Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.

THE UNIVERSITY OF WARWICK

FIRST YEAR EXAMINATION: Summer 2017 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 Permeability of free space µ₀ = 4π × 10⁻⁷ NA⁻²

Electricity and Magnetism Time allowed: 2 hours

ANSWER ALL QUESTIONS 1. a) {10}

Unseen question on a DC circuit is a stock question that student know they will have to tackle. This example builds upon the simpler examples given in the course and exercises, but it nonetheless straightforward to solve when applying Maxwell loops correctly. Students may run into difficulty due to over-complexity when using basic Kirchhoff rules instead.

We name the three Maxwell loops as written below and to have their currents flowing

in a counter-clockwise direction. {2}

(i) Loop 1 (overε₁): ε/R = 2I1− I2− I3 {1}

(ii) Loop 2 (overε₂): ε/R = −I₁+ 2I₂− I₃ {1}

(iii) Loop 3: 0= −I1− I2+ 4I3 {1}

Subtract (ii) from (i): 3I₁− 3I2 = 0 =⇒ I1 = I2. {1}

Solve (iii) for I₃: 0 = −2I1+ 4I3 =⇒ I3 = I1/2. {1}

Use (i) to find I₁: ε/R = 2I1− I1− I1/2 = I1/2 =⇒ I1 = 2ε/R.

And also I₂ = 2ε/R, I3 = ε/R. {1}

The voltage drops are:

∆V₁= |I1−I2|R = 0 , ∆V2= |I2−I3|R =ε, ∆V3 = |I1−I3|R =ε , ∆V4= |I3|2R = 2ε. {2}

b) {6}

This question has been covered in the lectures, but it has not been used in an exam before. This will test the student’s basic understanding of time-varying circuits and impedances.

Consider a circuit with an AC source with voltage V(t) = V0cos(ωt) and a capacitor with capacitance C. The charge on the capacitor is Q= CV . {1}

The current follows as I= dQ/dt. {1}

Hence,

dQ

dt = I = CdV

dt = −CωV₀sin(ωt) . {1}

Since cos(ωt+π/2) = − sin(ωt), we see that the current is ahead of the voltage in

phase byπ/2. {1}

Using complex notation with ˜V(t) = V0e^jωt, we have:

d ˜Q

dt = ˜I = CdV

dt = jωC ˜V . {1}

Applying Ohm’s Law, we find the complex impedance to be:

˜

c) This particular LRC circuit has not been covered in the lectures. However, its analy- sis follows the same procedure as taught.

(i) {3}

The resistor and inductor in series: Z^∗ = R + jωL. {1}

The equivalent impedance for the circuit:

Z˜_eq = 1

1

Z^∗+ jωC = R+ jωL

(1 −ω²LC) + jωRC . {2}

(ii) {2}

The current follows from Ohm’s Law ˜V = ˜ZI:˜ {1}

I˜= V˜_in

Z˜^∗ = V˜_in

R+ jωL {1}

(iii) {4}

We multiply denominator and nominator with R− jωL: {1}

I˜ = V˜_in Z˜_eq

= R+ j(ωR²C−ωL(1 −ω²LC))
V˜_in R²+ω²L²

= R + jωL²C ω²−_LC¹ + ^R_L
2

)] ˜V_in R²+ω²L² , {2}

The imaginary part of ˜Iis zero for the resonance frequency:

ω = s

1

LC− R L

2

. {1}

2. a) {5}

This basic question sets up the rest of Question 2 and is asking the student to write down correctly the basic equation extensively seen in the lectures.

The electric flux through a surface enclosing charges is linearly proportional to the

total charge enclosed by the surface: {1}

‹

S

E.dS = Q_encl.

ε₀

(i) Left-hand side us electric flux through a surface enclosing Q_encl.. {1}

(ii) E: electric field vector {1}

(iii) Q_encl.: total charge enclosed by volume whose surface is S {1}

(iv) ε₀: permittivity of free space {1}

(v) dS points outwards from normal to surface and it’s magnitude corresponds to

surface area (alternate mark) {1}

b) {7}

A basic question that, although not directly covered in the lectures, is a natural ex- tension that had been covered for other geometries in the lectures. It is a good test of the basic understanding of applying Gauss’s Law and within the abilities of all students. Last used in 2012-2013 exam.

(i) Symmetry arguments: Electric field is expected to be in the (cylindrical) radial direction and constant at constant radius from the axis. Hence, E= E(r) ˆr. {1}

(ii) Take a Gaussian surface to be a cylinder of radius r aligned with the shell. The

Gaussian cylinder had length L. {1}

(iii) We calculate the electric flux to be:

Φ_E =

¨

curv

E.dS +

¨

left

E.dS +

¨

right

E.dS .

Using dS_curv= ˆrdScurv, dS_left= −ˆzdSleftdS_right= ˆzdSright, together with symme- try arguments, we find

Φ_E = E(r) 2πr L. {2}

(iv) The enclosed charge is given by Q_encl=λLif r≥ R and Q_encl= 0 if r < R. {1}

(v) Applying Gauss’s Law: E(r) 2πr = λ/ε₀if r≥ R, zero otherwise. {1}

(vi) Summarising:

E =

( _λ

2πε0rˆr r≥ R

0 r< R , {1}

c) {7}

and tests the student’s ability to apply the calculation of electrostatic potential. Last used in 2012-2013 exam.

To calculate the potential difference:

V_a−Vb =

b

ˆ

a

E.dl , {1}

where we choose to integrate along a radial path from rato r_b: V_a−Vb =

ˆ _rb

ra

Edr, {1}

since E= E(r)ˆr and ˆr.dl = dr. {1}

We choose the reference potential to be at r_b = r0, i.e. V_b= 0. Also, ra= r and

Va= V (r). {1}

The potential difference for r≥ R:

V(r) = ˆ r0

r

E(r^′) dr^′ = ˆ r0

r

λ

2πε₀r^′dr^′ = λ 2πε₀ ^ln

r₀ r

. {2}

It is clear that r₀has to be a finite, non-zero radius in order for the logarithm to remain

finite. (alternate mark) {1}

For r< R:

V(r) = ˆ r0

r

E(r^′) dr^′ = ˆ r0

R

λ

2πε₀r^′dr^′ = λ 2πε₀ ^ln

r₀ R

. {1}

d) This is a more challenging unseen question that builds on the previous results and will differentiate the students. Last used in 2012-2013 exam.

(i) {2}

The same number of charges, but opposite in sign, accumulate on the inside of

the conductor, i.e. at r= a. {1}

This leaves the same number of charges, but positive, on the outer edge of the conductor, i.e. at r= b. Hence,λ_a= −λ andλ_b=λ. {1}

(ii) {4}

i. r< a:

V(r) = λ 2πε₀ ^ln

ar₀ br



or V(r) = λ 2πε₀ ^ln

r₀ r

, {2}

ii. a≥ r ≥ b:

V(r) = λ 2πε₀ ^ln

r₀ b



or V(r) = λ 2πε₀ ^ln

r₀ a

 , {1}

iii. r> b:

V(r) = λ 2πε₀ ^ln

r₀ r



or V(r) = λ 2πε₀ ^ln

 br₀ ar



, {1}

3. a) {3}

A basic question asking to write down the Lorentz force.

F = q E + q v × B , {3}

b) {8}

This has been covered in the lectures and is a natural first application of the Lorentz force.

The kinetic energy of a charged particle is mv²/2. {1}

The rate of change of the kinetic energy is:

d dt

 1 2mv²



= 1 2m d

dt(v.v) = mdv

dt.v = q(v × B).v = 0 . {2}

The motion is split into parallel and perpendicular motion w.r.t the magnetic field direction. i.e. v= v_k+ v_⊥ with v_k× B = 0 and v_⊥.B = 0. {2}

The charge’s equation of motion becomes mdv_k

dt = 0 , mdv_⊥

dt = qv⊥× B ,

which shows that there is no acceleration along the magnetic field, i.e. constant

parallel motion. {1}

Also, we see that the force is both perpendicular to the velocity and the magnetic field vectors. The orbit is circular with a balance between the force and the centripetal force:

|q|v_⊥B = mv²_⊥

R =⇒ R = mv_⊥

|q|B.

where R is the radius of the circle. {1}

The combination of circular motion perpendicular to the magnetic field and constant

motion along it is a helical orbit. {1}

c) Unseen question, last used in 2007-2008 exam. Part (a) has been covered in the lectures but the application to the Hall effect in questions (b) and (c) is unseen. There is not much mathematical manipulation necessary and tests the student’s physical understanding of the lecture material.

(i) {4}

In time∆t the free charges will have travelled a distance v_d∆t. {1}

The volume covering that distance and the cross-section: V = v_d∆td². {1}

The total number of free charges in the volume: ∆Q= |−e|nV = ned²v_d∆t. {1}

The current is the rate of change of charge with respect to time: I = ∆Q/∆t =

ned²v_d. {1}

(ii) {8} Let the conductor be oriented in the z-direction and the magnetic field in the x-direction. For a charge moving at speed v_d in the z-direction will feel the magnetic Lorentz force:

The charges move in the negative y-direction and accumulate at the right side of the conductor. This sets up a potential difference between the left and right

sides. {1}

The resulting electric field wants to return the charges. Hence, there is an equi- librium between electric and magnetic forces:

E+ v × B = 0 =⇒ E = −v_dB ˆy, {1}

With the potential difference between the two plates at distance d, V = Ed, and

the expression for the current, we have {1}

V = vdB d = I B

ned . {1}

{2}

(iii) {2}

Negative conductive charges are deflected to the negative y-direction, setting up

an electric field E∼ −ˆy. {1}

Positive conductive charges are deflected to the positive y-direction, setting up an electric field E∼ ˆy. Hence by measuring the direction of the electric field the nature of the conductive charges and the type of semiconductor is revealed. {1}

4. a) {6}

A standard textbook question to set up the rest of the question.

Amp`ere’s Law is given as:

˛

C

B.dl = µ₀I_encl, {2}

(i) dl - vector element of the closed path of integration. {1}

(ii) I_encl - current enclosed by the closed path. {1}

(iii) The right-hand rule where thumb is in the direction of the current and the fingers

indicate the direction of the magnetic field. {2}

b) {2}

Again a standard textbook question to set up the rest of the question.

The total outward magnetic flux through a closed surface is zero as the same amount of magnetic field that enters the volume also leaves it, magnetic field lines have no

start of finish, i.e. {1}

‹

S

B.dS = 0 . {1}

c) This question is covered in the lectures, but not seen in recent exams. It was last used in 2007-2008 exam. It is a good test of a student’s understanding of Amp`ere’s Law and the Solenoidal Condition.

(i) {4}

We choose a path C as a circle concentric with axis of solenoid and radius r.

{1}

We apply Amp`ere’s Law to path C for which dl= dl ˆϕ: {1}

˛

C

B.dl =

˛

C

B_ϕdl = Bϕ 2π

ˆ

0

rdϕ = 2πrB_ϕ = µ₀I_encl. {1}

Because the current flowing through the cross-section of C is zero, I_encl = 0.

Hence, B_ϕ = 0. {1}

(ii) {4}

We apply the Solenoidal Condition over a cylindrical surface concentric with the solenoid axis. We evaluate:

‹

S

B.dS =

¨

left

B. (dS (−ˆz)) +

¨

right

B. (dS ˆz) +

¨

side

B. (dS ˆr) . {1}

Because the magnetic field entering the left-hand side leaves the right-hand side,

Hence,

‹

S

B.dS =

¨

side

B. (dS ˆr) = Br

¨

side

dS = Br2πrL, {1}

Since the total flux is zero, we require B_r= 0. {1}

(iii) {9}

To investigate the magnetic field outside the solenoid we apply Amp`ere’s Law to a rectangular path in the plane of constantϕoutside of the solenoid: {1}

˛

C

B.dl =

b

ˆ

a

B.dl +

c

ˆ

b

B.dl +

d

ˆ

c

B.dl +

a

ˆ

d

B.dl .

=

L

ˆ

0

B_z(r1)dz +

r2

ˆ

r1

B_rdr+

0

ˆ

L

B_z(r2)dz +

r1

ˆ

r2

B_rdr.

{2}

Since B_r= 0, we have

˛

C

B.dl = (Bz(r1) − Bz(r2)) L = µ₀I_encl. {1}

The enclosed current through the surface of C is zero, hence B_z(r1) = Bz(r2). {1}

As r₁ and r₂ are chosen arbitrarily and the magnetic field should be zero at infnity, we require B_z= 0 outside the solenoid. {1}

To investigate the magnetic field inside the solenoid we choose a rectangu- lar path in the plane of constant ϕ overlapping the solenoid wires, and apply

Amp`ere’s Law: {1}

Following the same analysis as before we find:

(Bz(r1) − Bz(r2)) L = µ₀I_encl ,

where B_z(r2) = 0 and Iencl= NI = nLI. {1}

Thus, B_z(r1) = µ₀nLI. This result is true for all values of r₁≤ a. Hence,

B =

( µ₀nI ˆz r≤ a

0 r> a . {1}