Last Updated: April 2026
JEE Main Magnetic Effects of Current 2027 sits firmly inside the high-yield electromagnetism cluster — a topic that contributes 2–3 questions every shift in JEE Main Physics. Combined with Electrostatics, Current Electricity, and EM Induction, magnetism builds the spine of Class 12 Physics weightage. The good news: the entire chapter reduces to four laws (Biot–Savart, Ampere, Lorentz, Cyclotron) and a handful of geometry-specific results. Memorise those, and 90% of JEE problems become substitution exercises. This guide walks you through the formula sheet, three worked examples, 35 problem categories, and a 10-MCQ quiz at the bottom.
JEE Main Weightage of Magnetic Effects
Looking at the last 6 years of NTA papers, “Magnetic Effects of Current and Magnetism” averages 2 questions per shift, almost always one numerical (force/field calculation) and one conceptual (current loop, solenoid, cyclotron). Combined with EM Induction (typically 1 more question), the magnetism block is worth 12 marks every paper — about 10% of the Physics section. You cannot afford to neglect it.
The Four Pillars of the Chapter
- Biot–Savart Law: field due to a current element
- Ampere’s Circuital Law: field of high-symmetry currents (wire, solenoid, toroid)
- Lorentz Force: force on a moving charge / current-carrying wire
- Magnetic Dipole / Torque: behaviour of current loops in a field
Almost every JEE question is a substitution into one of these four frameworks.
Master Formula Sheet
| Configuration | Magnetic Field B | Notes |
|---|---|---|
| Biot–Savart (element) | dB = (μ₀/4π) · I (dl × r̂)/r² | Vector — direction by right-hand rule |
| Infinite straight wire | B = μ₀I/(2πr) | Tangent to circle around wire |
| Finite wire (angles α, β at ends) | B = μ₀I/(4πr) · (sinα + sinβ) | r is perpendicular distance |
| Centre of circular loop | B = μ₀I/(2R) | Along axis |
| Loop, axial point distance x | B = μ₀IR²/[2(R²+x²)^(3/2)] | Reduces to centre formula at x=0 |
| Solenoid (inside, long) | B = μ₀nI | n = turns per metre |
| Solenoid end | B = μ₀nI/2 | Half of inside value |
| Toroid | B = μ₀NI/(2πr) | Inside the toroid only |
| Lorentz force on charge | F = q(v × B) | Always perpendicular to v ⇒ no work |
| Force on wire | F = I(L × B); F = BIL sinθ | θ between L and B |
| Force per unit length (parallel wires) | F/L = μ₀I₁I₂/(2πd) | Same dir attract; opposite repel |
| Radius of circular path | r = mv/(qB) | For v ⊥ B |
| Cyclotron frequency | f = qB/(2πm) | Independent of v (non-relativistic) |
| Helical pitch | p = (2πm/qB) · v cosθ | v has component along B |
| Magnetic moment of loop | M = NIA | Direction: right-hand rule |
| Torque on loop | τ = M × B; τ = MB sinθ | Aligns M with B |
| Potential energy | U = −M·B | Minimum when M ∥ B |
The Three Right-Hand Rules You Must Memorise
Rule 1 (Straight wire): Thumb along current; curled fingers give field direction.
Rule 2 (Loop): Curl fingers along current; thumb gives M (and field at centre).
Rule 3 (Cross product F = qv × B): Fingers point along v, curl toward B; thumb = F (for positive charge; reverse for negative).
Worked Example 1: Field at Centre of a Square Loop
“A wire is bent into a square of side a carrying current I. Find B at the centre.”
Each side acts as a finite wire with α = β = 45°. Distance from centre to each side = a/2. Field due to one side:
B₁ = (μ₀I/4π)(1/(a/2))(sin45° + sin45°) = (μ₀I/4π)(2/a)(√2) = μ₀I√2/(2πa).
Four sides add (same direction):
B = 4 × μ₀I√2/(2πa) = 2√2 μ₀I/(πa).
Worked Example 2: Charged Particle in a Magnetic Field
“A proton (m=1.67×10⁻²⁷ kg, q=1.6×10⁻¹⁹ C) enters a 0.5 T field with v=2×10⁶ m/s perpendicular to B. Find radius and time period.”
r = mv/(qB) = (1.67×10⁻²⁷ × 2×10⁶)/(1.6×10⁻¹⁹ × 0.5) = 4.18×10⁻² m ≈ 4.2 cm.
T = 2πm/(qB) = (2π × 1.67×10⁻²⁷)/(1.6×10⁻¹⁹ × 0.5) = 1.31×10⁻⁷ s ≈ 131 ns.
Worked Example 3: Force Between Two Parallel Wires
“Two long parallel wires 5 cm apart carry currents 4 A and 6 A in the same direction. Find force per metre.”
F/L = μ₀I₁I₂/(2πd) = (4π×10⁻⁷ × 4 × 6)/(2π × 0.05) = (4×10⁻⁷ × 24)/0.1 = 9.6×10⁻⁵ N/m, attractive.
35 Practice Problem Categories
- B at centre of circular loop, R given
- B on axis of loop at distance x
- B at centre of semicircular loop
- B at centre of arc subtending angle θ
- B at corner of square loop
- B at midpoint of finite straight wire
- B at end of finite wire
- Two long wires — field at midpoint
- Two long wires — point of zero field
- Solenoid — field inside vs outside
- Toroid — field at mean radius
- Helmholtz coils — field at centre
- Force on a charge moving in uniform B (perpendicular)
- Velocity selector E/B condition
- Cyclotron — energy after N revolutions
- Cyclotron — maximum kinetic energy
- Charge entering at angle θ to B — helical path
- Charge entering crossed E and B
- Force per unit length between two parallel wires
- Three coplanar parallel wires — force on middle
- Force on a wire bent into a loop
- Force on a semicircular wire in uniform B
- Torque on rectangular loop in uniform B
- Potential energy of a magnetic dipole
- Work done in rotating dipole through angle θ
- Galvanometer sensitivity — voltage and current
- Conversion of galvanometer to ammeter (shunt)
- Conversion of galvanometer to voltmeter (series resistance)
- Bar magnet and circular loop equivalence
- Field due to current in a tube (Ampere’s law)
- Field due to current in a coaxial cable
- Magnetic field at a point from infinite sheet of current
- Time period of oscillation of bar magnet
- Combined dipole/coil system — net moment
- Force on a current-carrying spring (extension/compression)
Common Traps NTA Loves
Trap 1: Sign of cross product. A negative charge moving in +x in field +z experiences force in +y, not −y. Always re-derive with right-hand rule for negative charges by reversing the result.
Trap 2: Solenoid “n” vs “N”. n = turns per metre; N = total turns. Many students plug N into B = μ₀nI and get answers off by a factor of length.
Trap 3: Field at end of solenoid. It is half the inside value, not the full value. Useful in problems where particle exits the solenoid axially.
Trap 4: Magnetic field does no work. A common conceptual question. Since F = qv × B is always perpendicular to v, B can change direction but never speed of a charge. Speed change always implies an electric field is also present.
FAQ
Q1. How much weightage does Magnetic Effects carry in JEE Main 2027?
About 6–8 marks per shift (2 direct + 1 in EM induction). Roughly 10% of the Physics section.
Q2. Is the cyclotron derivation important?
Yes — the cyclotron frequency f = qB/(2πm) is asked at least once every 2–3 years. Also memorise the resonance condition for AC across the dees.
Q3. Do I need to memorise the Biot–Savart vector form?
Yes. JEE often gives a current configuration that is not in the standard set (e.g., L-shaped wire) where you must integrate from scratch. The vector form is the only way.
Q4. How does this chapter overlap with EM Induction?
EM Induction starts where this chapter ends — once you can compute B for any geometry, induction problems become time-derivatives of flux Φ = B·A. Master magnetism first.
Q5. What’s the fastest way to solve “force on a curved wire” problems?
A wire of any shape carrying current I in a uniform B experiences the same force as a straight wire connecting its endpoints carrying the same current. This shortcut saves enormous time.
Internal Links — Build Your Electromagnetism Stack
- JEE Physics Electrostatics 2027 — prerequisite for E×B problems
- JEE Main Current Electricity 2027 — companion chapter
- JEE Main Physics — Chapter Weightage
- Top 20 Important Chapters for JEE Main
- JEE Main vs JEE Advanced
Take the 10-MCQ Magnetic Effects Quiz
Test your formula recall and right-hand rules. Aim for 8/10 in 12 minutes.
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The Velocity Selector: A Recurring NTA Favourite
A charged particle moving with velocity v through perpendicular E and B experiences zero net force when qE = qvB ⇒ v = E/B. This is the velocity selector condition, used in mass spectrometers and Thomson’s e/m experiment. JEE Main has asked at least 3 questions on this exact configuration in the last 6 years. Memorise the criterion and the geometry: E typically points up, B into the page, particle moves rightward.
Galvanometer to Ammeter / Voltmeter Conversion
This sub-topic alone has yielded 4 NTA questions since 2019. Memorise:
- Ammeter: Galvanometer + low-resistance shunt S in parallel. Shunt S = (Iₘ Rₘ)/(I − Iₘ), where Iₘ is full-scale deflection current.
- Voltmeter: Galvanometer + high-resistance R in series. R = (V/Iₘ) − Rₘ.
Ideal ammeter has zero resistance, ideal voltmeter has infinite resistance. Both questions are pure substitution once you remember the topology.
Two-Week Magnetism Mastery Plan
Days 1–3: Read NCERT Chapter 4 (Class 12) cover-to-cover. Solve every example. Get comfortable with vector cross product and the right-hand rule.
Days 4–7: Move to HC Verma Vol-2 (Chapters 34–35) or DC Pandey. Solve 8 problems daily, prioritising Biot–Savart applications and force-on-loop problems.
Days 8–11: Previous-year JEE Main questions (2019–2025) on magnetism. Time yourself — under 90 seconds per problem.
Days 12–14: Mixed mock with EM Induction. Magnetism + induction together in a 30-minute timed sprint, 12 questions, target 10/12.
What to Skip (Be Strategic)
JEE Main rarely asks the following — you can de-prioritise them in the final 30 days:
- Para/dia/ferromagnetic susceptibility numerical (Class 12 magnetism chapter, separate from this one)
- Curie’s law derivation
- Earth’s magnetism — declination/dip beyond the basic definitions
Focus instead on Biot–Savart computations, parallel wire force, cyclotron, and torque on a loop. These four sub-topics yield 90% of the JEE Main magnetism marks.
Conclusion: Magnetic Effects of Current is the highest-leverage chapter in the EM cluster. Memorise the 17-row formula sheet, drill the three right-hand rules until they are automatic, and solve all 35 categories above. With this preparation, you will lock in the 6–8 magnetism marks every JEE Main shift — and these are exactly the marks that separate a 95 percentile from a 99 percentile finish.