Tag: electromagnetic induction

Quizzes (6) Questions (197)

Questions Related to electromagnetic induction

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

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in secondary in the step-up transformer is $1:10$. If the power to the consumer has to be supplied at $200\ V$, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is:

  1. $200:1$
  2. $150:1$
  3. $100:1$
  4. $50:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

 An inductor of inductance $100\ mH$ is connected in series with a resistance, a variable capacitance and an AC source of frequency $2.0\ kHz$; The value of the capacitance so that maximum current may be drawn into the circuit. 

  1. 50 nF

  2. 60 nF

  3. 63 nF

  4. 79 nF
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l}{X _L} = Lw = {10^{ - 1}} \times 2\pi  \times 2 \times {10^3}\{X _L} = 4\pi  \times {10^2}\Z = \sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} \i = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} }}\for,{i _{\max }}\{X _L} = {X _C}\\therefore {X _C} = Lw = \dfrac{1}{{Cw}}\C = \dfrac{1}{{{w^2}L}} = \dfrac{1}{{{{10}^{ - 1}} \times 4{\pi ^2} \times 4 \times {{10}^6}}}\ = \dfrac{{{{10}^{ - 5}}}}{{16{\pi ^2}}} = 63nF\end{array}$

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

$5 \mathrm { mV }$ is induced in a coil, when current in another nearby coil changes by $5 \mathrm { A }$ in $0.1$sec. The mutual inductance between the two coils will be

  1. $0.1 \mathrm { H }$
  2. $0.2 \mathrm { H }$
  3. $0.1 \mathrm { mH }$
  4. $0.2 \mathrm { mH }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The induced emf is given by e = M * (di / dt). Rearranging for mutual inductance M gives M = e / (di / dt). Substituting e = 5 mV = 5 x 10^-3 V and di/dt = 5 A / 0.1 s = 50 A/s yields M = (5 x 10^-3) / 50 = 0.1 x 10^-3 H = 0.1 mH.

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

In mutual induction 
A: when current in one coil increases, induced current in neighbouring coil flows in the opposite direction
B: When current in one coil decreases, induced current in neighbouring coil flows in the opposite direction

  1. A is true, B is false

  2. A and B are false

  3. A and B are true

  4. A is false, B is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lenz's Law states that the induced current will flow in a direction that opposes the change in magnetic flux. If current in the primary coil increases, the induced current in the secondary coil creates a magnetic field opposing the increase (opposite direction). If current decreases, the induced current creates a field to support the flux (same direction). Thus, A is true and B is false.

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

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $10^{-3}$ Wb to link with A and a flux per turn of $0.8 \times 10^{-3}$ Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :

  1. 5/4

  2. 1/1.6

  3. 1.6

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $\dfrac{L _1}{L _2}=\dfrac{200*10^{-3}}{400*0.8*10^{-3}}=1/1.6$

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

Two concentric coils each of radius equal to $2\pi\ cm$ are placed at right angles to each other. $3$ Ampere and $4$ ampere are the currents flowing in each coil respectively. The magnetic induction in $Weber/m^{2}$ at the centre of the coils will be ($\mu _{0}=4\pi \times 10^{-7}\ Wb/A-m$)

  1. $12\times 10^{-5}$
  2. $10^{-5}$
  3. $5\times 10^{-5}$
  4. $7\times 10^{-5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Magnetic Induction at centre of coil is
$B=\dfrac {\mu _0}{2r} \sqrt {I _1^2 +I _2^2}\quad I _1=3\ A$
$I _2=4\ A$
$=\dfrac {4\pi\times 10^{-7}}{2\times \dfrac {2\pi}{100}}\times \sqrt {3^2 +4^2}$
$=\dfrac {4\pi \times 10^{-5}\times 5}{2\times 2\pi}$
$=5\times 10^{-5} \dfrac {wb}{m^2}$
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

  1. 2.56 H

  2. 0.32 H

  3. 0.64 H

  4. 1.28 H
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First, find the resistance of the bulb: R = V^2 / P = 60^2 / 10 = 360 ohms. The operating current is I = P / V = 10 / 60 = 1/6 A. When connected to 100V, the impedance Z = V_source / I = 100 / (1/6) = 600 ohms. Since Z^2 = R^2 + Xl^2, 600^2 = 360^2 + Xl^2. Xl^2 = 360000 - 129600 = 230400. Xl = 480 ohms. Since Xl = 2 * pi * f * L, 480 = 2 * 3.14 * 60 * L. L = 480 / 377 = 1.273 H, which rounds to 1.28 H.

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

Two coaxial coils are very close to each other and their mutual inductance is $5mH$. If a current $50sin{500t}$ is passed in one of the coils then the peak value of induced emf in the secondary coil will be

  1. $5000V$
  2. $500V$
  3. $150V$
  4. $125V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to principle of mutual inductance, flux induced in coil  is equa to the current flowing in coil 1

$\phi _{2}= Mi _{1}$
By Faraday's laws,
$\dfrac{d\phi _{2}}{dt}=M\dfrac{di _{1}}{dt}$ = EMF
$\therefore EMF = 5\times 10^{-3}\dfrac{d}{dt}50 sin 500t$
$\therefore EMF = 125 cos500t$
So, maximum vaue of EMF would be 125 V