Electricity and Magnetism

Electricity and magnetism carry about 7 questions (our estimate), second only to mechanics.

Seven sub-areas share them: electrostatics and capacitors, DC circuits, thermoelectricity, magnetic force and fields, magnetic materials, induction, and AC.

Most numerical questions are a single formula; Section B ones chain two (a cell with internal resistance, an RLC power factor, a charge in a magnetic field).

5.1 Electrostatics

Coulomb's law: F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}, k=9\times10^9 N m² C⁻²; in a medium divide by \varepsilon_r.

Two charges of 1 µC at 1 m: 9\times10^{-3} N. Electric field E=\frac{F}{q}=\frac{kQ}{r^2} (lines start on positive charges, never cross, are perpendicular to conductors' surfaces); between plates E=\frac{V}{d}.

Dipole: p=qd; torque pE\sin\theta; energy -pE\cos\theta; axial field \frac{2kp}{r^3}, equatorial \frac{kp}{r^3}.

Gauss's law: \oint\vec E\cdot d\vec A=\frac{q_{enc}}{\varepsilon_0}.

Results: infinite line \frac{\lambda}{2\pi\varepsilon_0r}; infinite sheet \frac{\sigma}{2\varepsilon_0}; conductor surface \frac{\sigma}{\varepsilon_0}; charged sphere outside \frac{kQ}{r^2}, inside a conductor 0, inside a uniformly charged solid sphere \frac{kQr}{R^3}.

Potential V=\frac{kQ}{r} (scalar; adds algebraically); E=-\frac{dV}{dr}; work W=q(V_A-V_B); potential energy of two charges \frac{kq_1q_2}{r}; equipotential surfaces are perpendicular to field lines. 1 eV =1.6\times10^{-19} J. Point-charge zero field between like charges 4\muC and 1\muC 30 cm apart is at 20 cm from the larger one.

Capacitors. C=\frac{Q}{V}; parallel plate C=\frac{\varepsilon_0A}{d} (with a dielectric K\frac{\varepsilon_0A}{d}); sphere 4\pi\varepsilon_0R; cylindrical, spherical shells.

Series: \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}, same charge; parallel: C=C_1+C_2, same voltage. Energy U=\frac{1}{2}CV^2=\frac{Q^2}{2C}=\frac{1}{2}QV; energy density \frac{1}{2}\varepsilon_0E^2.

Dielectric (inserted with the battery off): C\to KC, V\to\frac{V}{K}, E\to\frac{E}{K}, Q constant, U\to\frac{U}{K}; (battery on: V same, Q\to KQ). Polarisation reduces the field inside the dielectric; displacement D=\varepsilon E.

Types: paper, mica, electrolytic (polarised), variable. Sharing charge between capacitors: common V=\frac{C_1V_1+C_2V_2}{C_1+C_2}, with an energy loss.

5.2 DC circuits

Ohm's law V=IR; resistivity R=\rho\frac{l}{A} (\rho rises with T for metals, falls for semiconductors); drift velocity v_d=\frac{I}{nAe}; conductivity \sigma=\frac{1}{\rho}. R_t=R_0(1+\alpha t).

Series R=\sum R_i; parallel \frac{1}{R}=\sum\frac{1}{R_i} (6 Ω ∥ 3 Ω = 2 Ω). Power P=VI=I^2R=\frac{V^2}{R}; a 100 W, 220 V bulb has R=484\ \Omega and, on 110 V, gives 25 W. Energy =Pt; H=I^2Rt (Joule's law).

Cell: \varepsilon=V+Ir; current I=\frac{\varepsilon}{R+r}; terminal voltage V=\varepsilon-Ir (cell of 2 V, r=0.5 Ω, R=4.5 Ω: 1.8 V). Maximum power transfer when R=r. Series of n cells n\varepsilon, nr; parallel \varepsilon, \frac{r}{n}.

Kirchhoff: junction rule \sum I=0 (charge conservation); loop rule \sum\varepsilon=\sum IR (energy). Wheatstone bridge balances when \frac{P}{Q}=\frac{R}{S}.

Meter bridge, potentiometer (compare emfs, \frac{\varepsilon_1}{\varepsilon_2}=\frac{l_1}{l_2}; measures without drawing current).

Galvanometer (moving coil, \theta\propto I): ammeter by a parallel shunt S=\frac{I_gG}{I-I_g}; voltmeter by a series multiplier R=\frac{V}{I}_g-G; an ideal ammeter has zero resistance, an ideal voltmeter infinite.

Ohmmeter uses a cell with the galvanometer and a series resistor (zero at right end).

5.3 Thermoelectric effect

Seebeck effect: a loop of two different metals with junctions at different temperatures carries an emf, \varepsilon=aT+\frac{1}{2}bT^2 (a thermocouple).

The neutral temperature is where \frac{d\varepsilon}{dT}=0, the inversion temperature is 2T_n-T_c. A thermopile (several couples in series) detects small radiation.

Peltier effect: current through a junction absorbs or releases heat (\Pi), the reverse of Seebeck; it makes thermoelectric coolers. Thomson effect: heat is absorbed or released along a single conductor with a temperature gradient.

Applications: thermometry of very high and low temperatures, refrigeration.

5.4 Magnetic effect of current

Force on a charge: \vec F=q\vec v\times\vec B (no work done); in a perpendicular field the path is a circle of radius r=\frac{mv}{qB}, period T=\frac{2\pi m}{qB} (independent of speed: the cyclotron); with \vec E also: Lorentz q(\vec E+\vec v\times\vec B).

A proton at 10^6 m s⁻¹ in 0.1 T: r\approx10 cm. Force on a conductor: F=BIl\sin\theta. Torque on a coil: \tau=NIAB\sin\theta (moving-coil galvanometer); magnetic moment m=NIA.

Hall effect: V_H=\frac{BI}{nqt} gives the sign and density of the carriers and measures B.

Biot–Savart: dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}, \mu_0=4\pi\times10^{-7} T m A⁻¹.

Long straight wire B=\frac{\mu_0I}{2\pi r} (5 A at 0.1 m: 10^{-5} T); centre of a circular loop \frac{\mu_0NI}{2R}; on the axis \frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}; solenoid B=\mu_0nI (n turns per metre), toroid \frac{\mu_0NI}{2\pi r}.

Ampère's law: \oint\vec B\cdot d\vec l=\mu_0I_{enc}. Parallel wires: force per length \frac{\mu_0I_1I_2}{2\pi d}, attractive for currents in the same direction (defines the ampere). 10 A and 20 A at 10 cm: 4\times10^{-4} N m⁻¹.

5.5 Magnetic properties of matter

Earth's magnetism: declination, dip (inclination) \delta, horizontal component B_H; \tan\delta=\frac{B_V}{B_H}; dip is 0 at the magnetic equator and 90^\circ at the poles.

Materials: diamagnetic (\chi small negative, weakly repelled: Bi, Cu, water); paramagnetic (\chi small positive, \chi\propto\frac{1}{T} Curie: Al, Pt); ferromagnetic (\chi large positive, domains, lose ferromagnetism above the Curie temperature: Fe, Co, Ni). \mu=\mu_0\mu_r, \mu_r=1+\chi, B=\mu_0(H+M).

Hysteresis: the B–H loop; retentivity, coercivity; the loop area is the energy lost per cycle per volume. Soft iron (narrow loop, low retentivity) is for transformer cores and electromagnets; steel (wide loop) for permanent magnets.

5.6 Electromagnetic induction

Faraday: \varepsilon=-N\frac{d\Phi}{dt}, \Phi=BA\cos\theta (weber); Lenz's law (the sign: the induced current opposes the change; conservation of energy).

Motional emf \varepsilon=Blv (a rod of 0.5 m at 4 m s⁻¹ in 0.2 T: 0.4 V); rotating rod \frac{1}{2}B\omega l^2. Induced charge \frac{\Delta\Phi}{R}. Eddy currents (braking, induction furnace, laminated cores to reduce them).

AC generator: coil of N turns, area A, rotating at \omega: \varepsilon=NBA\omega\sin\omega t.

Self-inductance \varepsilon=-L\frac{dI}{dt}, \Phi=LI; solenoid L=\mu_0n^2Al; energy U=\frac{1}{2}LI^2 (2 H, 0 to 5 A in 0.1 s: emf 100 V, energy 25 J). Mutual inductance \varepsilon_2=-M\frac{dI_1}{dt}.

Transformer (ideal): \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}; step-up raises voltage and lowers current; losses: copper, iron (eddy, hysteresis), flux leakage; efficiency \frac{P_s}{P_p}.

High-voltage transmission lowers I^2R loss.

5.7 Alternating currents

I=I_0\sin\omega t; I_{rms}=\frac{I_0}{\sqrt{2}}=0.707I_0 (220 V mains has a peak of 311 V; Nepal's supply is 220 V, 50 Hz); mean value over a half cycle \frac{2I_0}{\pi}.

Reactances X_L=\omega L (the voltage leads current by 90^\circ), X_C=\frac{1}{\omega C} (the voltage lags by 90^\circ); resistor in phase. Series LCR: Z=\sqrt{R^2+(X_L-X_C)^2}, \tan\phi=\frac{X_L-X_C}{R}.

Resonance at \omega_0=\frac{1}{\sqrt{LC}} (L=1 mH, C=1\,\muF: 5 kHz): Z=R, current maximum, phase zero. Quality factor Q=\frac{\omega_0L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}} (sharpness of resonance).

Power P=V_{rms}I_{rms}\cos\phi; power factor \cos\phi=\frac{R}{Z} (R = 30 Ω, X_L=60, X_C=20: Z=50, \cos\phi=0.6); a pure L or C draws no average power (wattless current).

Choke coil lowers current with little loss; the LC circuit oscillates with f=\frac{1}{2\pi\sqrt{LC}}.

(a) A Van de Graaff generator at a science museum

(b) Largest potential before air sparks

Chapter summary

• E=\frac{kQ}{r^2}, V=\frac{kQ}{r}; Gauss's law gives symmetric fields.
• Series capacitors share charge; parallel share voltage; energy \frac{1}{2}CV^2.
• Terminal voltage is \varepsilon-Ir; shunts go in parallel, multipliers in series.
• Moving charges in B go in circles; parallel currents attract.
• Para-, dia- and ferro-magnetism differ in sign and size of \chi.
• Induced emf opposes the change (Lenz); transformers trade voltage for current.
• RLC: Z, resonance at \frac{1}{2\pi\sqrt{LC}} and power factor \frac{R}{Z}.

Practice questions

Set A — Recall (12 questions)

1. The force between two 1 µC charges 1 m apart in air isa9 Nb1 Nc9\times10^{-3} Nd9\times10^{-6} N
2. Capacitors of 2 µF and 3 µF connected in series have a combined capacitance ofa1.2 µFb5 µFc1.5 µFd6 µF
3. The energy stored in a 10 µF capacitor charged to 100 V isa0.5 Jb5 Jc0.5 mJd0.05 J
4. The equivalent resistance of 6 Ω and 3 Ω in parallel isa9 Ωb2 Ωc3 Ωd4.5 Ω
5. The resistance of a 220 V, 100 W bulb isa484 Ωb2.2 Ωc22000 Ωd220 Ω
6. The magnetic field 0.1 m from a long straight wire carrying 5 A isa10^{-6} Tb10^{-4} Tc10^{-5} Td2\times10^{-6} T
7. A 0.5 m conductor carrying 4 A is perpendicular to a 0.2 T field. The force on it isa0.1 Nb0.8 Nc1.6 Nd0.4 N
8. A coil of 100 turns has its flux changed by 0.01 Wb in 0.1 s. The induced emf isa1 Vb10 Vc100 Vd0.1 V
9. A transformer has 100 primary and 1000 secondary turns; 220 V is applied to the primary. The secondary voltage isa2200 Vb22 Vc22000 Vd220 V
10. The rms value of an AC of peak voltage 311 V is nearlya311 Vb440 Vc220 Vd155 V
11. The resonant frequency of a series LC circuit with L=1 mH and C=1\ \muF is nearlya159 Hzb1 kHzc31.6 kHzd5033 Hz
12. A material with a small positive magnetic susceptibility isadiamagneticbparamagneticcferromagneticdsuperconducting

Answers and explanations — Set A

1. (c) — 9\times10^9\times10^{-12}/1.

2. (a) — \frac{2\times3}{2+3}=1.2.

3. (d) — \frac{1}{2}\times10^{-5}\times10^4=0.05 J.

4. (b) — \frac{6\times3}{9}=2.

5. (a) — R=\frac{V^2}{P}=\frac{48400}{100}.

6. (c) — \frac{\mu_0I}{2\pi r}=\frac{2\times10^{-7}\times5}{0.1}.

7. (d) — BIl=0.2\times4\times0.5.

8. (b) — N\frac{\Delta\Phi}{\Delta t}=100\times0.1.

9. (a) — 220\times\frac{1000}{100}.

10. (c) — \frac{311}{\sqrt{2}}.

11. (d) — \frac{1}{2\pi\sqrt{10^{-9}}}=5033 Hz.

12. (b) — paramagnetic; diamagnetic \chi<0, ferromagnetic \chi\gg1.

Set B — Understanding (8 questions)

1. Charges +4\ \muC and +1\ \muC are 30 cm apart. The electric field is zero on the line between them at a distance from the 4\ \muC charge ofa10 cmb15 cmc25 cmd20 cm
2. The potential at a distance of 0.9 m from a 2 µC point charge isa2\times10^4 Vb2\times10^3 Vc2\times10^5 Vd1.8\times10^4 V
3. A parallel-plate capacitor has plate area 100 cm² and separation 1 mm in air (\varepsilon_0=8.85\times10^{-12}). Its capacitance isa8.85 pFb88.5 pFc885 pFd8.85 nF
4. A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 4.5 Ω resistor. The terminal voltage isa2.0 Vb1.6 Vc1.8 Vd0.2 V
5. A galvanometer of 100 Ω and full-scale current 1 mA is to read 1 A. The shunt needed is abouta0.1 Ωb1 Ωc10 Ωd100 Ω
6. A coil of 50 turns, area 0.02 m² carries 2 A in a 0.5 T field. The maximum torque isa0.5 N mb2 N mc10 N md1 N m
7. A 2 H inductor's current rises from 0 to 5 A in 0.1 s. The average induced emf isa10 Vb100 Vc50 Vd200 V
8. In a series circuit R=30\ \Omega, X_L=60\ \Omega, X_C=20\ \Omega. The power factor isa0.8b0.5c0.6d1

Answers and explanations — Set B

1. (d) — \frac{x}{0.3-x}=\sqrt{\frac{4}{1}}=2, x=0.2 m.

2. (a) — \frac{9\times10^9\times2\times10^{-6}}{0.9}.

3. (b) — \frac{8.85\times10^{-12}\times10^{-2}}{10^{-3}}=8.85\times10^{-11} F.

4. (c) — I=\frac{2}{5}=0.4 A; V=2-0.4\times0.5=1.8 V.

5. (a) — S=\frac{0.001\times100}{0.999}\approx0.1\ \Omega.

6. (d) — NIAB=50\times2\times0.02\times0.5.

7. (b) — L\frac{\Delta I}{\Delta t}=2\times50.

8. (c) — Z=\sqrt{30^2+40^2}=50; \cos\phi=\frac{30}{50}.

Set C — Application (6 questions)

1. A 100 W, 220 V lamp is run on 110 V. Its power isa50 Wb100 Wc12.5 Wd25 W
2. In a balanced Wheatstone bridge P=10\ \Omega, Q=20\ \Omega, R=30\ \Omega. The fourth arm S isa15 Ωb40 Ωc60 Ωd20 Ω
3. Two long parallel wires 10 cm apart carry 10 A and 20 A in the same direction. The force per unit length isa4\times10^{-4} N m⁻¹, attractiveb4\times10^{-4} N m⁻¹, repulsivec2\times10^{-4} N m⁻¹, attractived4\times10^{-5} N m⁻¹, attractive
4. A proton (m=1.67\times10^{-27} kg) moves at 10^6 m s⁻¹ perpendicular to a 0.1 T field. The radius of its path is nearlya1 cmb10 cmc1 md0.1 mm
5. A series RLC circuit has R=10\ \Omega, L=0.1 H and C=100\ \muF. Its quality factor is nearlya31.6b316c3.16d0.316
6. A 0.5 m rod moves at 4 m s⁻¹ perpendicular to a 0.2 T field. The emf across its ends isa0.1 Vb4 Vc0.8 Vd0.4 V

Answers and explanations — Set C

1. (d) — P\propto V^2, \frac{1}{4}\times100.

2. (c) — \frac{P}{Q}=\frac{R}{S}: S=\frac{RQ}{P}=60.

3. (a) — \frac{2\times10^{-7}\times10\times20}{0.1}=4\times10^{-4}; same-direction currents attract.

4. (b) — r=\frac{mv}{qB}=\frac{1.67\times10^{-27}\times10^6}{1.6\times10^{-19}\times0.1}=0.104 m.

5. (c) — \omega_0=316 rad s⁻¹, Q=\frac{\omega_0L}{R}=\frac{31.6}{10}.

6. (d) — Blv=0.2\times0.5\times4.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. State Coulomb's law, define electric field and potential, and explain capacitors in series and parallel with the energy stored. [10]
2. State Kirchhoff's laws; explain internal resistance, the Wheatstone bridge and the potentiometer, with numericals. [10]
3. Explain electromagnetic induction, the transformer and the series LCR circuit with resonance and power factor. [10]

(a) Capacitors

(b) Transformer

(c) Van de Graaff

(d) Galvanometer

(e) Lightning

Model answer D1: Electrostatics and capacitors (10 marks)

Introduction. Coulomb's law: the force between two point charges in vacuum is F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}, along the line joining them, with \frac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}.

For 2\ \mu\text{C} and 3\ \mu\text{C} at 0.3\ \text{m}: F = 9\times10^9\times\frac{6\times10^{-12}}{0.09} = 0.6\ \text{N}.

Field and potential.

1. Electric field E = F/q_0 = \frac{kQ}{r^2} (unit N C⁻¹ or V m⁻¹).
2. Potential V = \frac{kQ}{r} (a scalar); E = -\frac{dV}{dr}.
3. Gauss's law: \oint E\cdot dA = \frac{Q}{\varepsilon_0}; inside a charged conducting sphere the field is zero.
4. Potential energy of two charges U = \frac{kq_1q_2}{r}.

Capacitor. C = \frac{Q}{V}; a parallel-plate capacitor has C = \frac{\varepsilon_0A}{d}, and a dielectric of constant K multiplies it by K.

Arrangement	Equivalent capacitance	Charge and voltage
Series	\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}	Same Q, voltages add
Parallel	C = C_1 + C_2	Same V, charges add

For 4\ \mu\text{F} and 12\ \mu\text{F}: series = \frac{4\times12}{16} = 3\ \mu\text{F} and parallel = 16\ \mu\text{F}.

Energy stored. U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV. For the 3\ \mu\text{F} combination charged to 12\ \text{V}: U = \frac{1}{2}\times3\times10^{-6}\times144 = 216\ \mu\text{J}. The energy density is \frac{1}{2}\varepsilon_0E^2.

Conclusion. Charges create fields and potentials, and capacitors store energy in the field between their plates.

Model answer D2: DC circuits (10 marks)

Introduction. Ohm's law (V = IR) is not enough for a network, so Kirchhoff's laws are used.

Kirchhoff's laws.

1. Junction (current) law: the sum of currents entering a junction equals the sum leaving (\sum I = 0), which expresses conservation of charge.
2. Loop (voltage) law: around a closed loop \sum E = \sum IR, which expresses conservation of energy.

Internal resistance. A cell of emf E and internal resistance r gives I = \frac{E}{R + r} and a terminal voltage V = E - Ir. For E = 12\ \text{V}, r = 1\ \Omega, R = 5\ \Omega: I = 2\ \text{A} and V = 10\ \text{V}. Maximum power goes to the load when R = r.

Wheatstone bridge. Four resistances P, Q, R, S are balanced when no current flows in the galvanometer: \frac{P}{Q} = \frac{R}{S}. For P = 10\ \Omega, Q = 15\ \Omega, R = 20\ \Omega: S = \frac{RQ}{P} = 30\ \Omega.

The meter bridge is the practical form, using S = R\,\frac{100 - l}{l}.

Potentiometer. A long uniform wire carries a steady current. A cell is balanced at length l where the potential difference equals its emf, so no current is drawn.

Uses: (a) compare emfs \frac{E_1}{E_2} = \frac{l_1}{l_2}; (b) find the internal resistance r = R\left(\frac{l_1}{l_2} - 1\right). It is more accurate than a voltmeter.

Series and parallel. Series: R = R_1 + R_2; parallel: \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}.

Instrument	Principle	Advantage
Wheatstone bridge	Balanced ratio	Accurate R
Potentiometer	Null deflection	No current drawn from the cell
Voltmeter	Deflection	Draws some current

Conclusion. Kirchhoff's laws solve any network, and null-deflection methods give accurate measurements.

Model answer D3: Electromagnetic induction, transformer and AC circuits (10 marks)

Introduction. Faraday's law: the induced emf is E = -\frac{d\Phi}{dt}, and Lenz's law (the minus sign) says that the induced current opposes the change that produces it, so energy is conserved.

Transformer. It has primary and secondary coils on a soft-iron core, and works only with AC. \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} (ideal).

A step-down transformer from 11{,}000\ \text{V} to 230\ \text{V} has turns ratio \frac{11000}{230} = 47.8. Losses: copper loss, eddy currents (laminated core), hysteresis and flux leakage.

Series LCR circuit. The reactances are X_L = \omega L and X_C = \frac{1}{\omega C}, so Z = \sqrt{R^2 + (X_L - X_C)^2} and \tan\phi = \frac{X_L - X_C}{R}. Numerical: R = 40\ \Omega, L = 0.2\ \text{H}, C = 100\ \mu\text{F}, 230\ \text{V}, 50\ \text{Hz}.

Quantity	Working	Value
X_L	2\pi\times50\times0.2	62.8\ \Omega
X_C	\frac{1}{2\pi\times50\times10^{-4}}	31.8\ \Omega
Z	\sqrt{40^2 + 31.0^2}	50.6\ \Omega
I_{rms}	230/50.6	4.54\ \text{A}
Phase angle	\tan^{-1}(31.0/40)	37.8^\circ (voltage leads)
Power factor	R/Z	0.79
Power	VI\cos\phi = I^2R	826\ \text{W}

Resonance. When X_L = X_C, Z = R (minimum), the current is maximum and the power factor is 1. The resonant frequency is f_0 = \frac{1}{2\pi\sqrt{LC}}; here f_0 = 35.6\ \text{Hz}. This is how a radio is tuned to a station.

Conclusion. Changing flux induces emf, transformers use it to change voltage, and in an AC circuit the average power depends on the power factor.