Electromagnetism
Electromagnetism studies the magnetic effects of moving charges and the forces magnetic fields exert on currents and individual moving charges. The PMDC MDCAT 2026 syllabus highlights three areas: magnetic flux, magnetic flux density, and the motion of a charged particle in a magnetic field. Expect 1-2 MCQs per paper.
Magnetic Flux Density
The magnetic flux density (also called the magnetic field B) at a point is defined by the force experienced by a current-carrying conductor or a moving charge there. Its SI unit is the tesla (T): 1 T = 1 N A−1 m−1 = 1 kg s−2 A−1.
Force on a current-carrying conductor
F = BIL sinθ
where I is the current, L is the length of the wire in the field, and θ is the angle between B and the wire. The force is maximum when the wire is perpendicular to B (θ = 90°) and zero when parallel.
| Source | Formula for B | Notes |
|---|---|---|
| Long straight wire | B = μ0I / (2πr) | r = perpendicular distance; circles around wire (right-hand rule) |
| Centre of circular loop | B = μ0I / (2r) | r = loop radius; perpendicular to plane |
| N-loop coil | B = μ0NI / (2r) | N times the single-loop value |
| Long solenoid (inside) | B = μ0nI | n = turns per metre; uniform along axis |
| Toroid | B = μ0NI / (2πr) | r = mean radius of the torus |
| Bar magnet (axial, far) | B = (μ0/4π) · 2M/r³ | M = magnetic moment (informal MDCAT note) |
The constant μ0 = 4π × 10−7 T m A−1 is the permeability of free space.
Magnetic Flux
The magnetic flux through a surface is the total number of field lines crossing it:
Φ = B · A = BA cosθ
where θ is the angle between B and the area vector (normal to the surface). SI unit: weber (Wb); 1 Wb = 1 T m2.
- If the surface is perpendicular to B (θ = 0), Φ is maximum: Φ = BA.
- If parallel to B (θ = 90°), Φ = 0.
- For a coil of N turns, flux linkage = NΦ.
Motion of Charged Particle in Magnetic Field
A charge q moving with velocity v in a magnetic field B feels the Lorentz force:
F = q v × B |F| = qvB sinθ
The force is always perpendicular to v, so it changes direction but not speed. Magnetic forces do no work on a free charge.
Three special cases
- v parallel to B: F = 0. The charge moves in a straight line; B has no effect.
- v perpendicular to B: The force provides centripetal acceleration. The charge moves in a circle.
- v at an angle to B: The component along B is unchanged; the perpendicular component drives circular motion. Result: a helical path.
Circular motion — key formulas
For v perpendicular to B, equating qvB to mv2/r:
r = mv/(qB)
The angular frequency and period are
ω = qB/m, T = 2πm/(qB) = 2π/ω
Crucially, the period is independent of v. This is the principle of the cyclotron: faster particles trace bigger circles but take exactly the same time to complete a half-loop.
Crossed E and B fields select a single speed: only particles with v = E/B pass undeflected (electric and magnetic forces cancel). Selected ions then enter a region with B alone and curve with radius r = mv/(qB), allowing m/q to be measured precisely.
Worked MCQs
Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — it's where the deeper concept lives.
Q1. A charged particle moving in a uniform magnetic field experiences a force that is:
F = q v × B. The cross-product is perpendicular to both v and B, so the force never does work and the speed stays constant. The force is maximum when v is perpendicular to B and zero when parallel.
Q2. The radius of the circular path of a charged particle in a magnetic field is given by:
qvB = mv2/r ⇒ r = mv/(qB). The period T = 2πm/(qB) is independent of v — same time for any speed, just different circles.
Q3. SI unit of magnetic flux is:
Flux Φ is measured in webers (Wb = T m2). Flux density B is measured in teslas. Henry is the unit of inductance; gauss is a CGS unit (104 G = 1 T).
Q4. A solenoid has 500 turns per metre carrying 2 A. The field inside is approximately:
B = μ0nI = 4π × 10−7 × 500 × 2 = 4π × 10−4 ≈ 1.26 × 10−3 T. The field is uniform along the length and zero outside (long-solenoid approximation).
Q5. An electron and a proton enter the same magnetic field perpendicularly with the same kinetic energy. Which has the smaller circular radius?
Same KE: mv2/2 same, so mv = √(2mE). Therefore r = mv/(qB) = √(2mE)/(qB) ∝ √m. Electron has smaller mass, hence smaller radius. (Charges have equal magnitude.)
Quick Recap
- Magnetic flux: Φ = BA cosθ (Wb).
- Flux density (field): B in tesla; F = BIL sinθ on a wire.
- Solenoid: B = μ0nI; long straight wire: B = μ0I/(2πr).
- Lorentz force: F = qv × B; magnitude qvB sinθ; does no work.
- v ⊥ B ⇒ circle of radius r = mv/(qB); period T = 2πm/(qB), independent of v.
- v at angle to B ⇒ helical motion.
- Velocity selector: undeflected if v = E/B.