Biot-Savart & Ampere's Law: JEE Main Guide

Magnetic effects of current is one of the most formula-rich chapters in JEE Main Physics, contributing three to five questions per session. The good news: once you understand which law to apply (Biot-Savart for irregular geometries, Ampere's law for symmetric ones), the calculations become routine. This guide maps out every standard configuration JEE tests and the fastest route to each answer.

Biot-Savart Law: The Fundamental Tool

The Biot-Savart law gives the magnetic field dB produced by an infinitesimal current element Idl at a point P separated by displacement r: dB = (μ₀/4π) × (Idl × r̂)/r². The direction is given by the right-hand rule on the cross product (fingers along dl, curl toward r̂, thumb gives dB). The magnitude: dB = (μ₀/4π) × (I dl sinθ)/r² where θ is the angle between dl and r.

Key results to memorise: (1) Infinite straight wire: B = μ₀I/2πd (d = perpendicular distance). (2) Finite wire from angle α to β: B = (μ₀I/4πd)(sinα + sinβ). (3) Circular loop at centre: B = μ₀I/2R. (4) Circular arc of angle θ at centre: B = μ₀Iθ/4πR. The circular-arc result is JEE's favourite — a wire bent into a semicircle gives B = μ₀I/4R at the centre. For the broader magnetism context, see our magnetic effects of current guide.

Ampere's Circuital Law: Symmetry Shortcuts

Ampere's law: ∮ B⃗ · dl⃗ = μ₀I_enclosed. It is powerful when symmetry makes B constant along a closed loop (Amperian loop). Choose the Amperian loop so that B is either parallel or perpendicular to dl everywhere, and constant in magnitude where parallel. Then ∮ B·dl = B × (circumference) = μ₀ × (enclosed current).

Infinite solenoid: B = μ₀nI inside (n = turns per metre), zero outside. Toroid: B = μ₀NI/2πr inside the toroid (r = radius from centre), zero elsewhere. Infinite plane of current (surface current density K): B = μ₀K/2 on each side. The solenoid and toroid results appear directly in JEE questions — know them cold. Take a free mock test on magnetic effects to test your application speed.

Force Between Current-Carrying Conductors

Two parallel wires carrying currents I₁ and I₂ separated by distance d exert force per unit length F/L = μ₀I₁I₂/2πd. Parallel currents attract; antiparallel currents repel. This is the basis of the SI definition of the ampere (historically). JEE uses this in direct force questions and in torque-on-loop problems. Magnetic moment of a current loop: m = NIA (N turns, current I, area A). Torque on loop in field B: τ = m × B = mB sinθ. Potential energy: U = −m·B = −mB cosθ. These appear in exam questions about galvanometers and compass needles in uniform fields.

Field Inside a Long Hollow Cylinder and Solenoid Variants

For a long solid cylinder of radius R carrying current I uniformly distributed: inside (r < R), B = μ₀Ir/2πR²; outside (r > R), B = μ₀I/2πr. The field increases linearly inside and decreases as 1/r outside — a graph question favourite. For a hollow cylinder (current only on surface), B = 0 inside and B = μ₀I/2πr outside. JEE 2024 had a question using the "inside cylinder" result directly. A finite solenoid has fringe effects at the ends: the field at each end is exactly half the central field (B_end = μ₀nI/2). Apply this when the problem says "at the end of a solenoid." For related content on electromagnetic induction see our electromagnetic induction guide and for force on moving charges see our magnetic force guide.