JEE Physics — Magnetism And Emi Complete Chapter Guide

Chapter Overview & Weightage

Magnetism + EMI is one of the most reliable scoring chapters in JEE Main. Year after year, you can count on 3–4 questions from this combined chapter. Together with AC Circuits (which examiners treat as an extension of EMI), this block can easily fetch you 12–16 marks.

JEE Main Weightage Data (2019–2024)

Year	Questions	Marks	Topics Hit
2024	4	16	Biot-Savart, LCR, Self-inductance, Faraday’s
2023	3	12	Ampere’s Law, Mutual Inductance, AC power
2022	4	16	Force on wire, EMF in rotating coil, Inductors
2021	3	12	Biot-Savart, Lenz’s Law, LCR resonance
2020	4	16	Ampere’s Law, Self-inductance, AC circuits
2019	3	12	Magnetic force, Faraday’s Law, Impedance

Overall: 3–4 questions per paper, ~10–12% weightage. JEE Advanced tests deeper derivation-level understanding of the same topics.

The chapter splits naturally into three blocks: Sources of magnetic fields (Biot-Savart + Ampere), Electromagnetic Induction (Faraday + Lenz + Inductance), and AC Circuits. Treat them as one interconnected story — a changing B gives EMF (Faraday), EMF drives current through inductors (self-inductance), and inductors in AC circuits give you impedance.

Key Concepts You Must Know

Prioritized by how often they appear in PYQs:

Tier 1 — Appears almost every year:

Magnetic field due to a current-carrying wire (finite and infinite), circular loop, and solenoid using Biot-Savart
Ampere’s Circuital Law application to solenoids and toroids
Faraday’s Law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$ and how to calculate flux through non-trivial surfaces
Self-inductance of a solenoid: $L = \mu_0 n^2 Al$
Energy stored in an inductor: $U = \frac{1}{2}LI^2$
LCR series circuit — impedance, phase angle, resonance condition

Tier 2 — Every other year:

Force between two parallel current-carrying conductors (definition of Ampere)
Motional EMF: $\mathcal{E} = Bvl$ and power dissipated
Mutual inductance: $M = \frac{\mu_0 N_1 N_2 A}{l}$ for coaxial solenoids
Power factor in AC circuits: $\cos\phi = R/Z$
Transformer equation: $V_s/V_p = N_s/N_p$

Tier 3 — JEE Advanced territory:

RL circuit time constant ( $\tau = L/R$ ) transient analysis
LC oscillation frequency: $\omega = 1/\sqrt{LC}$
Moving rod in a magnetic field connected to a circuit (Galton board style)

Important Formulas

d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2}

Results you must memorize:

Configuration	Formula
Infinite straight wire	$B = \dfrac{\mu_0 I}{2\pi r}$
Finite wire (subtending angles $\alpha, \beta$ )	$B = \dfrac{\mu_0 I}{4\pi r}(\sin\alpha + \sin\beta)$
Circular loop at centre	$B = \dfrac{\mu_0 I}{2R}$
Circular arc (angle $\theta$ )	$B = \dfrac{\mu_0 I \theta}{4\pi R}$

When to use: Any problem that gives you a specific geometry and asks for B at a point. The infinite wire formula is used more than any other.

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

Results:

Solenoid (inside): $B = \mu_0 n I$ where $n$ = turns per unit length
Toroid (inside): $B = \dfrac{\mu_0 N I}{2\pi r}$
Toroid (outside): $B = 0$

When to use: Symmetric geometries — solenoid, toroid, infinite cylinder of current. When symmetry exists, Ampere’s Law is faster than Biot-Savart.

\mathcal{E} = -\frac{d\Phi_B}{dt}, \qquad \Phi_B = \int \vec{B} \cdot d\vec{A}

For a coil of N turns: $\mathcal{E} = -N\frac{d\Phi_B}{dt}$

Motional EMF (rod of length $l$ moving at velocity $v$ ): $\mathcal{E} = Bvl$

EMF in rotating coil: $\mathcal{E} = NBA\omega\sin(\omega t)$ → this is the AC generator equation.

\mathcal{E} = -L\frac{dI}{dt}, \qquad L_{solenoid} = \mu_0 n^2 A l = \frac{\mu_0 N^2 A}{l}

M = k\sqrt{L_1 L_2} \quad (k = \text{coupling coefficient} \leq 1)

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

When to use: Any problem involving “back-EMF”, current growth/decay in RL circuits, or energy stored.

Z = \sqrt{R^2 + (X_L - X_C)^2}, \qquad X_L = \omega L, \quad X_C = \frac{1}{\omega C}

Resonance: $\omega_0 = \dfrac{1}{\sqrt{LC}}$ , at resonance $Z = R$ (minimum impedance)

Power: $P_{avg} = V_{rms} I_{rms} \cos\phi$ , where $\tan\phi = \dfrac{X_L - X_C}{R}$

RMS values: $V_{rms} = \dfrac{V_0}{\sqrt{2}}$ , $I_{rms} = \dfrac{I_0}{\sqrt{2}}$

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 Shift 1 (Biot-Savart)

Question: A current of 2 A flows through a circular loop of radius 4 cm. The magnetic field at the centre is $B_1$ . Now the same wire is bent into a square loop of side $a$ . The current is still 2 A. The magnetic field at the centre of the square is $B_2$ . Find $B_1/B_2$ .

Why this is tricky: Students often mix up the formula for the centre of a square loop. The key is that the wire length is conserved.

The wire length is the same: circumference of circle = perimeter of square.

2\pi R = 4a \implies a = \frac{\pi R}{2} = \frac{\pi \times 4}{2} = 2\pi \text{ cm}

B_1 = \frac{\mu_0 I}{2R} = \frac{\mu_0 \times 2}{2 \times 0.04} = \frac{\mu_0}{0.04}

Each side of a square contributes. For a finite wire of length $a$ at perpendicular distance $d = a/2$ from centre, subtending angles $\pm 45°$ :

B_{one\ side} = \frac{\mu_0 I}{4\pi (a/2)} \times 2\sin 45° = \frac{\mu_0 I \sqrt{2}}{2\pi a}

Four sides: $B_2 = 4 \times \frac{\mu_0 I \sqrt{2}}{2\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}$

Substituting $a = 2\pi \times 0.01$ m (converting $2\pi$ cm):

\frac{B_1}{B_2} = \frac{\pi^2}{8\sqrt{2}} \approx \frac{9.87}{11.31} \approx 0.87

PYQ 2 — JEE Main 2023 (Faraday’s Law + Lenz)

Question: A square coil of side 10 cm has 20 turns and resistance 2 Ω. It is placed in a magnetic field $B = (0.5t^2 + 3t + 1)$ T perpendicular to the coil. Find the induced current at $t = 2$ s.

\frac{dB}{dt} = t + 3

At $t = 2$ s: $\frac{dB}{dt} = 2 + 3 = 5$ T/s

|\mathcal{E}| = N \cdot A \cdot \frac{dB}{dt} = 20 \times (0.1)^2 \times 5 = 20 \times 0.01 \times 5 = 1 \text{ V}

I = \frac{\mathcal{E}}{R} = \frac{1}{2} = 0.5 \text{ A}

Students forget to multiply by $N$ (number of turns). Each turn contributes separately to the total EMF. $\mathcal{E}_{total} = N \cdot \frac{d\Phi}{dt}$ , not just $\frac{d\Phi}{dt}$ .

PYQ 3 — JEE Main 2022 (LCR Resonance)

Question: In a series LCR circuit, $L = 5$ mH, $C = 80$ μF, $R = 40$ Ω. The source voltage is $V = 200\sin(1000t)$ V. Find (a) the resonant frequency, (b) the current amplitude at resonance.

From $V = 200\sin(1000t)$ , the driving angular frequency $\omega = 1000$ rad/s.

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 10^{-3} \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-7}}} = \frac{1}{2 \times 10^{-3.5}}

\omega_0 = \frac{1}{\sqrt{4 \times 10^{-7}}} = \frac{10^{3.5}}{\sqrt{4}} \approx 1581 \text{ rad/s}

Wait — $\omega \neq \omega_0$ , so the circuit is not at resonance.

$X_L = \omega L = 1000 \times 5 \times 10^{-3} = 5 \text{ }\Omega$ $X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 80 \times 10^{-6}} = 12.5 \text{ }\Omega$ $Z = \sqrt{40^2 + (5 - 12.5)^2} = \sqrt{1600 + 56.25} = \sqrt{1656.25} \approx 40.7 \text{ }\Omega$

I_0 = \frac{V_0}{Z} = \frac{200}{40.7} \approx 4.9 \text{ A}

At resonance, $X_L = X_C$ , $Z = R$ (minimum), and current is maximum. The phase angle $\phi = 0$ — voltage and current are in phase. This is the highest power-transfer condition.

Difficulty Distribution

Based on PYQ analysis from JEE Main 2019–2024 (roughly 20 papers):

Difficulty	Percentage	What it looks like
Easy	40%	Direct formula: B at centre of loop, resonance frequency, transformer ratio
Medium	45%	Two-step problems: motional EMF with circuit, Ampere’s Law + force
Hard	15%	Multi-concept: RL transient + energy, rotating coil + torque, complex AC circuits

JEE Advanced skews this distribution heavily toward Hard (60%+). In Advanced, expect proof-level questions on mutual inductance, RL circuits with differential equations, and magnetic field calculations in 3D geometries.

Expert Strategy

How toppers approach this chapter:

The standard mistake is treating Magnetism and EMI as separate chapters to memorize. Toppers see it as one physical idea: changing magnetic flux creates EMF, and current creates magnetic flux. Everything follows from this.

The “3-pass” approach for this chapter:

First pass: Master Biot-Savart results for standard geometries (wire, loop, solenoid) and Faraday’s Law. These are pure formula questions — guaranteed marks.
Second pass: LCR circuits. Learn to draw the phasor diagram. It converts every AC problem into a geometry problem with a right triangle.
Third pass: Inductance (self + mutual) and RL transients. This is where JEE Advanced goes deep. For Main, just know the energy formula and the solenoid formula.

For PYQ practice, prioritize:

JEE Main 2020–2024 (most similar pattern to current exam)
BITSAT previous years (excellent for medium difficulty Biot-Savart and AC)
HC Verma Chapter 35–38 (Exercises, not Concepts) for building calculation speed

Allocate roughly 12–15 hours for this combined chapter. Don’t skip AC Circuits — it’s consistently 1–2 questions in Main and examinees who skip it lose easy marks.

Common Traps

Trap 1: Sign confusion in Faraday’s Law

The negative sign ( $\mathcal{E} = -d\Phi/dt$ ) tells you the direction of induced EMF (Lenz’s Law), not the magnitude. For magnitude calculations, always use $|\mathcal{E}| = |d\Phi/dt|$ . Lenz’s Law (opposing the cause) tells you direction separately. Mixing these up wastes time in MCQs.

Trap 2: Confusing $X_L$ and $X_C$ dominance

When $X_L > X_C$ : circuit is inductive, current lags voltage by angle $\phi$ . When $X_C > X_L$ : circuit is capacitive, current leads voltage.

The phrase “current leads in a capacitor” confuses students into thinking the whole circuit is capacitive when $X_C > X_L$ . Just remember: ELI the ICE man (E leads I in inductors, I leads E in capacitors).

Trap 3: Applying Biot-Savart to a solenoid directly

Students integrate Biot-Savart for a solenoid and get confused. For a solenoid, use Ampere’s Law directly: $B = \mu_0 nI$ inside, $B \approx 0$ outside. Biot-Savart is for when Ampere’s Law doesn’t apply (no symmetry).

Trap 4: Using peak values in power formulas

$P = V_{rms} I_{rms} \cos\phi$ uses RMS values, not peak. If the question gives $V_0$ and $I_0$ , divide by $\sqrt{2}$ first. Many students write $P = V_0 I_0 \cos\phi$ directly and lose half marks. JEE setters specifically use peak values in the question to check this.

Trap 5: Forgetting the direction of force between parallel wires

Two wires carrying current in the same direction attract. Opposite directions repel. This is counterintuitive because two positive charges repel, but here it’s magnetic force, not electrostatic. The definition of 1 Ampere is based on this attraction force.