Physics — Electromagnetism for the JEE Main Exam

Common failure modes

These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.

• !Applying Gauss's law to asymmetric charge distributions where it does not simplify the integral
• !Wrong sign convention in Faraday's law (forgetting the minus sign, or misidentifying the "positive normal" direction)
• !Confusing electric field lines with equipotential surfaces — they are always perpendicular
• !Misapplying superposition: forgetting to vector-add field contributions from multiple sources
• !Treating magnetic flux through a loop incorrectly when the loop is tilted relative to the field

Study tips

• 1Drill Gauss's law for the five standard geometries (sphere, infinite line, infinite plane, cylindrical shell, spherical shell) until the result is immediate.
• 2Use Lenz's law as a sanity check after every induction calculation — the induced current must oppose the change that caused it.
• 3Memorise the Biot-Savart law result for a long straight wire and a circular loop; JEE Advanced builds composite-geometry problems from these.
• 4For capacitors: practice energy stored, charge redistribution when plates move, and dielectric insertion in a single multi-step problem.
• 5Sketch field lines and equipotentials for every E&M setup — it forces conceptual clarity before algebra.

Sample JEE Main Physics — Electromagnetism questions

These sample items mirror the format and difficulty of real JEE Main questions. Practice with thousands more on the free Koydo question bank.

1. 1

   A charge Q is enclosed in a spherical shell. According to Gauss's law, the electric flux through the shell is:

   • AQ / ε₀Correct
   • BQ / (4πε₀)
   • CQ / (4πε₀R²)
   • D4πR²Q / ε₀
   Why this answer?

   Illustrative JEE-style: Gauss's law states ∮E·dA = Q_enc / ε₀. The total flux through any closed surface enclosing charge Q is Q / ε₀, independent of the shape or size of the surface.

2. 2

   A magnetic flux through a coil changes from 5 Wb to 2 Wb in 0.1 s. The magnitude of the induced EMF is:

   • A0.3 V
   • B3 V
   • C30 VCorrect
   • D300 V
   Why this answer?

   Illustrative JEE-style: By Faraday's law, |EMF| = |ΔΦ/Δt| = |5 − 2| / 0.1 = 30 V for a single-turn coil.

3. 3

   Two parallel infinite plates with surface charge densities +σ and −σ face each other. The electric field between the plates is:

   • Aσ / ε₀Correct
   • Bσ / (2ε₀)
   • C2σ / ε₀
   • Dzero
   Why this answer?

   Illustrative JEE-style: Each plate contributes σ / (2ε₀). Between opposite plates the fields add (both point in the same direction), giving E = σ / ε₀. Outside the plates the fields cancel.