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Electromagnetism

Electromagnetism studies magnetic fields and the force a magnetic field exerts on a moving charge. The PMDC MDCAT 2026 syllabus highlights three areas: magnetic flux density, magnetic flux, and the motion of a charged particle in a magnetic field. Expect 1-2 MCQs per paper.

PMC Table of Specifications. Three subtopics: Magnetic Flux, Magnetic Flux Density, and Motion of Charged Particle in Magnetic Field.

Magnetic Flux Density

The magnetic flux density B (also called the magnetic field strength) at a point measures the strength of the magnetic field there. It is defined through the force felt by a charge q moving with speed v perpendicular to the field:

B = F / (qv)

So B is the force per unit charge per unit velocity. It is a vector quantity, directed along the field line at that point.

Units of magnetic flux density
FormExpressionComment
SI unittesla (T)Named after Nikola Tesla
From force on a charge1 T = 1 N C−1 (m s−1)−1B = F/(qv)
Equivalent1 T = 1 N A−1 m−1Since 1 C s−1 = 1 A
Base units1 T = 1 kg s−2 A−1Useful for dimension questions
From flux1 T = 1 Wb m−2B = Φ/A — "flux density"
CGS unit1 T = 104 gaussEarth's field ≈ 5 × 10−5 T

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.

Common trap. "Magnetic flux" (Φ, weber) and "magnetic flux density" (B, tesla) are not the same thing. Flux density is a per-unit-area quantity (B = Φ/A); flux is a property of a chosen surface. Examiners often swap units.

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

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, both ω and the period T are independent of v: a faster particle traces a bigger circle but takes exactly the same time to go round it.

Right-hand rule. Point fingers along v, curl them towards B; the thumb gives the direction of v × B. For a positive charge, F is along the thumb; for a negative charge (like an electron), reverse it.

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:

  • Always parallel to the field
  • Always parallel to the velocity
  • Perpendicular to both v and B
  • Maximum when v is parallel to B

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:

  • r = qB/(mv)
  • r = mv/(qB)
  • r = qvB/m
  • r = m/(qvB)

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:

  • Tesla
  • Weber
  • Henry
  • Gauss

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 square coil of side 0.20 m is placed with its plane perpendicular to a uniform magnetic field of 0.50 T. The magnetic flux through the coil is:

  • 0.10 Wb
  • 0.020 Wb
  • 0.20 Wb
  • Zero

Area A = 0.20 × 0.20 = 0.040 m2. The plane is perpendicular to B, so the area vector is parallel to B and θ = 0: Φ = BA cos0 = 0.50 × 0.040 = 0.020 Wb. Flux would be zero only if the plane contained the field lines.

Q5. An electron and a proton enter the same magnetic field perpendicularly with the same kinetic energy. Which has the smaller circular radius?

  • Proton
  • Electron
  • Both equal
  • Cannot be determined

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

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