Electromagnetic Induction and AC JEE Notes PDF: Download Now

Electromagnetic Induction and Alternating Currents JEE Notes

Electromagnetic induction (EMI) is the production of an electromotive force (emf) in a conductor whenever the magnetic flux linked with it changes. The chapter splits naturally into two halves: induction (Faraday, Lenz, inductance) and alternating current (sinusoidal sources, impedance, power). JEE asks direct concept questions, derivations of standard results, and numericals that couple EMI with mechanics or circuit theory.

Key Definitions

• Magnetic flux through a surface: $$\Phi_B = \int_S \vec{B} \cdot d\vec{S}$$
• Induced emf: emf produced by a change in magnetic flux linkage. For a single loop, Faraday's law gives: $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$ Its magnitude is: $$|\mathcal{E}| = \left|\frac{d\Phi_B}{dt}\right|$$
• Lenz's law: The induced current's magnetic effect opposes the change in flux that produced it. This is reflected by the negative sign in Faraday's law and is consistent with energy conservation.
• Self-inductance: For a coil with N turns and flux proportional to current: $$L = \frac{N\Phi_B}{I}$$ The SI unit is henry (H).
• Mutual inductance: $$M = \frac{N_2\Phi_{21}}{I_1} = k\sqrt{L_1L_2}$$ Here, the coefficient of coupling satisfies: $$0 \leq k \leq 1$$
• Sinusoidal alternating current and voltage: $$I(t) = I_0\sin(\omega t)$$ $$\mathcal{E}(t) = \mathcal{E}_0\sin(\omega t + \phi)$$ The angular frequency is: $$\omega = 2\pi f$$

$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$ $$I_{\mathrm{rms}} = \frac{I_0}{\sqrt{2}}$$ $$Z_{\mathrm{LCR}} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

EMI problems often involve a straight conductor moving in a field, a rotating loop in a uniform magnetic field, or an LCR resonance curve. The following sections explain these cases in detail.

Faraday's Law, Lenz's Law and Motional EMF

A. Faraday's Quantitative Law

For a single loop:

$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$

For a coil with N turns, each linked with the same magnetic flux:

$$\mathcal{E} = -N\frac{d\Phi_B}{dt}$$

The minus sign represents Lenz's opposition principle.

B. Sources of Flux Change That Appear in JEE

1. Changing the magnitude of the magnetic field, such as switching an electromagnet on or off.
2. Changing orientation: a coil rotating in a uniform magnetic field produces a sinusoidal emf. With an appropriate choice of initial orientation: $$\mathcal{E}(t) = NBA\omega\sin(\omega t)$$
3. Changing the area linked with the magnetic field, such as a sliding conductor on rails.
4. Composite cases, such as a circular loop whose radius varies with time.

C. Motional EMF in a Straight Conductor

A straight rod moving in a uniform magnetic field develops a motional emf given by:

$$\mathcal{E} = (\vec{v} \times \vec{B}) \cdot \vec{\ell}$$

When the rod, velocity and magnetic field are mutually perpendicular, the magnitude becomes:

$$|\mathcal{E}| = B\ell v$$

The polarity follows from the magnetic force on positive charges. For a closed circuit, the induced current direction can also be checked using Lenz's law.

D. Induced Electric Field

Even in empty space, a changing magnetic field sets up a non-conservative electric field. For a fixed closed path:

$$\oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}$$

This is the integral form of Faraday's law.

Worked Example 1: Sliding Wire Frame

A rectangular frame measuring 0.2 m × 0.3 m enters a uniform 0.5 T magnetic field at 4 m/s. The field is perpendicular to the frame, and the velocity is perpendicular to its 0.2 m side. Find the induced emf while entering the field.

1. The length sweeping across the field boundary is 0.2 m: $$\frac{dA}{dt} = \ell v = 0.2 \times 4 = 0.8\,\mathrm{m^2\,s^{-1}}$$
2. The induced emf magnitude is: $$|\mathcal{E}| = B\frac{dA}{dt} = 0.5 \times 0.8 = 0.4\,\mathrm{V}$$

Answer: 0.4 V

Worked Example 2: Rotating Coil Generator

A 100-turn coil with an area of 250 cm² rotates at 50 Hz in a uniform 0.1 T magnetic field. Its axis of rotation lies in the coil's plane and is perpendicular to the magnetic field. Find the rms emf.

1. Convert the area: $$A = 250 \times 10^{-4} = 2.5 \times 10^{-2}\,\mathrm{m^2}$$
2. Calculate the angular frequency: $$\omega = 2\pi f = 2\pi \times 50 = 100\pi\,\mathrm{rad\,s^{-1}}$$
3. Calculate the peak emf: $$\mathcal{E}_0 = NBA\omega$$ $$\mathcal{E}_0 = 100 \times 0.1 \times 2.5 \times 10^{-2} \times 100\pi \approx 78.5\,\mathrm{V}$$
4. Calculate the rms emf: $$\mathcal{E}_{\mathrm{rms}} = \frac{\mathcal{E}_0}{\sqrt{2}} = \frac{78.5}{\sqrt{2}} \approx 55.5\,\mathrm{V}$$

Answer: 55.5 V

Study Tip

After revising Faraday's and Lenz's laws, practise at least 20 previous-year numericals involving different sources and loops from the JEE Advanced Previous Papers. Write the magnetic flux as a function of time first, then differentiate it.

Self Inductance, Mutual Inductance and Energy Stored in Magnetic Field

A. Self Inductance

Common geometries:

• Long air-core solenoid: $$L = \mu_0 n^2 A\ell$$ Here, n is the number of turns per unit length: $$n = \frac{N}{\ell}$$
• Air-core toroid with a thin cross-section: $$L \approx \frac{\mu_0 N^2 A}{2\pi r_{\mathrm{mean}}}$$ This approximation assumes that the radial thickness is small compared with the mean radius.
• Thin circular wire loop: $$L \approx \mu_0 r\left[\ln\left(\frac{8r}{a}\right) - 2\right]$$ Here, r is the loop radius and a is the wire radius. This approximation neglects internal inductance and assumes: $$a \ll r$$

B. Mutual Inductance

For two long, coaxial air-core solenoids of the same length, with turn densities n₁ and n₂:

$$M = \mu_0 n_1 n_2 A\ell$$

Here, A is the common flux area, approximately the cross-sectional area of the inner solenoid.

C. Energy Considerations

The energy stored in a linear inductor carrying current I is:

$$U = \frac{1}{2}LI^2$$

The magnetic energy density in vacuum is:

$$u_B = \frac{B^2}{2\mu_0}$$

D. RL Growth and Decay

For growth from zero current under a constant source voltage:

$$I_{\infty} = \frac{V_s}{R}$$

For decay, let the initial current be I_i. The inductor voltage below uses the passive sign convention:

$$v_L = L\frac{dI}{dt}$$

Use the total resistance in the relevant current path. Growth and decay time constants are equal only if the resistance is the same in both cases.

Worked Example 3: Solenoid Inductance

A solenoid is 50 cm long, has a radius of 2 cm and contains 2,000 turns. It is filled with iron of constant relative permeability 1,500. Find its inductance, neglecting end effects.

1. Turns per unit length: $$n = \frac{2000}{0.5} = 4000\,\mathrm{m^{-1}}$$
2. Cross-sectional area: $$A = \pi r^2 = \pi(0.02)^2 \approx 1.257 \times 10^{-3}\,\mathrm{m^2}$$
3. Inductance: $$L = \mu_0\mu_r n^2 A\ell$$ $$L = (4\pi \times 10^{-7})(1500)(4000)^2(1.257 \times 10^{-3})(0.5)$$ $$L \approx 18.95\,\mathrm{H} \approx 19\,\mathrm{H}$$

Answer: 19 H

Worked Example 4: Mutual Inductance Coupling

Two coils have self-inductances of 2 mH and 8 mH, with a coefficient of coupling of 0.6. Calculate their mutual inductance.

$$M = k\sqrt{L_1L_2}$$

$$M = 0.6\sqrt{(2 \times 10^{-3})(8 \times 10^{-3})}\,\mathrm{H}$$

$$M = 2.4 \times 10^{-3}\,\mathrm{H} = 2.4\,\mathrm{mH}$$

Answer: 2.4 mH

Revision Hack

Create your own mini formula sheet and cross-check it with the Physics PDF on JEE Formula Sheets. Check unit conversions between H, mH and μH carefully.

Alternating Current: RMS Value, Phasor Analysis, LCR Circuits and Resonance

A. Sinusoidal Sources and Averages

• Peak to rms: $$I_{\mathrm{rms}} = \frac{I_0}{\sqrt{2}}$$ $$\mathcal{E}_{\mathrm{rms}} = \frac{\mathcal{E}_0}{\sqrt{2}}$$
• Average current over the positive half-cycle: $$I_{\mathrm{avg,half}} = \frac{2I_0}{\pi}$$
• Average current over a complete cycle: $$I_{\mathrm{avg,cycle}} = 0$$

B. Pure Components

In the table below, v and i represent instantaneous voltage and current. The phase column describes current relative to voltage.

C. Series LCR Circuit

Impedance magnitude:

$$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

Phase angle of the source voltage relative to the current:

$$\tan\phi = \frac{\omega L - \frac{1}{\omega C}}{R}$$

Current amplitude:

$$I_0 = \frac{\mathcal{E}_0}{Z}$$

D. Resonance

Series resonance occurs when the inductive and capacitive reactances are equal:

$$X_L = X_C$$

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

At resonance, impedance is minimum and current is maximum:

$$Z = R$$

$$I_{0,\mathrm{res}} = \frac{\mathcal{E}_0}{R}$$

The quality factor describes the sharpness of resonance:

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

The angular-frequency bandwidth between the half-power points is:

$$\Delta\omega = \omega_2 - \omega_1 = \frac{R}{L}$$

E. Power in AC Circuits

Average power:

$$P_{\mathrm{avg}} = \mathcal{E}_{\mathrm{rms}} I_{\mathrm{rms}}\cos\phi$$

For a series LCR circuit, the power factor is:

$$\cos\phi = \frac{R}{Z}$$

Worked Example 5: Capacitive Reactance

An ideal 5 μF capacitor is connected to a 230 V rms, 50 Hz AC supply. Find the rms current.

1. Capacitive reactance: $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}$$ $$X_C = \frac{1}{2\pi \times 50 \times 5 \times 10^{-6}} \approx 636.6\,\Omega$$
2. RMS current: $$I_{\mathrm{rms}} = \frac{\mathcal{E}_{\mathrm{rms}}}{X_C} = \frac{230}{636.6} \approx 0.36\,\mathrm{A}$$

Answer: 0.36 A rms

Worked Example 6: Resonance Current

A series circuit has a resistance of 10 Ω, an inductance of 0.05 H and a capacitance of 200 μF. It is driven at resonance by an AC source with a peak voltage of 100 V. Find the current amplitude and average power.

1. Resonant angular frequency: $$\omega_0 = \frac{1}{\sqrt{LC}}$$ $$\omega_0 = \frac{1}{\sqrt{0.05 \times 200 \times 10^{-6}}} \approx 316.2\,\mathrm{rad\,s^{-1}}$$
2. Impedance at resonance: $$Z = R = 10\,\Omega$$
3. Current amplitude: $$I_0 = \frac{\mathcal{E}_0}{Z} = \frac{100}{10} = 10\,\mathrm{A}$$
4. RMS current: $$I_{\mathrm{rms}} = \frac{10}{\sqrt{2}} \approx 7.07\,\mathrm{A}$$
5. At resonance, the power factor is unity: $$\cos\phi = 1$$ $$P_{\mathrm{avg}} = \mathcal{E}_{\mathrm{rms}} I_{\mathrm{rms}} = \frac{100}{\sqrt{2}} \times \frac{10}{\sqrt{2}} = 500\,\mathrm{W}$$

Answer: 10 A peak current and 500 W average power

Practice Pointer

Practise RL and RC graphs alongside AC phase plots. Explore the circuit questions in JEE Mains Previous Papers to improve your graph interpretation skills.

Important Formulas and Results at a Glance

40-Second Memory Drill

Cover the formula column and try to recall each expression. Then attempt timed practice sets from JEE Questions focused on EMI and AC to strengthen retention under pressure.

JEE Important Points, Common Mistakes and Quick Revision

• Sign conventions: Distinguish between induced emf and the voltage across an inductor under the passive sign convention: $$\mathcal{E}_{\mathrm{ind}} = -L\frac{dI}{dt}$$ $$v_L = L\frac{dI}{dt}$$
• Flux linkage counts turns: Remember to include the number of turns in coil questions: $$\lambda = N\Phi_B$$
• Resonance: Series LCR resonance gives minimum impedance and maximum source current. For an ideal parallel RLC circuit, resonance gives maximum impedance and minimum source current.
• Phase calculations: Keep angle units consistent. In sinusoidal expressions, angular frequency multiplied by time gives an angle in radians: $$\theta = \omega t$$
• Time constant intuition: During RL growth from zero current, the current reaches approximately 63% of its final value after one time constant: $$I(\tau) = I_{\infty}(1 - e^{-1}) \approx 0.632I_{\infty}$$
• Half-power frequencies: For a series LCR circuit driven at constant voltage amplitude, the current amplitude at each half-power frequency is: $$I_0(\omega_1) = I_0(\omega_2) = \frac{I_{0,\mathrm{res}}}{\sqrt{2}}$$ The bandwidth is: $$\omega_2 - \omega_1 = \frac{R}{L}$$
• Unit check: Inductance multiplied by the rate of change of current has units of voltage: $$\mathrm{H} \cdot \mathrm{A} \cdot \mathrm{s}^{-1} = \mathrm{V}$$
• Phasor diagrams: Take the current as the reference and draw the resistor voltage along it. The inductor voltage leads the current by 90°, while the capacitor voltage lags it by 90°.

Use this section for a quick revision, then practise mixed objective questions from our JEE Mains Online Coaching dashboard to check your understanding.