Electromagnetic Induction Formulas For JEE 2026
Electromagnetic Induction and Alternating Current (AC) are important chapters in JEE Physics and are known for being relatively scoring if prepared well. These topics explain how a changing magnetic field can generate an electric current and how alternating current behaves in different types of circuits. In this chapter, students study key concepts such as Faraday’s law of electromagnetic induction, Lenz’s law, motional EMF, and self and mutual inductance. Understanding these ideas helps students see how many real-life electrical devices and systems actually work.
The Alternating Current (AC) section further introduces concepts like AC circuits, resonance, power factor, and transformers, which are frequently asked in the exam. In JEE Mains, students can usually expect 2–3 questions each year from these topics, making them an important part of the syllabus. Since many questions depend on remembering the correct formulas and applying them quickly, regular practice is very helpful. For faster revision and better accuracy during preparation, students can also refer to a well-organized JEE Mains Physics Formula PDF, which makes it easier to revise key formulas before the exam.
Magnetic Flux Formula
Imagine holding a wire loop in front of a fan. The amount of "wind" passing through the loop depends on the loop's area, the wind speed, and the angle at which you hold it. Magnetic flux is the analogous quantity for magnetic field lines passing through a surface — it tells us "how much" magnetic field threads through a given area.
Magnetic Flux ($$\Phi_B$$): The total magnetic field passing through a surface. It depends on the field strength, the area of the surface, and the angle between the field and the surface normal.
Magnetic Flux
$$$\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta$$$where:
- $$B$$ = magnitude of the magnetic field (in tesla, T)
- $$A$$ = area of the surface (in m$$^2$$)
- $$\theta$$ = angle between $$\vec{B}$$ and the area normal $$\hat{n}$$
SI unit of magnetic flux: weber (Wb) = T·m$$^2$$
Worked Example
A circular loop of radius 10 cm is placed in a uniform magnetic field of 0.5 T. The plane of the loop makes an angle of $$30^\circ$$ with the field. Find the magnetic flux.
The angle between $$\vec{B}$$ and the normal to the loop is $$\theta = 90^\circ - 30^\circ = 60^\circ$$.
Area: $$A = \pi r^2 = \pi (0.1)^2 = 0.01\pi$$ m$$^2$$
$$\Phi_B = BA\cos\theta = 0.5 \times 0.01\pi \times \cos 60^\circ = 0.5 \times 0.01\pi \times 0.5$$
$$\Phi_B = 0.0025\pi \approx 7.85 \times 10^{-3}$$ Wb
Tip: When the problem says "the plane makes angle $$\alpha$$ with $$\vec{B}$$", the angle between $$\vec{B}$$ and the normal is $$(90^\circ - \alpha)$$. Be very careful with this distinction — it is a common JEE trap.
Faraday's Law of Electromagnetic Induction
In 1831, Michael Faraday discovered that a changing magnetic flux through a loop produces an electric current in the loop — even without a battery! This is called electromagnetic induction. The key insight is: it is not the flux itself but the change in flux that creates an EMF (voltage).
Induced EMF: The voltage generated in a conductor or loop due to a change in the magnetic flux passing through it.
Faraday's Law
The induced EMF in a coil of $$N$$ turns is:
$$$\mathcal{E} = -N\frac{d\Phi_B}{dt}$$$where:
- $$\mathcal{E}$$ = induced EMF (in volts, V)
- $$N$$ = number of turns in the coil
- $$\frac{d\Phi_B}{dt}$$ = rate of change of magnetic flux
The negative sign indicates the direction of the induced EMF (Lenz's law — see next section).
Lenz's Law
The negative sign in Faraday's law has deep physical meaning. Lenz's law states that the induced current always flows in a direction that opposes the change that caused it. If the flux through a loop is increasing, the induced current creates a magnetic field to oppose the increase (i.e., in the opposite direction). This is nature's way of conserving energy.
Lenz's Law — Direction Rule
- If flux is increasing ⇒ induced current opposes the external field (creates field in the opposite direction)
- If flux is decreasing ⇒ induced current supports the external field (creates field in the same direction)
Use the right-hand rule: curl the fingers in the direction of induced current; the thumb gives the direction of the induced magnetic field.
Worked Example
A bar magnet is pushed toward a coil with its north pole facing the coil. What is the direction of the induced current?
As the north pole approaches, the flux through the coil increases (more field lines enter the coil).
By Lenz's law, the induced current must oppose this increase, so the coil must act like a magnet with its north pole facing the approaching magnet (to repel it).
Using the right-hand rule, the current flows anticlockwise when viewed from the side facing the approaching magnet.
Motional EMF Formulas
When a conductor moves through a magnetic field, the free electrons inside it experience a magnetic force that pushes them to one end, creating a potential difference. This is called motional EMF — it is a direct consequence of Faraday's law.
Motional EMF — Straight Conductor
A rod of length $$l$$ moving with velocity $$v$$ perpendicular to a uniform magnetic field $$B$$:
$$$\mathcal{E} = Blv$$$where:
- $$B$$ = magnetic field (T)
- $$l$$ = length of the conductor (m)
- $$v$$ = velocity of the conductor (m/s)
If the rod moves at angle $$\theta$$ to the field:
$$$\mathcal{E} = Blv\sin\theta$$$Motional EMF — Rotating Rod
A rod of length $$l$$ rotating about one end in a perpendicular magnetic field with angular velocity $$\omega$$:
$$$\mathcal{E} = \frac{1}{2}Bl^2\omega$$$Worked Example
A metallic rod of length 1 m moves at 5 m/s perpendicular to a field of 0.4 T. The rod is part of a closed circuit with resistance 2 $$\Omega$$. Find the induced EMF and current.
$$\mathcal{E} = Blv = 0.4 \times 1 \times 5 = 2$$ V
$$I = \frac{\mathcal{E}}{R} = \frac{2}{2} = 1$$ A
Worked Example
A conducting rod of length 0.5 m rotates about one end in a plane perpendicular to a field of 0.2 T at 300 rpm. Find the EMF.
$$\omega = \frac{2\pi \times 300}{60} = 10\pi$$ rad/s
$$\mathcal{E} = \frac{1}{2}Bl^2\omega = \frac{1}{2} \times 0.2 \times (0.5)^2 \times 10\pi = 0.25\pi \approx 0.785$$ V
Self-Inductance Formulas
When current flows through a coil, it creates a magnetic field. If the current changes, the flux through the coil itself changes, inducing an EMF that opposes the change. This property of a coil to oppose changes in its own current is called self-inductance.
Self-Inductance ($$L$$): A measure of how effectively a coil opposes changes in current through itself. A coil with large $$L$$ strongly resists current changes.
Self-Inductance
The flux linkage through the coil due to its own current:
$$$N\Phi_B = LI$$$The self-induced EMF (back EMF):
$$$\mathcal{E} = -L\frac{dI}{dt}$$$where:
- $$L$$ = self-inductance (in henry, H)
- $$I$$ = current through the coil (A)
- $$\frac{dI}{dt}$$ = rate of change of current (A/s)
Self-Inductance of a Solenoid
$$$L = \mu_0 n^2 Al = \frac{\mu_0 N^2 A}{l}$$$where:
- $$\mu_0 = 4\pi \times 10^{-7}$$ T·m/A (permeability of free space)
- $$n = N/l$$ = number of turns per unit length
- $$A$$ = cross-sectional area of the solenoid
- $$l$$ = length of the solenoid
- $$N$$ = total number of turns
Worked Example
A solenoid has 500 turns, length 25 cm, and radius 2 cm. Find its self-inductance.
$$A = \pi r^2 = \pi(0.02)^2 = 4\pi \times 10^{-4}$$ m$$^2$$
$$L = \frac{\mu_0 N^2 A}{l} = \frac{4\pi \times 10^{-7} \times (500)^2 \times 4\pi \times 10^{-4}}{0.25}$$
$$L = \frac{4\pi \times 10^{-7} \times 250000 \times 4\pi \times 10^{-4}}{0.25} = \frac{4\pi^2 \times 10^{-4}}{0.25} \approx 1.58 \times 10^{-3}$$ H $$= 1.58$$ mH
Mutual Inductance Formula
When two coils are placed near each other, a changing current in one coil changes the magnetic flux through the second coil, inducing an EMF in it. This coupling between two coils is called mutual inductance.
Mutual Inductance ($$M$$): A measure of how much flux linkage is produced in one coil due to current in a nearby coil.
Mutual Inductance
Flux linkage in coil 2 due to current $$I_1$$ in coil 1:
$$$N_2 \Phi_{21} = M I_1$$$EMF induced in coil 2:
$$$\mathcal{E}_2 = -M\frac{dI_1}{dt}$$$For two coaxial solenoids (one inside the other):
$$$M = \mu_0 n_1 n_2 A l$$$where $$A$$ is the cross-sectional area of the inner solenoid and $$l$$ is the common length.
Note: Mutual inductance depends on the geometry, orientation, and distance between the coils. The relation $$M = k\sqrt{L_1 L_2}$$ holds, where $$k$$ is the coupling coefficient ($$0 \le k \le 1$$). For tightly wound coils on the same core, $$k \approx 1$$.
Energy Stored in an Inductor Formula
Just as a capacitor stores energy in its electric field, an inductor stores energy in its magnetic field when current flows through it. Building up the current requires work against the back EMF.
Energy in an Inductor
$$$U = \frac{1}{2}LI^2$$$where:
- $$U$$ = energy stored (in joules, J)
- $$L$$ = self-inductance (H)
- $$I$$ = current through the inductor (A)
Energy density (energy per unit volume) in a magnetic field:
$$$u = \frac{B^2}{2\mu_0}$$$Worked Example
An inductor of 200 mH carries a current of 5 A. Find the energy stored.
$$U = \frac{1}{2}LI^2 = \frac{1}{2} \times 0.2 \times 25 = 2.5$$ J
AC Generator — Working and EMF Formula
An AC generator (alternator) converts mechanical energy into alternating electrical energy. It works on the principle of electromagnetic induction — a coil rotated in a magnetic field experiences a continuously changing flux, producing a sinusoidally varying EMF.
EMF of an AC Generator
A coil of $$N$$ turns, area $$A$$, rotating with angular velocity $$\omega$$ in a field $$B$$:
$$$\mathcal{E} = \mathcal{E}_0 \sin(\omega t)$$$where the peak EMF is:
$$$\mathcal{E}_0 = NBA\omega$$$- The frequency of AC: $$f = \frac{\omega}{2\pi}$$, time period $$T = \frac{1}{f} = \frac{2\pi}{\omega}$$
- The EMF alternates between $$+\mathcal{E}_0$$ and $$-\mathcal{E}_0$$
RMS and Peak Values
Since AC voltage and current continuously change, we use root mean square (RMS) values to describe their effective (equivalent DC) values.
RMS and Peak Value Relations
| Quantity | Peak Value | RMS Value |
|---|---|---|
| Voltage | $$V_0$$ | $$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \approx 0.707\, V_0$$ |
| Current | $$I_0$$ | $$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707\, I_0$$ |
| Average value | — | $$V_{\text{avg}} = \frac{2V_0}{\pi} \approx 0.637\, V_0$$ |
Important: The 220 V supply in Indian homes is the RMS value. Peak voltage $$= 220\sqrt{2} \approx 311$$ V.
Tip: In JEE, when a problem says "an AC source of 100 V", it means $$V_{\text{rms}} = 100$$ V unless explicitly stated otherwise.
AC Circuits — Resistor, Inductor, Capacitor
AC circuit analysis involves understanding how resistors, inductors, and capacitors behave individually when connected to an AC source. Each component has a different relationship between voltage and current.
Pure Resistive Circuit (R only)
AC Through a Resistor
$$$V = V_0 \sin\omega t, \quad I = I_0 \sin\omega t$$$- Voltage and current are in phase (phase difference = $$0$$)
- $$I_0 = \frac{V_0}{R}$$
Pure Inductive Circuit (L only)
An inductor opposes changes in current. In an AC circuit, it creates a "resistance-like" effect called inductive reactance.
Inductive Reactance ($$X_L$$): The opposition offered by an inductor to AC current. It increases with frequency — at high frequencies, the inductor blocks more current.
AC Through an Inductor
$$$V = V_0 \sin\omega t, \quad I = I_0 \sin\!\left(\omega t - \frac{\pi}{2}\right)$$$- Current lags voltage by $$\frac{\pi}{2}$$ (or $$90^\circ$$)
- Inductive reactance: $$X_L = \omega L = 2\pi f L$$ (in ohms)
- $$I_0 = \frac{V_0}{X_L}$$
Pure Capacitive Circuit (C only)
A capacitor stores charge and releases it alternately in an AC circuit. It offers opposition to AC called capacitive reactance, which decreases with frequency.
Capacitive Reactance ($$X_C$$): The opposition offered by a capacitor to AC current. At high frequencies, charge has less time to build up, so the capacitor offers less opposition.
AC Through a Capacitor
$$$V = V_0 \sin\omega t, \quad I = I_0 \sin\!\left(\omega t + \frac{\pi}{2}\right)$$$- Current leads voltage by $$\frac{\pi}{2}$$ (or $$90^\circ$$)
- Capacitive reactance: $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$ (in ohms)
- $$I_0 = \frac{V_0}{X_C}$$
Tip: Remember the phase relationships with the mnemonic "CIVIL": In a Capacitor, I leads V; In an inductor (L), V leads I.
Series LCR Circuit
When a resistor ($$R$$), inductor ($$L$$), and capacitor ($$C$$) are connected in series with an AC source, the total opposition to current flow is called impedance. The voltages across the three components are not in phase, so they don't add up arithmetically — we must use a phasor diagram.
Impedance ($$Z$$): The total effective opposition to AC current in a circuit, measured in ohms ($$\Omega$$). It combines resistance and reactance.
Series LCR Circuit
Impedance:
$$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$$Current:
$$$I_0 = \frac{V_0}{Z}, \quad I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$$$Phase difference between voltage and current:
$$$\tan\phi = \frac{X_L - X_C}{R}$$$- If $$X_L > X_C$$: circuit is inductive, voltage leads current ($$\phi > 0$$)
- If $$X_C > X_L$$: circuit is capacitive, current leads voltage ($$\phi < 0$$)
- If $$X_L = X_C$$: circuit is at resonance, $$\phi = 0$$
Voltage Relations in Series LCR
- $$V_R = IR$$ (in phase with $$I$$)
- $$V_L = IX_L$$ (leads $$I$$ by $$90^\circ$$)
- $$V_C = IX_C$$ (lags $$I$$ by $$90^\circ$$)
- $$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$
Worked Example
A series LCR circuit has $$R = 40\;\Omega$$, $$L = 0.5$$ H, $$C = 50\;\mu$$F, connected to a 200 V, 50 Hz AC source. Find the impedance, current, and phase angle.
$$X_L = 2\pi f L = 2\pi \times 50 \times 0.5 = 50\pi \approx 157.1\;\Omega$$
$$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 50 \times 10^{-6}} = \frac{1}{0.005\pi} \approx 63.7\;\Omega$$
$$Z = \sqrt{40^2 + (157.1 - 63.7)^2} = \sqrt{1600 + 8715.6} = \sqrt{10315.6} \approx 101.6\;\Omega$$
$$I_{\text{rms}} = \frac{200}{101.6} \approx 1.97$$ A
$$\tan\phi = \frac{157.1 - 63.7}{40} = \frac{93.4}{40} = 2.335 \Rightarrow \phi \approx 66.8^\circ$$
Since $$X_L > X_C$$, the circuit is inductive and voltage leads current.
Resonance in LCR Circuit — Frequency Formula
At a special frequency called the resonant frequency, the inductive reactance exactly equals the capacitive reactance ($$X_L = X_C$$). At this point, they cancel each other, the impedance is minimum (equal to $$R$$), and the current is maximum.
Resonance Conditions
At resonance: $$X_L = X_C \Rightarrow \omega L = \frac{1}{\omega C}$$
$$$\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}$$$At resonance:
- Impedance is minimum: $$Z = R$$
- Current is maximum: $$I_{\text{max}} = \frac{V}{R}$$
- Phase angle $$\phi = 0$$ (voltage and current in phase)
- $$V_L = V_C$$ (they cancel but can individually be very large)
Quality Factor ($$Q$$): A measure of the sharpness of resonance. Higher $$Q$$ means a sharper peak and more selective circuit.
Quality Factor
$$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}}$$$At resonance: $$V_L = V_C = QV$$ (where $$V$$ is the source voltage).
Bandwidth: $$\Delta\omega = \frac{R}{L} = \frac{\omega_0}{Q}$$
Worked Example
An LCR circuit has $$L = 20$$ mH, $$C = 5\;\mu$$F, and $$R = 10\;\Omega$$. Find the resonant frequency and $$Q$$-factor.
$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times 5 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-7}}} = \frac{1}{2\pi \times 3.162 \times 10^{-4}}$$
$$f_0 \approx \frac{1}{1.987 \times 10^{-3}} \approx 503$$ Hz
$$Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{20 \times 10^{-3}}{5 \times 10^{-6}}} = \frac{1}{10}\sqrt{4000} = \frac{63.2}{10} = 6.32$$
Tip: At resonance, $$V_L$$ and $$V_C$$ can each be $$Q$$ times the source voltage. This is called voltage magnification. JEE may ask for $$V_L$$ or $$V_C$$ at resonance — just multiply the source voltage by $$Q$$.
Power in AC Circuits
In AC circuits, not all the power delivered by the source is "used up" — some is stored and returned by inductors and capacitors. Only the resistor dissipates power as heat. The power factor tells us what fraction of the total power is actually consumed.
Power Factor: The ratio of true (average) power consumed to the apparent power delivered. It equals $$\cos\phi$$, where $$\phi$$ is the phase angle between voltage and current.
Power in AC Circuits
Instantaneous power: $$P = VI$$ (varies with time)
Average (true) power:
$$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = I_{\text{rms}}^2 R$$$Apparent power:
$$$P_{\text{apparent}} = V_{\text{rms}} I_{\text{rms}}$$$Power factor:
$$$\cos\phi = \frac{R}{Z}$$$| Circuit | Power Factor | Power Consumed |
|---|---|---|
| Pure R | $$\cos\phi = 1$$ | Maximum: $$V_{\text{rms}} I_{\text{rms}}$$ |
| Pure L | $$\cos\phi = 0$$ | Zero (wattless current) |
| Pure C | $$\cos\phi = 0$$ | Zero (wattless current) |
| LCR at resonance | $$\cos\phi = 1$$ | Maximum: $$V_{\text{rms}}^2 / R$$ |
Note: In a purely inductive or purely capacitive circuit, the average power consumed is zero. The current in such circuits is called wattless current. Energy is alternately stored in the field and returned to the source.
Worked Example
A series LCR circuit draws 2 A (rms) from a 200 V, 50 Hz source. The power consumed is 200 W. Find the power factor and impedance.
$$\cos\phi = \frac{P}{V_{\text{rms}} I_{\text{rms}}} = \frac{200}{200 \times 2} = 0.5$$
$$\phi = 60^\circ$$
$$Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{200}{2} = 100\;\Omega$$
$$R = Z\cos\phi = 100 \times 0.5 = 50\;\Omega$$
Transformer Formulas — Step Up and Step Down
A transformer is a device that changes (transforms) AC voltage from one level to another using mutual induction. It consists of two coils (primary and secondary) wound on a common iron core. It works only with AC — not DC, because DC does not produce changing flux.
Transformer Relations
For an ideal transformer (no energy loss):
$$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k$$$where:
- $$V_p, V_s$$ = primary and secondary voltages
- $$N_p, N_s$$ = number of turns in primary and secondary coils
- $$I_p, I_s$$ = primary and secondary currents
- $$k$$ = transformation ratio
| Type | Condition | Effect |
|---|---|---|
| Step-up | $$N_s > N_p$$ ($$k > 1$$) | Increases voltage, decreases current |
| Step-down | $$N_s < N_p$$ ($$k < 1$$) | Decreases voltage, increases current |
Efficiency: $$\eta = \frac{P_{\text{output}}}{P_{\text{input}}} = \frac{V_s I_s}{V_p I_p} \times 100\%$$
Energy losses in a real transformer include:
- Copper loss: Heat produced in the coil wires due to their resistance ($$I^2R$$)
- Iron/Core loss: Eddy currents and hysteresis in the iron core
- Flux leakage: Not all flux from the primary links with the secondary
Worked Example
A transformer has 200 turns in the primary and 1000 turns in the secondary. If the primary voltage is 220 V and it draws 5 A, find the secondary voltage and current (assume ideal).
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = 220 \times \frac{1000}{200} = 1100$$ V
$$V_p I_p = V_s I_s \Rightarrow I_s = \frac{220 \times 5}{1100} = 1$$ A
This is a step-up transformer: voltage increased from 220 V to 1100 V, but current decreased from 5 A to 1 A.
Tip: A transformer does not change the frequency of AC. If the input is 50 Hz, the output is also 50 Hz. Also, $$P_{\text{input}} = P_{\text{output}}$$ for an ideal transformer, so power is conserved — you cannot get "free" energy.
Summary of Key Formulas
Quick Reference
| Quantity | Formula |
|---|---|
| Magnetic flux | $$\Phi_B = BA\cos\theta$$ |
| Faraday's law | $$\mathcal{E} = -N\frac{d\Phi_B}{dt}$$ |
| Motional EMF (rod) | $$\mathcal{E} = Blv$$ |
| Motional EMF (rotating rod) | $$\mathcal{E} = \frac{1}{2}Bl^2\omega$$ |
| Self-inductance (solenoid) | $$L = \frac{\mu_0 N^2 A}{l}$$ |
| Energy in inductor | $$U = \frac{1}{2}LI^2$$ |
| Peak EMF of AC generator | $$\mathcal{E}_0 = NBA\omega$$ |
| Impedance (series LCR) | $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ |
| Resonant frequency | $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$ |
| Power factor | $$\cos\phi = \frac{R}{Z}$$ |
| Transformer ratio | $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ |
Electromagnetic Induction Formulas For JEE 2026: Conclusion
Electromagnetic Induction and Alternating Current are important topics in JEE Physics because they explain how changing magnetic fields produce electric currents and how AC circuits behave in practical systems. Understanding Electromagnetic Induction Formulas For JEE 2026 such as Faraday’s law, motional EMF, inductance, and resonance helps students solve many numerical problems quickly and accurately. Since these topics regularly appear in the JEE exam, mastering the formulas and concepts is very beneficial during preparation.
For effective revision, students should regularly practice problems and review key formulas. A well-structured JEE Mains Physics Formula PDF can be extremely useful for quick revision before exams and while solving practice questions. By understanding the concepts clearly and revising formulas consistently, students can improve their problem-solving speed and score better in JEE Physics.
