Electromagnetism

Question 1

(a) Integrate charge density over shells $dq = \rho(r)\,4\pi r^2\,dr$: $Q = \int_0^R \rho_0\left(1-\frac{r}{R}\right)4\pi r^2\,dr = 4\pi\rho_0\left[\frac{R^3}{3} - \frac{R^3}{4}\right] = 4\pi\rho_0\cdot\frac{R^3}{12} = \frac{\pi\rho_0 R^3}{3}.$ (setup 1, integration 1, result 1) → $Q = \dfrac{\pi \rho_0 R^3}{3}$.

(b) Enclosed charge within $r$: $Q_{enc}(r) = 4\pi\rho_0\int_0^r\left(r'^2 - \frac{r'^3}{R}\right)dr' = 4\pi\rho_0\left[\frac{r^3}{3} - \frac{r^4}{4R}\right].$ (1) Gauss: $E\cdot 4\pi r^2 = Q_{enc}/\varepsilon_0$ (1), so $E(r) = \frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3} - \frac{r^2}{4R}\right].$ (algebra 1, final 1)

(c) At $r=R$: $E(R) = \dfrac{\rho_0}{\varepsilon_0}\left[\dfrac{R}{3}-\dfrac{R}{4}\right] = \dfrac{\rho_0 R}{12\varepsilon_0}$. (1) Point charge field at $R$: $\dfrac{Q}{4\pi\varepsilon_0 R^2} = \dfrac{\pi\rho_0 R^3/3}{4\pi\varepsilon_0 R^2} = \dfrac{\rho_0 R}{12\varepsilon_0}$. (1) Equal — as expected since all enclosed charge acts as if central. (1)

Question 2

(a) Charge conserved: $Q = C_1 V_0 = 4\times100 = 400\,\mu\mathrm{C}$. (1) Parallel combined $C = C_1+C_2 = 10\,\mu\mathrm{F}$. (1) $V = Q/C = 400/10 = 40\,\mathrm{V}$. (1)

(b) Before: $U_i = \tfrac12 C_1 V_0^2 = \tfrac12(4\times10^{-6})(100)^2 = 0.020\,\mathrm{J}$. (2) After: $U_f = \tfrac12 C V^2 = \tfrac12(10\times10^{-6})(40)^2 = 0.0080\,\mathrm{J}$. (2)

(c) Energy lost $= 0.012\,\mathrm{J}$; dissipated as heat (resistance of connecting wires) and/or EM radiation during the transient charge redistribution. (2) Charge IS conserved ($400\,\mu\mathrm{C}$ before and after); only energy is not conserved in the ideal-wire limit. (1)

Question 3

(a) Only the leading side of length $a$ cuts flux; $\mathrm{EMF} = Bav = 0.40\times0.20\times3.0 = 0.24\,\mathrm{V}$. (3)

(b) $I = \mathrm{EMF}/R = 0.24/0.50 = 0.48\,\mathrm{A}$. (2) Flux into page is decreasing → induced current opposes decrease → maintains into-page flux → clockwise. (1)

(c) Force on current-carrying side in field: $F = BIa = 0.40\times0.48\times0.20 = 0.0384\,\mathrm{N}$. External force must balance this (opposing motion), so $F_{ext}=0.038\,\mathrm{N}$. (2) Check: $P = I^2R = (0.48)^2(0.50) = 0.1152\,\mathrm{W}$; $P/v = 0.1152/3.0 = 0.0384\,\mathrm{N} = F$. ✓ (2)

Question 4

(a) $I(2) = 4.0\times2.0 = 8.0\,\mathrm{A}$. $B = \mu_0 n I = (4\pi\times10^{-7})(5000)(8.0) = 5.03\times10^{-2}\,\mathrm{T}$. (3)

(b) Faraday (loop of radius $r$ inside): $\oint \vec E\cdot d\vec l = -\dfrac{d\Phi}{dt}$. $B(t)=\mu_0 n I(t)$, $\dfrac{dB}{dt}=\mu_0 n\,\dfrac{dI}{dt} = (4\pi\times10^{-7})(5000)(4.0) = 2.513\times10^{-2}\,\mathrm{T/s}$. (1) For $r<r_s$: $E(2\pi r) = \pi r^2\,\dfrac{dB}{dt}$ → $E = \dfrac{r}{2}\dfrac{dB}{dt}$. (1) $E = \dfrac{0.010}{2}(2.513\times10^{-2}) = 1.26\times10^{-4}\,\mathrm{V/m}$. (2)

(c) For $r>r_s$, enclosed flux uses area $\pi r_s^2$: $E(2\pi r) = \pi r_s^2\dfrac{dB}{dt}$ → $E = \dfrac{r_s^2}{2r}\dfrac{dB}{dt}$. (1) $E = \dfrac{(0.02)^2}{2(0.04)}(2.513\times10^{-2}) = 1.26\times10^{-4}\,\mathrm{V/m}$. (2) (Numerically equal here because $r_s^2/r = 0.01 = r$ at $r=0.01$; coincidence of chosen values.)

(d) Faraday's law (Maxwell–Faraday): $\oint \vec E\cdot d\vec l = -\frac{d}{dt}\int \vec B\cdot d\vec A.$ (equation 1, identification 1)

Question 5

(a) $B_0 = E_0/c = 60/(3\times10^8) = 2.0\times10^{-7}\,\mathrm{T}$. (2)

(b) $S_{avg} = \dfrac{E_0^2}{2\mu_0 c} = \dfrac{(60)^2}{2(4\pi\times10^{-7})(3\times10^8)} = \dfrac{3600}{0.7540} = 4.77\,\mathrm{W/m^2}$. (3)

(c) Power $= S_{avg}\times A = 4.77\times2.0 = 9.55\,\mathrm{W}$. (2) Absorbing surface: force $= P/c = 9.55/(3\times10^8) = 3.18\times10^{-8}\,\mathrm{N}$. (1)

[
{"claim":"Q1a total charge Q = pi*rho0*R^3/3","code":"r,R,rho0=symbols('r R rho0',positive=True); Q=integrate(rho0*(1-r/R)*4*pi*r**2,(r,0,R)); result = simplify(Q - pi*rho0*R**3/3)==0"},
{"claim":"Q2a common voltage 40V and energy 0.020->0.008 J","code":"C1,V0,C2=4e-6,100,6e-6; Q=C1*V0; V=Q/(C1+C2); Ui=0.5*C1*V0**2; Uf=0.5*(C1+C2)*V**2; result = abs(V-40)<1e-9 and abs(Ui-0.02)<1e-9 and abs(Uf-0.008)<1e-9"},
{"claim":"Q3 EMF 0.24V, I 0.48A, F 0.0384N equals P/v","code":"B,a,v,R=0.40,0.20,3.0,0.50; emf=B*a*v; I=emf/R; F=B*I*a; P=I**2*R; result = abs(emf-0.24)<1e-9 and abs(I-0.48)<1e-9 and abs(F-P/v)<1e-12"},
{"claim":"Q5 B0=2e-7T, Savg approx 4.77, force approx 3.18e-8","code":"E0,c,mu0=60,3e8,4*pi*1e-7; B0=E0/c; S=E0**2/(2*mu0*c); Pw=S*2.0; Fr=Pw/c; result = abs(B0-2e-7)<1e-12 and abs(float(S)-4.77)<0.05 and abs(float(Fr)-3.18e-8)<1e-9"}
]