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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.

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

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.

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.

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.

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.

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)

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)

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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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.

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