Electricity and Magnetism for the Physics Olympiad — NSEP and INPhO

Weightage: Electromagnetism is the second large area in NSEP and INPhO. The questions favour symmetry arguments, energy methods and clever circuit reduction over long algebra. Confirm the current syllabus and pattern with HBCSE.

1. Electrostatics from symmetry

Coulomb's law gives $F=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}$, and the potential of a point charge is $V=\frac{q}{4\pi {\epsilon }_{0}r}$. Superposition adds fields as vectors and potentials as scalars, so potentials are usually easier.

Gauss's law: $\oint \mathbf{E}\cdot d\mathbf{A}=Q_{enc}/{\epsilon }_0$. It is useful only when symmetry fixes the field's direction and magnitude on a surface.

Inside a conductor in equilibrium, $\mathbf{E}=0$, the potential is constant, and any charge sits on the surface. A dipole of moment $p$ has an on-axis field $\frac{2p}{4\pi {\epsilon }_0{r}^3}$, falling as $r^{-3}$.

The method of images. A point charge $q$ at distance $d$ from a grounded infinite conducting plane produces the same field in front of the plane as the charge together with an image $-q$ at distance $d$ behind it. The force on the charge is $\frac{q^2}{4\pi {\epsilon }_0(2d{)}^2}$, and the induced charge on the plane totals $-q$.

2. Capacitors and energy

A capacitor stores $C=Q/V$. For a parallel plate, $C={\epsilon }_{0}A/d$, and a dielectric of constant $\kappa$ multiplies it by $\kappa$. In parallel, $C=C_1+C_2$, and in series $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$.

The stored energy is $U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}$, with energy density $\frac{1}{2}{\epsilon }_0{E}^2$.

Worked example. Two identical capacitors, one charged to $V$ and one uncharged, are connected in parallel. The charge is shared equally, so each has $V/2$. The initial energy is $\frac{1}{2}C{V}^2$ and the final energy is $2\times \frac{1}{2}C(V/2{)}^2=\frac{1}{4}C{V}^2$. Half of the energy is lost, as heat and radiation in the wire, whatever its resistance.

When a dielectric slab is inserted with the battery connected, $V$ is fixed, $C$ rises and charge flows in. With the battery disconnected, $Q$ is fixed, $V$ falls and the slab is drawn in.

3. Circuits

Reduce circuits by symmetry. Points at the same potential by symmetry can be joined without changing the currents, and a branch carrying no current can be removed. For a cube of resistors $R$ per edge, the resistance between opposite corners is $\frac{5}{6}R$, and between adjacent corners $\frac{7}{12}R$.

For an RC circuit charging through $R$, $q=CV\left(1-e^{-t/RC}\right)$, with time constant $\tau =RC$. For an RL circuit, $I=\frac{V}{R}\left(1-e^{-Rt/L}\right)$ with $\tau =L/R$. Kirchhoff's laws and the maximum power transfer at $R_{load}=R_{source}$ complete the toolkit.

4. Magnetic fields and forces

A charge $q$ in a field $\mathbf{B}$ feels $\mathbf{F}=q\mathbf{v}\times \mathbf{B}$, which does no work. For $\mathbf{v}\perp \mathbf{B}$ the path is a circle of radius $r=\frac{mv}{qB}$ and period $T=\frac{2\pi m}{qB}$, independent of speed. If $\mathbf{v}$ has a component along $\mathbf{B}$, the path is a helix.

Fields to know:

• Long straight wire: $B=\frac{{\mu }_{0}I}{2\pi r}$.
• Centre of a circular loop: $B=\frac{{\mu }_{0}I}{2R}$.
• Long solenoid: $B={\mu }_{0}nI$ inside, and zero outside.

Ampere's law: $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{enc}$. The force on a current element is $Id\mathbf{l}\times \mathbf{B}$, and two parallel wires with currents in the same direction attract.

The magnetic moment of a loop is $m=IA$, and the torque in a field is $\tau =mB\sin \theta$.

5. Electromagnetic induction

Faraday's law: $\mathcal{E}=-\frac{d\Phi }{dt}$, and Lenz's law gives the sign: the induced current opposes the change in flux.

Motional emf in a rod of length $l$ moving at $v$ perpendicular to $B$ is $\mathcal{E}=Blv$. A rod of length $l$ rotating about one end at angular speed $\omega$ in a perpendicular field has $\mathcal{E}=\frac{1}{2}B\omega l^2$.

Worked example. A rod of length $0.5$ m rotates at $10$ rad/s in a field of $0.2$ T. Then $\mathcal{E}=\frac{1}{2}(0.2)(10)(0.25)=0.25$ V.

When a rod slides on rails and is pulled at constant speed $v$, the power spent against the magnetic force equals the electrical power $\frac{B^2{l}^2{v}^2}{R}$, which shows energy conservation.

Self-inductance gives $\mathcal{E}=-LdI/dt$ and stored energy $\frac{1}{2}L{I}^2$. For a solenoid, $L={\mu }_0{n}^{2}A\ell$.

6. Alternating current

For a series circuit with $R$, $L$ and $C$ driven at $\omega$, the impedance is $Z=\sqrt{R^2+(\omega L-1/\omega C{)}^2}$. Resonance occurs at ${\omega }_0=1/\sqrt{LC}$, with the quality factor $Q={\omega }_{0}L/R$. Average power is $VI\cos \varphi$. A transformer changes the voltage in the ratio of turns and conserves power in the ideal case.

7. Method

Look for symmetry first, then an energy or conservation argument, then algebra. For fields, ask whether Gauss or Ampere applies. For circuits, ask whether symmetry or a time constant gives the answer without solving equations. Check limits, such as $d\to \infty$ or $R\to 0$.

Common traps

• Using Gauss's law without enough symmetry.
• Thinking a magnetic force changes kinetic energy. It does not.
• Forgetting the minus sign in Lenz's law.
• Missing the energy lost when charged capacitors are connected.
• Using the sheet field $\sigma /2{\epsilon }_0$ at a conductor surface, where the field is $\sigma /{\epsilon }_0$.

Memory aids

• "Potentials add as scalars": superposition.
• "Image is minus q behind the plane": conducting plane.
• "Half of the energy is lost": connecting equal capacitors with one charged.

Summary

Electrostatic problems are solved by symmetry, Gauss's law and potentials, with images for conductors. Capacitors store energy, and connecting charged capacitors loses energy.

Circuits yield to symmetry and time constants. Magnetic forces do no work, and induction problems follow Faraday and Lenz, with energy conservation as the check.

Exam protocol

• Look for symmetry before computing.
• Prefer potentials to fields when adding contributions.
• Check energy conservation in induction problems.
• Confirm the syllabus and pattern with HBCSE.