Magnetic effects of current and magnetism for JEE Main

This unit gives JEE Main a steady stream of questions because it mixes three skills: finding a field (Biot–Savart or Ampère), finding a force (on a charge or a wire), and applying it (circular motion of a charge, a galvanometer, a loop in a field). The formulas are short. The errors come from directions, so get comfortable with the right-hand rules before anything else.

The official 2026 syllabus lists the Biot–Savart law for a circular loop; Ampère's law for a long straight wire and a solenoid; forces on moving charges in magnetic and electric fields; the force on a current-carrying conductor and between parallel currents; torque on a current loop; the moving coil galvanometer, its sensitivity and conversion to an ammeter or voltmeter; and the magnetism part: the current loop as a magnetic dipole, the bar magnet as an equivalent solenoid, the field of a magnetic dipole, torque on it, and para-, dia- and ferromagnetic materials with the effect of temperature.

Biot–Savart law

A small element dl carrying current I produces, at distance r,

dB = (μ₀/4π) × I dl sin θ / r², with μ₀/4π = 10⁻⁷ T m/A.

The direction is given by dl × r̂: curl the fingers of your right hand along the current and your thumb points along the field inside a loop.

A straight wire whose extension passes through the point gives zero field there, because sin θ = 0 for every element. That is how "semicircle plus straight leads" questions reduce to the arc alone.

Ampère's law

∮B · dl = μ₀I_enclosed. Like Gauss's law, it is always true but only useful when symmetry makes B constant along the loop.

• Long straight wire: B × 2πr = μ₀I, so B = μ₀I/(2πr) outside. Inside a solid wire of radius R carrying uniform current, B = μ₀Ir/(2πR²).
• Long solenoid: a rectangular loop with one side inside gives B = μ₀nI (n = turns per metre), uniform inside and nearly zero outside. At either end the field is about half of this.

A toroid (B = μ₀NI/2πr inside) follows from the same method, although the syllabus names only the wire and solenoid.

Forces on charges and currents

The Lorentz force is F = q(E + v × B). The magnetic part is perpendicular to v, so it does no work and never changes the speed.

A charge entering a uniform B at right angles moves in a circle:

• radius r = mv/(qB) = p/(qB) = √(2mK)/(qB)
• period T = 2πm/(qB), independent of speed and radius

If v makes an angle θ with B, the path is a helix with pitch (v cos θ) × T. If v is parallel to B, there is no force at all.

With crossed E and B fields, a charge passes undeflected when qE = qvB, so v = E/B. This is a velocity selector.

A straight conductor of length L carrying current I in a field B feels F = ILB sin θ. Two long parallel wires a distance d apart exert F/L = μ₀I₁I₂/(2πd) on each other. Like currents attract; opposite currents repel.

This force gave the older definition of the ampere: two wires 1 m apart, each carrying 1 A, feel 2 × 10⁻⁷ N per metre. Since 2019 the SI defines the ampere by fixing the value of e instead.

Torque on a loop and the galvanometer

A loop of N turns and area A carrying I has magnetic moment m = NIA, normal to its plane. In a uniform field:

• τ = mB sin θ = NIAB sin θ
• net force zero
• potential energy U = −mB cos θ

In a moving coil galvanometer, a radial field keeps sin θ = 1, so NIAB = kφ, where k is the torsion constant and φ the deflection. Therefore φ = (NAB/k) I.

• Current sensitivity: φ/I = NAB/k
• Voltage sensitivity: φ/V = NAB/(kG), where G is the coil's resistance

Doubling the number of turns doubles the current sensitivity, but it also doubles G, so the voltage sensitivity stays the same.

An ideal ammeter has zero resistance and an ideal voltmeter infinite resistance.

Magnetism and magnetic materials

A bar magnet's field looks like a solenoid's, and a current loop behaves like a magnetic dipole. For a dipole of moment m at distance r (large compared with its size):

• on the axis: B = (μ₀/4π)(2m/r³)
• on the equatorial line: B = (μ₀/4π)(m/r³), opposite to m
• torque τ = mB sin θ and energy U = −mB cos θ, exactly as for an electric dipole

Above its Curie temperature, a ferromagnet becomes paramagnetic.

Worked problems

Problem 1: proton and alpha particle. A proton and an alpha particle enter the same uniform field at right angles. Find the ratio of radii (proton : alpha) if they have (a) the same speed and (b) the same kinetic energy.

(a) r ∝ m/q: (1/1) : (4/2) = 1 : 2.
(b) r ∝ √m/q: (√1/1) : (√4/2) = 1 : 1, so the radii are equal.

Problem 2: semicircle with leads. A long wire comes in along a straight line, bends into a semicircle of radius 5 cm, and leaves along the same line. It carries 10 A. Find B at the centre of the semicircle.

The straight parts point through the centre and contribute nothing. For the arc (θ = π):
B = μ₀I/(4R) = (4π × 10⁻⁷ × 10)/(4 × 0.05) = 2π × 10⁻⁵ T ≈ 6.3 × 10⁻⁵ T.

Problem 3: galvanometer conversions. A galvanometer has G = 50 Ω and full-scale current 2 mA. Convert it to (a) a 10 V voltmeter and (b) a 2 A ammeter.

(a) R = 10/0.002 − 50 = 5,000 − 50 = 4,950 Ω in series.
(b) S = (0.002 × 50)/(2 − 0.002) = 0.1/1.998 ≈ 0.05 Ω in parallel.

Problem 4 (numerical answer): parallel wires. Two long parallel wires 10 cm apart carry 5 A and 10 A in the same direction. Find the force per metre on each, in units of 10⁻⁵ N/m.

F/L = 2 × 10⁻⁷ × 5 × 10/0.1 = 10⁻⁴ N/m = 10 × 10⁻⁵ N/m, attractive.

Practice set

1. Find the field at the centre of a semicircular arc of radius R carrying current I.
2. Does the period of a charged particle circling in a uniform field depend on its speed?
3. A long solenoid has 1,000 turns per metre and carries 2 A. Find B inside.
4. A charged particle enters a uniform magnetic field moving parallel to it. Describe its path.
5. A 100-turn circular coil of radius 10 cm carries 1 A in a field of 0.5 T. Find the maximum torque on it.
6. An electron passes undeflected through crossed fields E = 10⁴ V/m and B = 0.02 T. Find its speed.
7. A paramagnetic sample has susceptibility χ at 300 K. Estimate its susceptibility at 600 K.
8. The number of turns of a galvanometer coil is doubled, and its resistance doubles with it. What happens to its current and voltage sensitivity?

Answers

1. μ₀I/(4R).
2. No: T = 2πm/(qB).
3. 4π × 10⁻⁷ × 1,000 × 2 = 8π × 10⁻⁴ ≈ 2.5 × 10⁻³ T.
4. A straight line, because v × B = 0.
5. NIAB = 100 × 1 × π(0.1)² × 0.5 = 0.5π ≈ 1.57 N m.
6. v = E/B = 5 × 10⁵ m/s.
7. χ ∝ 1/T, so χ/2.
8. Current sensitivity doubles; voltage sensitivity is unchanged.

What to do next

• Draw the field directions for a wire, a loop and a solenoid until the right-hand rules are automatic.
• Solve 15 "arc plus straight wires" problems, adding each segment's contribution with its direction.
• Practise radius and period questions for protons, deuterons and alpha particles at equal speed, momentum and kinetic energy.
• Move on to electromagnetic induction, and revise circular motion for charged-particle problems.