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JEE Important Chapter - Electromagnetic Induction and Alternating Current

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Electromagnetic Induction and Alternating Current for JEE

Electromagnetic Induction and Alternating Current for JEE

Electromagnetic Induction and Alternating Current is an important section as far as JEE is concerned. The Electromagnetic induction section explains what is electromagnetic induction along with advanced problems related to it. There are many concepts and problems related to AC circuits that are discussed in this article.


Electromagnetic induction begins with laws governing the principle of electromagnetic induction and moves on to self-induction and mutual induction. The self-inductance, mutual inductance and energy stored in an inductor are very important in this section along with the application of electromagnetic induction and types of electromagnetic induction problems.


In alternating current, we first deal with pure resistive, inductive and capacitive AC circuits and their phasor diagrams. Then phasor diagrams and impedance of RLC circuit are discussed along with various AC circuit problems.


Let us see what is electromagnetic induction, and the important concepts of alternating current  and alternating current formulas needed for JEE Main and JEE advanced with solved examples.


Important Topics of Electromagnetic Induction and Alternating Current

  • Faraday’s Law of electromagnetic induction

  • Lenz’s law

  • Motional emf

  • Self Induction and Mutual Induction

  • Combination of Inductance

  • Growth current in LR circuit

  • Purely resistive, capacitive and inductive AC circuits

  • Inductive reactance and capacitive reactance

  • Series RLC circuit


Important Concepts of Electromagnetic Induction and Alternating Current

Name of the Concept

Key Points of the Concept

  1. Faraday’s Law of electromagnetic induction

  • According to the first law of electromagnetic induction, whenever the magnetic flux linked with the coil changes(ɸ) changes with time(t), an emf is induced in it.

  • According to the second law of electromagnetic induction, the induced emf is directly proportional to the rate of change of magnetic flux.

$e\varpropto \dfrac{d\phi}{dt}$

  1. Lenz’s law

  • Lenz’s law gives the direction of induced current or induced emf and it is based on the law of conservation of energy

  • Lenz’s law states that the direction of induced emf or induced current is in such a way that it opposes the cause that produces it.

$e=- \dfrac{d\phi}{dt}$ 

  • The minus shows that the induced emf opposes the change in magnetic flux.

  1. Motional emf

  • Consider a conductor rod of length l moving with a velocity (v) perpendicular to the uniform magnetic field (B). Then the motional emf induced in the conductor is given by,

$e=Blv$


(Image will be uploaded soon)



  1. Self Induction and Mutual Induction

  • Due to self-induction, an emf is induced in the coil whenever there is a change in current in the coil. The emf induced in the coil is given b,

$e=- L\dfrac{di}{dt}$

  • Sefl inductance(L) is the property of the coil to resist the change in current in the coil and its unit is Henry(H)

(Image will be uploaded soon)

  • Due to mutual induction, an emf is induced in the secondary coil due to the change in current in the first coil.

$E_2=- M\dfrac{dI_1}{dt}$

  • Coefficient of mutual inductance (M) depends on both coils


  1. Combination of Inductance

  • When two coils having inductances L1 and L2 are connected in series and mutual inductance is neglected, then equivalent inductance is given by the formula,

$L_{eq}=L_1+L_2$

  • If the mutual inductance is considered, then

$L_{eq}=L_1+L_2\pm 2M$

  • If two inductances are connected in parallel, then

$\dfrac{1}{L_{eq}}=\dfrac{1}{L_{1}}+\dfrac{1}{L_{2}}$



  1. Growth current in LR circuit

  • When the switch is closed at t=0, then-current through the inductor at any time is given by,

$I=I_0(1-e^{\dfrac{-t}{\tau}})$


  • The steady current (I0) is the maximum constant  current passing through the current after some time.

  • The value of the time constant is given by,

$\tau=\dfrac{L}{R}$

  1. Purely resistive, capacitive and inductive AC circuits

  • For a purely resistive circuit, voltage and current are in the same phase.

$V=V_0\sin(\omega t)$

$I=I_0\sin(\omega t)$

Where V0 and I0 are the peak voltage and current respectively with angular frequency ⍵

  • Power dissipated in the purely resistive circuit is given by,

$P=V_{rms}\cdot I_{rms}$

  • For purely inductive circuit, the current is lagging behind the voltage by 900.

$V=V_0\sin(\omega t)$

$I=I_0\sin(\omega t-\dfrac{\pi}{2})$

  • Power dissipated in the purely inductive circuit is zero

  • For a purely capacitive circuit, the current is leading the voltage by 900.

$V=V_0\sin(\omega t)$

$I=I_0\sin(\omega t+\dfrac{\pi}{2})$

  • Power dissipated in a purely capacitive circuit is zero

Er


  1. Inductive reactance and capacitive reactance

  • Inductive reactance(xL) is the resistance offered by an inductor against the AC current.

$X_L=L\omega $

  • Capacitive reactance(xC) is the resistance offered by a capacitor against the AC current.

$X_C=\dfrac{1}{C\omega} $

  1. Series RLC circuit

  • In a series RLC circuit, resistor, capacitor and inductor are connected in series.

$V=V_0\sin(\omega t)$

$I=I_0\sin(\omega t-\phi )$


  • Power factor of the series RLC circuit is the cosine of the phase difference between the voltage and current.

$\text{power factor}=\cos\phi=\dfrac{R}{Z}$

$\cos\phi=\dfrac{\text{Apparent power}}{\text{True power}}$

  • Impedance (Z) is the opposition or resistance offered by the RLC circuit against AC current

$Z=\sqrt{R^2+(X_L- X_C)^2}$

  1. Resonance in series RLC circuit

  • The current in the RLC circuit becomes maximum when the inductive reactance and capacitive reactance are equal. 

  • The formula for resonance frequency is given by,

$f=\dfrac{1}{2\pi\sqrt{LC}} $

  • At resonance, the impedance of the RLC circuit becomes equal to the resistance of the circuit.


List of Important Formulae

Sl. No

Name of the Concept

Formula

1.

Emf induced in a coil rotating in a magnetic field(B)

$e=\text{NBA}\omega \cos (\omega t)$

Where N is the number of turns, A is the area of cross-section and ⍵ is the angular velocity of the rotational motion of the coil


2.

Self-inductance of a solenoid having a length (l) and N number of turns

$L=\dfrac{\mu_0N^2A}{l} $

3.

Self-inductance of a circular coil having radius R

$L=\dfrac{\mu_0\pi N^2R}{2} $

4.

Frequency of LC oscillation

$f=\dfrac{1}{2\pi\sqrt{LC}} $

5.

Energy stored in an Inductor 

$E=\dfrac{1}{2}LI^2 $

6.

RMS current and RMS voltage

$I_{rms}=\dfrac{I_0}{\sqrt{2}}$

$V_{rms}=\dfrac{V_0}{\sqrt{2}}$

7.

Relation between mutual inductance and self inductances of two coils

$M=K\sqrt{L_1L_2}$

Where k is the coupling factor

8.

Quality factor of a series resonance circuit

$Q=\dfrac{L\omega_0}{R} $

$Q=\dfrac{1}{C\omega_0R} $

9.

Transformer ratio

$K=\dfrac{N_s}{N_p} $

$K=\dfrac{V_s}{V_p} $


Solved Examples

  1. A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms-1, at right angles to the horizontal component of the earth’s magnetic field of 0.3 × 10-4 Wb/m2. The value of the induced emf in wire is

Ans:  

Given,

Length of the horizontal wire, l = 10 m

Horizontal component of earth’s magnetic field, B = 0.3 × 10-4 Wb/m2

Here, the rod and horizontal component of earth’s magnetic field are perpendicular to each other. We can calculate the induced emf using the motional emf formula.

$e=Blv$

$e=0.3\times10^{-4}\times10\times 5$

$e=1.5\times10^{-3}~V$

Therefore, the value of induced emf in the wire is 1.5 × 10-3 V


Key Point: While calculating the motional emf induced in a conductor, the magnetic field perpendicular to the length of the conductor has to be considered.


  1. The potential difference across the resistance, capacitance and inductance is 80 V, 40 V and 100 V respectively in a series RLC circuit.

Ans: 

Given,

Voltage across the resistance, VR = 80 V 

Voltage across the inductance, VL = 100 V 

Voltage across the capacitance, VC = 40 V

The net voltage across the RLC circuit is given by, 

$V_{net}=\sqrt{{V_R}^2+(V_L- V_C)^2}$

$V_{net}=\sqrt{{80}^2+(100- 40)^2}$

$V_{net}= \sqrt{{80}^2+60^2}$

$V_{net}= 100 \text{ V}$

The power factor of the RLC circuit is calculated by the formula,

$\cos\phi=\dfrac{V_R}{V_{net}}$

$\cos\phi=\dfrac{80}{100}$

$\cos\phi=0.8$

Therefore, the power factor of the RLC circuit is 0.8


Key Point:  The net voltage across RLC circuit is equal to the phasor sum of voltage across resistance, inductance and capacitance.


Previous Year Questions from JEE Paper

  1. A resonance circuit having inductance and resistance 2 × 10-4 H and 6.28 Ω respectively oscillate at 10 MHz frequency. The value of the quality factor of this resonator is________. (JEE 2021)

Ans:  

Given,

Resistance of the circuit, R = 6.28 Ω

Inductance of the circuit, L = 2 × 10-4 H

The quality factor of the resonator is calculated using the formula,

$Q=\dfrac{L\omega_0}{R} $

$Q=\dfrac{2\times10^{-4}\times2\pi\times10\times10^{6}}{6.28} $

$Q=2000$


Trick: Quality factor can be obtained by the ratio of inductive capacitance or inductive reactance to the resistance.


  1. A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______ J.(JEE 2021)

Ans:

Given. The voltage across inductance, V = 3t volts

Inductance of the coil, L = 2H

The current in the coil that depends on time(t) can be obtained using the formula given by,

$e=L\dfrac{di}{dt}$

$3t=L\dfrac{di}{dt}$

$\int di=\dfrac{1}{L}\int3t. dt$

$i=\dfrac{3t^2}{2L}$

Energy stored in the coil is given by,

$E=\dfrac{1}{2}LI^2 $

$E=\dfrac{1}{2}L\left(\dfrac{3t^2}{2L}\right)^2 $

$E=\dfrac{1}{2}\times2\times\left(\dfrac{3\times 4^2}{2\times2}\right)^2 $

$E=144\text{ J}$

Therefore, energy stored in the capacitor is 144 J


Trick:  When an inductor is connected to variable voltage, then induced emf is equal to the applied voltage.


Practice Questions

  1. In a series RL circuit(R=1.5Ω, L=3H) and DC voltage=1V. Find the current at T=2 s.(Ans: 0.4 A)

  2. A coil of resistance 30 Ω and inductive reactance 20 Ω at 50 Hz frequency. If an AC source of 200 V, 100 Hz is connected across the coil, the current in the coil will be  (Ans: 4 A)


Conclusion

In this article, we discussed important topics and important formulas of electromagnetic induction and alternating current from the JEE point of view. Students must make sure that they do not miss any of the above important topics to obtain a good score in the JEE Main exam.

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FAQs on JEE Important Chapter - Electromagnetic Induction and Alternating Current

FAQ

1. How many questions are asked from electromagnetic induction and alternating current in JEE?

About 1-2 questions from this chapter are asked for JEE. It corresponds to around 8 marks in the JEE Main exam. Most probably, numerical problems will be asked in this section.

2. What should we concentrate whiles studying for alternating current in the JEE exam?

More emphasis should be given to problems involving AC circuits while dealing with alternating current. But you can easily solve these problems with proper practice and conceptual understanding.

3. How to get a good score in JEE Main to get into the best NITs?

First of all, learn all the concepts and formulas with proper understanding. Then do all the problems in NCERT textbooks and other study materials. Then, move to the last 20 years' previous JEE papers to have a better understanding of the test pattern and important topics.