11.2 Electricity and Magnetism

Key Takeaways

  • NMAT electricity and magnetism items reward Coulomb’s law direction and magnitude sense, electric field and potential intuition, Ohm’s law, series/parallel resistor networks, power P = IV, qualitative magnetic force ideas, and Faraday induction conceptually
  • Like charges repel and opposite charges attract; Coulomb force scales as q₁q₂/r² and is enormous compared with gravity for everyday charged objects
  • Current is charge flow; V = IR for ohmic devices; series resistors share current and add resistances; parallel resistors share voltage and combine as reciprocals
  • Magnetic fields exert forces on moving charges and currents; induced emf arises from changing magnetic flux (Faraday), with Lenz’s law opposing the change
  • Circuit problems are won by redrawing the network, identifying series vs parallel groups, and checking units of volts, amperes, ohms, and watts
Last updated: August 2026

11.2 Electricity and Magnetism on NMAT Physics

Within CEM NMAT Physics (30 items, ~30 minutes), electricity and magnetism is a dense scoring region: charge interactions, circuit analysis, and qualitative field/induction ideas. College introductory premed depth means you should compute simple networks, not only name devices.

Quick frame: Electrostatics asks who pushes whom and how hard. Circuits ask how charge flows under voltage with resistance. Magnetism and induction ask what moving charges and changing flux do.

Electric charge and Coulomb’s law

Electric charge is quantized in elementary units (e ≈ 1.6 × 10⁻¹⁹ C) and conserved. Two types: positive and negative. Like charges repel; opposite charges attract.

Coulomb’s law (point charges or spherically symmetric charge distributions outside):

F=kq1q2r2F = k\frac{|q_1 q_2|}{r^2}

with k ≈ 9.0 × 10⁹ N·m²/C². Force is along the line joining charges; direction from the repulsion/attraction rule. Force on each charge is equal in magnitude and opposite in direction (Newton’s third law).

IdeaExam use
1/r² dependenceDouble distance → force falls by 4
Product of chargesDouble one charge → double force
Conductors vs insulatorsCharges free to move vs bound
GroundingCan supply or remove charge to neutralize

Worked example A
Two point charges +2.0 μC and +8.0 μC are 0.30 m apart in air. Magnitude of force:

F=(9.0×109)(2.0×106)(8.0×106)(0.30)2=(9.0×109)1.6×10110.091.6 NF = (9.0\times10^9)\frac{(2.0\times10^{-6})(8.0\times10^{-6})}{(0.30)^2} = (9.0\times10^9)\frac{1.6\times10^{-11}}{0.09} ≈ 1.6\ \text{N}

Direction: repulsive (both positive).

Electric field and electric potential (intro)

The electric field E at a point is the force per unit positive test charge:

E=Fq0F=qE\vec{E} = \frac{\vec{F}}{q_0} \quad \Rightarrow \quad \vec{F} = q\vec{E}

Field lines point away from positive charge and toward negative charge. For a point charge, E = k|q|/r² with radial direction.

Electric potential V (voltage) is electric potential energy per unit charge. Potential difference relates to work: moving charge q through ΔV changes PE by qΔV. In a uniform field, |ΔV| = Ed along a displacement parallel to the field over distance d.

High-yield contrast:

QuantityTypeRelated to
Force FVectorqE
Field EVectorForce on + test charge
Potential VScalarPE per charge
Potential energy UScalarqV (with reference)

Worked conceptual scenario B
A proton and an electron in the same E field experience forces of equal magnitude but opposite direction (opposite charge). Acceleration of the electron is much larger because of smaller mass.

Current, resistance, and Ohm’s law

Electric current I is the rate of charge flow: I = Δq/Δt, unit ampere (A). Conventional current direction is the direction positive charge would flow (from higher to lower potential through a resistor).

Ohm’s law for ohmic materials/devices:

V=IRV = IR

Resistance R (ohms, Ω) depends on material resistivity ρ, length L, and cross-sectional area A: R = ρL/A for a uniform wire.

Power delivered or dissipated:

P=IV=I2R=V2RP = IV = I^2 R = \frac{V^2}{R}

Unit watt (W). Higher current through a resistor means more heating (I²R).

Worked example C
A 12 V battery is connected to a single 6.0 Ω resistor (ideal wires).

I=V/R=12/6.0=2.0 AI = V/R = 12/6.0 = 2.0\ \text{A}

P=IV=24 WorP=V2/R=144/6=24 WP = IV = 24\ \text{W} \quad \text{or} \quad P = V^2/R = 144/6 = 24\ \text{W}

Series and parallel resistors

Series: same current through each; voltages add; equivalent resistance:

Rs=R1+R2+R3+R_s = R_1 + R_2 + R_3 + \cdots

Parallel: same voltage across each branch; currents add; equivalent resistance:

1Rp=1R1+1R2+1R3+\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots

For two resistors: R_p = R₁R₂/(R₁+R₂). Parallel equivalent is always less than the smallest branch resistance.

ConfigurationCurrentVoltageR_eq
SeriesSameDividesSum (larger)
ParallelDividesSameReciprocal sum (smaller)

Worked example D — mixed network
R₁ = 4.0 Ω and R₂ = 12.0 Ω in parallel, then that combination in series with R₃ = 3.0 Ω, across 12 V.

First parallel pair:

R12=4×124+12=3.0 ΩR_{12} = \frac{4\times12}{4+12} = 3.0\ \Omega

Total: R_eq = 3.0 + 3.0 = 6.0 Ω. Battery current I_total = 12/6.0 = 2.0 A (through R₃). Voltage across parallel pair: V₁₂ = I_total × R₁₂ = 2.0 × 3.0 = 6.0 V. Current through R₁: I₁ = 6.0/4.0 = 1.5 A; through R₂: 6.0/12 = 0.5 A (sum 2.0 A — check).

Worked example E — power comparison
Two identical resistors: more power from a battery when they are in parallel (lower R_eq → larger total current and P = V²/R_eq for fixed V) than in series.

Magnetic fields (qualitative)

Moving charges and currents produce magnetic fields. A magnetic field exerts force on a moving charge:

F=qvBsinθF = qvB\sin\theta

(magnitude form), maximum when velocity is perpendicular to B, zero when parallel. Direction by right-hand rule for positive charges (left-hand or reverse for negative). Currents in wires feel F = ILB sinθ in uniform B.

Key qualitative facts:

  • Magnetic field lines form closed loops (no magnetic monopoles in standard intro physics)
  • A compass needle aligns with the local field
  • Parallel currents attract; antiparallel currents repel (consistent with field-and-force rules)
  • Charged particle in uniform B (v ⟂ B) moves in a circle (centripetal force magnetic); helical path if v has a parallel component

Worked conceptual scenario F
An electron beam enters a region of B directed into the page with velocity to the right. Magnetic force is perpendicular to v, bending the path — never doing work if F ⟂ v (speed constant, direction changes).

Electromagnetic induction (Faraday conceptual)

Magnetic flux through a loop relates to how much field “threads” the area (Φ ~ BA cosθ for uniform field). Faraday’s law: induced emf equals the negative rate of change of flux:

E=dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}

Flux changes if B changes, area changes, or orientation changes. Lenz’s law: induced current creates a field that opposes the change in flux (conservation of energy — you do work against magnetic drag when you pull a loop out of a field with current flowing).

Applications at awareness level: generators, transformers (AC, changing flux in coils), induction cooktops, and medical devices that use changing fields carefully.

Worked conceptual scenario G
A bar magnet’s north pole moves toward a conducting loop face-on. Flux of “northward” field through the loop increases; induced current produces flux away from the magnet (loop face acts like a north pole repelling the approaching north) — Lenz opposition.

Worked conceptual scenario H
Steady current in a coil produces steady B and constant flux. Induced emf is zero until current (hence B) changes — or until the coil moves relative to a magnet.

Error traps checklist

TrapFix
Series R_eq smaller than eitherSeries adds; parallel reduces
P = I/VP = IV
Magnetic force on charge at restNeed motion (or effective current) relative to B
Induced emf needs only B presentNeeds changing flux
Coulomb force linear in rInverse square
Conventional current = electron velocity directionOpposite to electron drift in metals

Study protocol

  1. Redraw series/parallel reductions until R_eq is automatic.
  2. Practice three power formulas and when each is convenient.
  3. State Lenz’s law on every induction stem before calculating.
  4. Sketch field directions for point charges and simple current loops.

Section checkpoint

You are ready when you can: (1) compute Coulomb force magnitude and state attraction/repulsion, (2) relate E, F, and charge sign, (3) reduce mixed resistor networks and find branch currents, (4) compute circuit power, and (5) explain Faraday/Lenz with a magnet-and-loop story.

Test Your Knowledge

Two point charges of +3.0 μC and −3.0 μC are separated by distance r. If r is doubled, the magnitude of the electrostatic force between them becomes:

A
B
C
D
Test Your Knowledge

Three resistors of 2 Ω, 2 Ω, and 2 Ω are connected in parallel across an ideal 6 V battery. What is the total current drawn from the battery?

A
B
C
D
Test Your Knowledge

A 10 Ω resistor dissipates 40 W of power. What is the current through the resistor?

A
B
C
D
Test Your Knowledge

Which situation produces an induced emf in a closed conducting loop according to Faraday’s law?

A
B
C
D