Noesis
Level 2 — RecallElectromagnetism

Electromagnetism

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall / Standard textbook problems / Short derivations) Time limit: 30 minutes Total marks: 40

Use ε0=8.85×10−12 F/m\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{F/m}, k=1/(4πε0)=9.0×109 N m2/C2k = 1/(4\pi\varepsilon_0)=9.0\times10^{9}\ \mathrm{N\,m^2/C^2}, μ0=4π×10−7 T m/A\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A} where needed.

Q1. State the three fundamental properties of electric charge (quantization, conservation, and one more), giving one sentence each. (3 marks)

Q2. Two point charges q1=+3 μCq_1 = +3\ \mu\mathrm{C} and q2=−5 μCq_2 = -5\ \mu\mathrm{C} are placed 0.20 m0.20\ \mathrm{m} apart in vacuum. Calculate the magnitude of the electrostatic force between them and state whether it is attractive or repulsive. (4 marks)

Q3. Using Gauss's law, derive the expression for the electric field magnitude at a distance rr from an infinitely long straight wire carrying a uniform linear charge density λ\lambda. (5 marks)

Q4. Three capacitors of 2 μF2\ \mu\mathrm{F}, 3 μF3\ \mu\mathrm{F} and 6 μF6\ \mu\mathrm{F} are connected in series across a 12 V12\ \mathrm{V} battery. Find (a) the equivalent capacitance and (b) the total energy stored. (5 marks)

Q5. Define electric potential at a point. Write the expression for the potential due to a point charge qq at distance rr, and calculate the potential 0.30 m0.30\ \mathrm{m} from a +2 nC+2\ \mathrm{nC} charge. (4 marks)

Q6. A copper wire of cross-sectional area 1.0×10−6 m21.0\times10^{-6}\ \mathrm{m^2} carries a current of 2.0 A2.0\ \mathrm{A}. If the free-electron density is 8.5×1028 m−38.5\times10^{28}\ \mathrm{m^{-3}}, calculate the drift velocity of the electrons. (4 marks)

Q7. State Faraday's law of electromagnetic induction and Lenz's law. Explain briefly how Lenz's law is a statement of energy conservation. (4 marks)

Q8. A solenoid of length 0.50 m0.50\ \mathrm{m} has 20002000 turns and carries a current of 3.0 A3.0\ \mathrm{A}. Calculate the magnetic field magnitude inside the solenoid. (3 marks)

Q9. In an RC charging circuit with R=10 kΩR = 10\ \mathrm{k\Omega} and C=100 μFC = 100\ \mu\mathrm{F}, find (a) the time constant, and (b) the time taken for the capacitor to reach 63%63\% of its final charge. (4 marks)

Q10. Write down the value of the speed of light in terms of ε0\varepsilon_0 and μ0\mu_0, and verify numerically that c≈3×108 m/sc \approx 3\times10^{8}\ \mathrm{m/s}. (4 marks)

Answer keyMark scheme & solutions

Q1. (3 marks)

  • Quantization: charge exists in integer multiples of the elementary charge, q=neq = ne, e=1.6×10−19 Ce = 1.6\times10^{-19}\ \mathrm{C}. (1)
  • Conservation: total charge of an isolated system remains constant; charge is neither created nor destroyed. (1)
  • Additivity (third property): total charge is the algebraic sum of individual charges; charge is a scalar. (Also acceptable: charge is invariant / two kinds exist.) (1)

Q2. (4 marks) Coulomb's law: F=k∣q1q2∣r2F = k\dfrac{|q_1 q_2|}{r^2}. (1)

F=9.0×109×(3×10−6)(5×10−6)(0.20)2F = 9.0\times10^{9}\times\frac{(3\times10^{-6})(5\times10^{-6})}{(0.20)^2} (1)

F=9.0×109×15×10−120.04=9.0×109×3.75×10−10=3.375 NF = 9.0\times10^{9}\times\frac{15\times10^{-12}}{0.04} = 9.0\times10^{9}\times3.75\times10^{-10} = 3.375\ \mathrm{N} (1)

F≈3.4 NF \approx 3.4\ \mathrm{N}, attractive (opposite signs). (1)

Q3. (5 marks)

  • Choose a coaxial cylindrical Gaussian surface of radius rr, length LL. (1)
  • By symmetry E⃗\vec E is radial and constant on the curved surface; flux through end caps is zero. (1)
  • Flux: ∮E⃗⋅dA⃗=E(2πrL)\oint \vec E\cdot d\vec A = E(2\pi r L). (1)
  • Enclosed charge qenc=λLq_{enc} = \lambda L, so Gauss's law gives E(2πrL)=λLε0E(2\pi r L) = \dfrac{\lambda L}{\varepsilon_0}. (1)
  • E=λ2πε0r\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}} (directed radially outward for λ>0\lambda>0). (1)

Q4. (5 marks) (a) Series: 1Ceq=12+13+16=3+2+16=1 (μF)−1\dfrac{1}{C_{eq}} = \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} = \dfrac{3+2+1}{6} = 1\ (\mu\mathrm{F})^{-1}. (2) Ceq=1 μF.C_{eq} = 1\ \mu\mathrm{F}. (1)

(b) U=12CeqV2=12(1×10−6)(12)2U = \tfrac12 C_{eq}V^2 = \tfrac12 (1\times10^{-6})(12)^2. (1) U=12(1×10−6)(144)=7.2×10−5 J=72 μJ.U = \tfrac12(1\times10^{-6})(144) = 7.2\times10^{-5}\ \mathrm{J} = 72\ \mu\mathrm{J}. (1)

Q5. (4 marks)

  • Definition: electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point (against the field). (1)
  • Point charge: V=14πε0qr=kqrV = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r} = k\dfrac{q}{r}. (1)
  • V=9.0×109×2×10−90.30V = 9.0\times10^{9}\times\dfrac{2\times10^{-9}}{0.30}. (1)
  • V=9.0×109×6.667×10−9=60 VV = 9.0\times10^{9}\times6.667\times10^{-9} = 60\ \mathrm{V}. (1)

Q6. (4 marks) Drift velocity: I=nAevd⇒vd=InAeI = nAev_d \Rightarrow v_d = \dfrac{I}{nAe}. (1)

vd=2.0(8.5×1028)(1.0×10−6)(1.6×10−19)v_d = \frac{2.0}{(8.5\times10^{28})(1.0\times10^{-6})(1.6\times10^{-19})} (1)

Denominator =8.5×1028×1.6×10−25=1.36×104= 8.5\times10^{28}\times1.6\times10^{-25} = 1.36\times10^{4}. (1)

vd=2.01.36×104≈1.47×10−4 m/s.v_d = \frac{2.0}{1.36\times10^{4}} \approx 1.47\times10^{-4}\ \mathrm{m/s}. (1)

Q7. (4 marks)

  • Faraday's law: the induced EMF equals the negative rate of change of magnetic flux, E=−dΦdt\mathcal{E} = -\dfrac{d\Phi}{dt}. (1)
  • Lenz's law: the induced current flows in a direction that opposes the change in flux producing it. (1)
  • Energy conservation: the induced current opposes the change, so work must be done (e.g. against the magnetic force) to maintain the flux change; this mechanical/external work is converted into electrical energy. (1)
  • If the current instead aided the change, energy would be created from nothing — violating conservation. (1)

Q8. (3 marks) Inside a solenoid: B=μ0nIB = \mu_0 n I with n=N/L=2000/0.50=4000 turns/mn = N/L = 2000/0.50 = 4000\ \mathrm{turns/m}. (1) B=(4π×10−7)(4000)(3.0)B = (4\pi\times10^{-7})(4000)(3.0) (1) B=4π×10−7×1.2×104=1.508×10−2 T≈1.5×10−2 T.B = 4\pi\times10^{-7}\times1.2\times10^{4} = 1.508\times10^{-2}\ \mathrm{T} \approx 1.5\times10^{-2}\ \mathrm{T}. (1)

Q9. (4 marks) (a) τ=RC=(10×103)(100×10−6)=1.0 s\tau = RC = (10\times10^{3})(100\times10^{-6}) = 1.0\ \mathrm{s}. (2)

(b) Charge grows as q=q0(1−e−t/τ)q = q_0(1-e^{-t/\tau}). At 63%63\%, 1−e−t/τ=0.63⇒e−t/τ≈0.37⇒t=τ1-e^{-t/\tau}=0.63 \Rightarrow e^{-t/\tau}\approx0.37 \Rightarrow t=\tau. (1) t=1.0 s.t = 1.0\ \mathrm{s}. (1)

Q10. (4 marks)

  • c=1ε0μ0c = \dfrac{1}{\sqrt{\varepsilon_0\mu_0}}. (1)
  • ε0μ0=(8.85×10−12)(4π×10−7)=1.112×10−17 s2/m2\varepsilon_0\mu_0 = (8.85\times10^{-12})(4\pi\times10^{-7}) = 1.112\times10^{-17}\ \mathrm{s^2/m^2}. (1)
  • 1.112×10−17=3.335×10−9\sqrt{1.112\times10^{-17}} = 3.335\times10^{-9}. (1)
  • c=1/(3.335×10−9)=3.0×108 m/sc = 1/(3.335\times10^{-9}) = 3.0\times10^{8}\ \mathrm{m/s}. ✓ (1)
[
  {"claim":"Q2 Coulomb force is 3.375 N", "code":"k=9.0e9; q1=3e-6; q2=5e-6; r=0.20; F=k*q1*q2/r**2; result=abs(F-3.375)<1e-3"},
  {"claim":"Q4 equivalent series capacitance is 1 uF and energy 72 uJ", "code":"Ceq=1/(1/2+1/3+1/6); U=0.5*(1e-6)*12**2; result=abs(Ceq-1)<1e-9 and abs(U-72e-6)<1e-9"},
  {"claim":"Q6 drift velocity approx 1.47e-4 m/s", "code":"I=2.0; n=8.5e28; A=1.0e-6; e=1.6e-19; vd=I/(n*A*e); result=abs(vd-1.47e-4)<2e-6"},
  {"claim":"Q10 c from eps0 mu0 is about 3e8 m/s", "code":"eps0=8.85e-12; mu0=4*pi*1e-7; c=1/sqrt(eps0*mu0); result=abs(float(c)-3.0e8)<2e6"}
]