Tag: electromagnetic induction and alternating currents

Questions Related to electromagnetic induction and alternating currents

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 


Correct Option: C
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.)

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


Correct Option: D
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).

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


Correct Option: B
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. $P= \dfrac{{\pi}^2 {B _p}^2 d^2 f}{6k\rho D^2}$


Correct Option: A
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 .

A transformer core is laminated to 

  1. reduce hysteresis loss

  2. reduce eddy current loss

  3. reduce copper loss

  4. reduce all of the above loss


Correct Option: B
Explanation:

The transformer core is laminated to break eddy currents to form loop and hence reducing the power loss due to heat generated due to eddy current.


Answer-(B)

Eddy currents are used in

  1. electrolysis

  2. making a galvanometer dead beat

  3. electroplating

  4. to increase the sensitivity of galvanometer


Correct Option: B
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. 

The working of magnetic braking of trains is based on

  1. Steady current

  2. Eddy current

  3. Alternating current

  4. Pulsating current


Correct Option: B
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.

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


Correct Option: A
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.

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. $0.8$


Correct Option: B

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


Correct Option: C