THE UNIVERSITY OF WARWICK
FIRST YEAR EXAMINATION: Summer 2019 ELECTRICITY AND MAGNETISM
Answer 4 questions Time Allowed: 2 hours
Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
The nominal mark assigned to each part of a question is indicated by means of a bold figure enclosed by curly brackets, e.g. {2} , immediately following that part. This mark scheme is for guidance only and may be adjusted by the examiner.
Calculators may be used for this examination
The following information may be used:
Constants:
Permittivity of free space ε0 = 8.85 × 10−12 C2N−1m−2
Electricity and Magnetism Time allowed: 2 hours
ANSWER ALL QUESTIONS
a): book work, basic question to set the scene, b): unseen, but an AC circuit type question is asked every year, c): unseen, but a typical LRC circuit type question that is asked every year
1. a) {4}
(i) Junction rule: at a junction where several wires meet the total current is zero so
that no charges build up at the junction. {2}
(ii) Loop rule: Over a circuit loop that is closed the sum of voltages from sources of emf must equal the voltage drop across all other components. {2}
b) (i) {5}
i. There are two loops in the circuit and we associate a Maxwell loop to each. We choose the left loop to have a Maxwell current I1flowing clock-wise and
the right loop to have a Maxwell current flowing clock-wise as well. {1} ii. For the two loops, the Loop Rule for Maxwell loops gives:
ϵ1
R = 3I1− I2, ϵ2
R = −I1+ 3I2.
{2} iii. The system of algebraic equations is solved for the currents and we find:
I1 =
3ϵ1+ ϵ2
8R , I2 =
ϵ1+ 3ϵ2 8R . The currents across the resistors are:
IR1 = I1, IR2 = I2, IR3 = I1− I2 =
ϵ1− ϵ2 4R
(ii) {3} The potential drop across each resistor is:
V1 = 2R|I1| = 3ϵ1+ ϵ2 4 , V2 = 2R|I2| = ϵ1+ 3ϵ2 4 , V3 = R|I2− I1| = |ϵ2− ϵ1| 4 . (iii) {3}
The power dissipated in resistor R3is:
P = V3|I2− I1| = V32 R = R3(I2− I1) 2 = (ϵ2− ϵ1)2 16R . c) (i) {3}
The impendances of each element are ˜ZR= R, ˜ZC= 1/ jωCand ˜ZL= jωL. {1} The total impedance of three impendances in parallel:
˜ Ztot = !Z˜−1 R + ˜ZC−1+ ˜ZL−1 "−1 , {1} and we find ˜ Ztot = # 1 R+ j $ ωC− 1 ωL %&−1 . {1} (ii) {3}
We use Ohm’s Law to find the total current: ˜I= ˜Vin/ ˜Ztot {1}
Using the result above, the total current is: ˜ I = ˜Vin # 1 R+ j $ ωC− 1 ωL %& . {1} The amplitude of the current is:
| ˜I| = | ˜Vin| ' 1 R2+ $ ωC− 1 ωL %2 .
{1}
(iii) {4}
In the limitsω→ 0 andω → ∞, ˜I → ∞ {1}
For the value ofω= 1/√LC, the current reached a minimum at| ˜I| = | ˜Vin|/R. {1}
{2} All{4} may be rewarded if figure is correct and complete.
a) Basic question to set the scene and test understanding of vectors in the context of charges, b) Unseen exam question but that is a simple modification of a basic question covered in the course c) typical problem covered in the problem sheets, d) a bookwork question but more challenging
2. a) {5}
The Coulomb force of charge 2 acting on charge 1, in vector form, is: {2}
F21 = q1q2(r1− r2) 4πε0|r1− r2|3 = q1q2 4πε0|r1− r2|2 ˆr21 = q1q2 4πε0r2 ˆr21 where ˆr21 = r1− r2 |r1− r2|, r = |r 1− r2| (i) Drawing ˆr21 {1} (ii) Drawing F21 {1}
(iii) Drawing position vectors {1}
b) {8}
We define the position vectors of the charges:
⃗r1′ = d ˆy , ⃗r2′ = ⃗0 , ⃗r′3 = −d ˆy .
{2} We define the position vector of the test charge as⃗r0= a ˆx. {1}
The Coulomb force acting on the test charge is: ⃗F = qq0 4πε0 # (⃗r0−⃗r′1) |⃗r0−⃗r′1|3 + (⃗r0−⃗r ′ 2) |⃗r0−⃗r′2|3 + (⃗r0−⃗r ′ 3) |⃗r0−⃗r3′|3 & . {2} We fill in the position vectors:
⃗F = qq0 4πε0 # (a ˆx − d ˆy) |a ˆx − d ˆy|3+ axˆ |a ˆx|3+ (a ˆx + d ˆy) |a ˆx + d ˆy|3 & , = qq0 4πε0 # (a ˆx − d ˆy) (a2+ d2)3/2 + ˆ x a2 (a ˆx + d ˆy) (a2+ d2)3/2 & , = qq0 4πε0 # 2a (a2+ d2)3/2 + 1 a2 & ˆ x. {3} c) {6}
We make use of the fact in a conservative system work done equals the potential energy of the system, hence the work to bring a charge q2up to a distance a from a
charge q1is: {1}
W12 =
q1q2
4πε0a
(i) Bringing the first charge from infinity to the position (x,y) = (0,0) does not re-quire any work as there are no forces present acting on the charge, i.e.e W1=
0. {1}
(ii) Bringing the second charge from infinity to position (0,a), it has to overcome the potential energy difference from the interaction with the first charge. Hence the
work is: {1}
W2 = W12 =
q2 4πε0a
(iii) Bringing the third charge from infinity to position (a,0), it has to overcome the potential energy difference from the interaction with the first and second charges.
Hence the work is: {1}
W3 = W13+W23 = q2 4πε0a + q 2 4πε0 √ 2a
(iv) Bringing the fourth charge from infinity to position (a,a), it has to overcome the potential energy difference from the interaction with the first, second and third
charges. Hence the work is: {1}
W4 = W14+W24+W34 = q2 4πε0 √ 2a+ q2 4πε0a + q 2 4πε0a
(v) The total work of the system of charges is: {1} W = W12+W13+W23+W14+W24+W34 = q 2 4πε0a $ 1+ 1 +√1 2+ 1 √ 2+ 1 + 1 % = q 2 4πε0a ( 4+√2 ) d) {6}
The electrostatic field is found as the negative gradient of the electrostatic potential,
i.e. {1}
E = −∇V
We use the product rule of differentiation to write: {1} E = −∇V = − 1 4πε0 # (p.r) ∇$ 1 r3 % +∇(p.r) r3 &
We calculate each term separately.
(i) {2} ∇$ 1 r3 % = −3 r4∇r = −3 r4∇ * x2+ y2+ z2 = − 3 2r5 (2x∇x + 2y∇y + 2z∇z) = − 3 r5r
where we have used that ∇x= ˆx.
(ii) {1}
∇(p.r) = ∇ (pxx+ pyy+ pzz)
= px∇x+ py∇y+ pz∇z
= pxˆx+ pyˆy+ pzˆz = p
Combining the two terms we find the result: {1}
E = − 1 4πε0 # −3(p.r) r r5 + p r3 & = 1 4πε0r3 [3(p.ˆr) ˆr − p]
a) Basic question to set the scene, b) Problem covered in the lectures, except for part iii) which is an extension and test a student’s understanding of potentials.
3. a) {6}
The electric flux through a surface enclosing charges is linearly proportional to the
total charge enclosed by the surface: {1}
‹
S
E.dS = Qencl. ε0
(i) Left-hand side is electric flux through a surface enclosing Qencl.. {1} (ii) dS points outwards from normal to surface and it’s magnitude corresponds to
surface area {1}
(iii) E: electric field vector {1}
(iv) Qencl.: total charge enclosed by volume whose surface is S {1}
(v) ε0: permittivity of free space {1}
b) (i) {8}
The symmetry arguments are:
i. E is normal to the sheet. {1}
ii. E points away from the sheet ifσ > 0 {1}
iii. E is constant at fixed z {1}
We choose a Gaussian surface as a cylinder with area A at top and bottom. The
enclosed charge in the cylinder isσA. {2}
We evaluate the electric flux through three surfaces of the cylinder: {2} ‹ E.dA = ¨ top E.dA + ¨ bottom E.dA + ¨ side E.dA Thus, we find: ‹ E.dA = ¨ top E ˆz.ˆzdA + ¨ bottom (−E)ˆz.(−ˆz)dA + ¨ side E ˆz.ˆrdA = 2EA .
Equating left and right-hand sides of Gauss’s Law, we find: {1} E2A = σA
ε0 ⇒ E =
σ 2ε0
or in vector form: E = + σ 2ε0 ˆz z> 0 , −2εσ0 ˆz z< 0 , . (ii) {6}
The potential is found from {2}
Va−Vb = b
ˆ
a
E.dl ,
For points above the sheet (z> 0), with E = E ˆz and dl = ˆzdz
V(z) −Vb = b ˆ z Edz = b ˆ z σ 2ε0 dz= σ 2ε0(b − z) . {1} We choose the potential to be zero at a finite distance b. Then, {1}
V(z) = b ˆ z Edz = σ 2ε0(b − z) .
For points below the sheet (z< 0), we have to take into account that E = −E ˆz below the sheet. Hence,
V(z) = − 0 ˆ z Edz+ b ˆ 0 Edz = σ 2ε0 (b + z) . {2} (iii) {5}
The potential below the sheet (region 1), and above until the conductor (region
2) remains unchanged. {1}
The conducting slab is an equipotential volume, i.e. the value of the potential remains constant throughout, equal toσ(b − a)/2ε0. {1}
The potential should be a continuous function of z, also at z= a + d. {1}
V(z) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ σ 2ε0(b + z) z< 0 , region 1 σ 2ε0(b − z) 0≤ z < a , region2 σ 2ε0(b − a) a≤ z ≤ a + d , region3 σ 2ε0(b + d − z) z> a + d , region 4 .
a) Basic question to set the scene, b) and c) Problems covered in the lectures, but never in-cluded in an exam before, They are good test for the student’s understanding of inductors without challenging mathematics, d) Challenging problem, shown in the lecture notes but not necessarily discussed.
4. a) {6}
Faraday’s Law says that the induced emf in a closed loop equals minus the rate of change of magnetic flux through the loop:
ϵ = −dφB
dt , {2}
(i) φB=˜SB.dS - magnetic flux. {2}
(ii) ϵ - emf. {1}
(iii) The left-hand rule determines the emf direction (or minus the right-hand rule) {1}
b) {6}
For a circuit with two inductors in parallel, the potential drop across the two inductors
is the same. {1}
We apply the expression for the self-induced emf for an inductor, i.e. V = −LdI
dt ,
{1} and find the rate of change of the current across both inductors:
dI1 dt = − V L1 , dI2 dt = − V L2 . {1} The rate of change of the total current across both inductors is:
dI dt = dI1 dt + dI2 dt = − $ 1 L1 + 1 L2 % V . {1} A circuit with one inductor with a self-inductance Leq. is equivalent to the previous
circuit only ifL
dI dt = − V Leq. = − $ 1 L1 + 1 L2 % V , {1}
which leads to the result 1 Leq. = 1 L1 + 1 L2 . {1} c) (i) {4}
Faraday’s Law applied to each inductor is: ϵP = −NP dΦB dt , ϵS = −NS dΦB dt {2} Both inductors have the same value of magnetic flux pass through them. {1} We use this to eliminate the magnetic flux from the expressions above, leading to: ϵS = NS NP ϵP. {1} (ii) {4}
Energy conservation implies that the power in each circuit must always match. {1} PP = PS ⇒ ϵPIP = ϵSIS,
{1} This leads to the expression
IS =
ϵP ϵS
IP.
{1} Using the previous result forεS, we find
IS =
NP
NS
IP.
d) {5}
The general form of the motional emf is: {1}
dε = (v × B).dl .
Integrating both sides along the length of the rod leads to: {1}
ε =
L
ˆ
0
(v × B).dl ,
The velocity vector is in the azimuthal direction, i.e. v=v ˆθ. The magnetic field vector is into the page, i.e. B=-Bˆz. Also dl=ˆrdr. Hence, {1}
ε = − L ˆ 0 vBdr, or with v=ωr, ε = L ˆ 0 ωrBdr = −ωB L ˆ 0 rdr = −1 2ωB r 20 0 L 0 , {1} ε = −1 2ωBL 2. {1} An alternative answer (though not what was explicitly asked for) is starting from
Faraday’s law: {1}
ε = −dΦ dt .
where the magnetic flux dΦ=BdA with dA the area swept out by the rod in time dt.
Then, {1}
ε = −BdA dt
The area dA in terms of angle dθ {1}
dA = 1 2L
Then the emf is:
ε = −1 2BL
2dθ
dt
The rate of change of angle is the angular velocity: {1} dθ
dt = ω
Thus, we find the result: {1}
ε = −1 2ωBL
