Magnetic Effects of Current: Complete JEE Main Guide

Magnetic effects of current is among the most formula-dense chapters in JEE Main Physics, yet it is also among the most patterned. The same six or seven problem setups appear session after session: the field at the centre of a circular loop, the force between parallel currents, the torque on a current loop in a uniform field, and the galvanometer conversion problem. A student who masters these canonical setups and their variants will never be surprised by this chapter in the exam.

Biot-Savart Law and Its Applications

The Biot-Savart law gives the magnetic field dB = (μ₀/4π)(I dl × r̂)/r² at a point due to a current element. Its most important applications are: the field at the centre of a circular loop B = μ₀I/2R; the field at a point on the axis of a circular loop B = μ₀IR²/(2(R²+x²)^(3/2)); and the field due to a straight wire B = μ₀I/2πd (at perpendicular distance d from the wire). These three results account for 60% of Biot-Savart questions in JEE Main. Know them with their direction conventions.

For a semicircular wire or an arc of angle θ, the field at the centre scales linearly: B = μ₀Iθ/4πR. A question giving an arbitrary arc angle is simply a scaled version of the full circle result. The direction is determined by the right-hand rule: curl the fingers in the direction of current flow and the thumb points in the field direction at the centre. To test your field-calculation speed, take a free magnetism mock test.

Ampere's Law and the Solenoid

Ampere's law ∮B·dl = μ₀I_enclosed lets us find the field in symmetric current distributions without integrating Biot-Savart. The key results: inside a long solenoid B = μ₀nI (where n is turns per unit length); inside a toroid B = μ₀NI/2πr; outside a long solenoid B ≈ 0. For a straight wire, Ampere's law confirms B = μ₀I/2πr outside, and inside a solid conductor B = μ₀Ir/2πR² — linear with radius inside and falling as 1/r outside. The graph of B vs radial distance from a current-carrying conductor (linear inside, hyperbolic outside) is a direct graph-reading question in JEE.

Multiple solenoids or coaxial systems require superposition of the individual field vectors, accounting for direction. For fields pointing in the same direction, add magnitudes; for opposing fields, subtract. This vector superposition is the main source of errors in solenoid problems.

Force on Current and Torque on a Loop

A current-carrying conductor in a magnetic field experiences F = IL × B, giving F = BIL sinθ for a straight conductor. Two parallel currents attract if they flow in the same direction and repel if they flow in opposite directions; force per unit length is μ₀I₁I₂/2πd. JEE tests this through the definition of the ampere (the current that produces 2×10⁻⁷ N/m between wires 1 m apart) at least once per session.

A current loop (area A, N turns) in a uniform field experiences a torque τ = NIAB sinθ, where θ is the angle between the magnetic moment and the field. The potential energy is U = −NIAB cosθ, minimum (stable) when aligned, maximum (unstable) when anti-aligned. The moving-coil galvanometer uses this torque: at equilibrium, NIAB = kφ, giving deflection φ ∝ I. Converting a galvanometer to an ammeter requires a shunt (low resistance in parallel); to a voltmeter, a high resistance in series. These conversion problems are solved in two steps: find the full-scale deflection current, then apply the voltage divider or current divider. Our electrostatics guide and Session 2 analysis show why mastering this chapter pays dividends in every paper.