 0.16 A.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

An L-C-R series circuit with $100\omega$ resistance is connected to an A.C source of 200 V and angular frequency $300 rad\,s^{-1}$. When only the capacitor is removed, the current lags behind the voltage by $60^0$ . When only inductor is removed, the current leads the voltage by $60^0$. If all elements are connected , the current in the circuit is

  1. 0.5 A

  2. 1.5 A

  3. 2 A

  4. 2.5 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given R = 100. When C is removed (LR circuit), tan(60) = XL/R => XL = 100 * sqrt(3) = 173.2. When L is removed (RC circuit), tan(60) = XC/R => XC = 100 * sqrt(3) = 173.2. Since XL = XC, the circuit is at resonance when all elements are connected. At resonance, Z = R = 100. Current I = V/R = 200 / 100 = 2 A.