Danmarks Tekniske Universitet
Written examination date: 17 December 2021
Page 1 of 5 pages
Course title: Physics 2 – General Engineering
Course number: 10041
All aids allowed
Exam duration: 3 hours
The exam comprises 10 True/False questions and 3 problems that are assessed as a whole.
Weighting: 75%
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True or False
1) The magnitude of the electric field from a point charge decreases linearly with distance.
(F)
2) The electrostatic potential energy of three positive charges arranged in a given
configuration is the same as that of three negative charges in the same configuration.
(T)
3) The energy stored in a capacitor is independent of the electric potential difference
between the capacitor plates. (F)
4) The current is always the same through two resistors connected in parallel, regardless
of their individual resistances. (F)
5) If a moving electron is deflected by a magnetic field, the perpendicular force deflecting
it depends on the mass of the electron. (F)
6) At a party the total magnetic flux through any inflated balloon is always zero. (T)
7) If a magnetic field, applied perpendicular to a wire loop, is decreased, a magnetic field
in the same direction will be induced by currents in the wire. (T)
8) At a fixed frequency the inductive reactance of an inductor is doubled if the inductance
is doubled. (T)
9) In an electromagnetic wave in vacuum, the angle between the electric and magnetic
field vectors depends on the frequency of the wave. (F)
10) The diffraction minima for light passing through a single slit are spaced further apart
for optical light with wavelength 500 nm than for ultraviolet light with wavelength 200
nm. (T)
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Problem 1
A) A cylindrical shell of length L = 70 cm and radius R2 = 4.0 cm << L is given a charge of
Q = –0.7 µC = –7 x 10-7 C. Assume this charge is uniformly distributed on the outer surface.
Compute the electric field on the surface of the conductor.
B) A second cylinder with the same length but with a slightly smaller radius R1 = 3.7 cm is
given an equal but opposite charge +Q and inserted concentrically inside the larger one as
shown in Figure 1(a) (figure dimensions not to scale). An unknown dielectric material is placed
in the space between the two conductors, yielding a capacitance of C = 1.199 x 10-7 F. Use
this value and Table 1 to identify the unknown dielectric.
Material
Air
Teflon
Porcelain
Aluminium
oxide
Ethanol
Water
Strontium
titanate
Barium
titanate
Dielectric
constant κ
1.0006
2.1
7.0
8.5
15.2
80.0
240
3000
Table 1
C) Three of these capacitors are placed in a circuit as shown in Figure 1(b) and charged such
that the potential difference between points a and b is Vab = 80 V. At time t = 0, the switch S
is then closed. Calculate the equivalent capacitance of the circuit, the initial current at points
a and c, and the initial power delivered to the resistor.
D) If the equivalent capacitance in the circuit were instead 5 x 10-7 F, how long would it take
for the current at point a to drop to 10% of its initial value after the switch is closed?
Figure 1
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Solution:
A) Surface area of cylinder A = 2πRL = 0.176 m2 and σ = Q/A = –3.98 x 10-6 C/m2 → E = σ/ε0
= –449.4 kV/m (= –4.49 x 105 N/C).
B) Without dielectric: C0 = 2πε0L/ln(R2/R1) = 2πε0* 0.7 m / ln(4 cm/3.7 cm) = 4.995 x 10-10 F.
With dielectric: C = κC0 → κ = C/C0 = 1.199 x 10-7 F/4.995 x 10-10 F = 240.0 → mystery dielectric
is strontium titanate.
C) Capacitors connected in parallel → Ceq = 3C = 3*1.199 x 10-7 F = 3.597 x 10-7 F.
At t = 0:
Vab = IaR → Ia = 80 V/120 Ω = 0.667 A.
Junction rule applied twice: Ic = (2/3)Ia = 0.444 A.
Power to resistor: P = IaVab = 0.667 A * 80 V = 53.33 W.
D) Time constant τ = RCeq = 120 Ω * 5.0 x 10-7 F = 6.0 x 10-5 s.
Ia(t) = Ia(0)*exp(–t/τ) → exp(–t/τ) = 0.1 → t = –6.0 x 10-5 s *ln(0.1) = 1.382 x 10-4 s (~0.14 ms).
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Problem 2
Figure 2(a) shows a coil of conducting wire, held at an angle of θ = 45⁰ relative to a flat surface.
The coil has N = 320 windings and a radius of r = 10 cm. In the coil there is a current of
I = 2.4 A, with the direction as indicated in the figure. The magnetic field from the Earth is
assumed to be 50 µT upwards, perpendicular to the flat surface.
Figure 2
A) Find the size of the magnetic moment, µ, and of the torque, τ, on the coil.
B) A point, P, is in the centre of the plane of the coil. Ignoring the field from the Earth, how
large is the magnetic field component perpendicular to the flat surface?
C) The current in the coil is now set to zero, so only the Earth field is present. Now the coil is
dropped to the flat surface (so the angle is θ = 0⁰). What is the change in the magnetic flux
through the coil?
D) The inductance of the coil is measured to be 13.8 mH. It is now used in a circuit together
with a resistor with resistance R = 14.5  and a capacitor with capacitance C = 17.0 F. The
circuit is driven by an alternating EMF, as shown in Figure 2(b). At what angular frequency
does the circuit become resonant? What is the impedance at this frequency?
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Solution:
A) The magnetic moment is µ = NIA = 320 * 2.4 A * (0.1 m)2 *  = 24.12 Am2
The torque is  = µ  B = µ * B * sin(θ), where θ is the angle between the normal to the coil
and the magnetic field, so θ = 45⁰. Thus  = 24.12 Am2 * 50 µT * sin(45⁰) = 8.528 * 10-4 Nm.
B) The field from the coil is given by B = (µ0 * N * I) / (2 * r) = (4**10-7 Tm/A * 320 * 2.4 A) / (2
* 10 cm) = 4.83 mT. As the coil is at 45⁰ the vertical component is B/sqrt(2) = 3.41 mT.
C) The magnetic flux is  = N * B * A * cos θ. Initially the flux is init = N * B * A * cos(45⁰) =
320 * 50 µT * (0.1 m)2 *  * cos(45⁰) = 3.55*10-4 Wb. The final flux is final = N * B * A * cos(0⁰)
= 5.03*10-4 Wb, so the difference is 1.47*10-4 Wb.
D) XC = (C)-1 and XL = L. At resonance XC = XL so  = (sqrt(LC))-1 = 2064.60 rad/s. At
resonance the impedance Z = sqrt[R2 + (XL–XC)2] is just the resistance, as XL and XC cancel
out. So Z = 14.5 .
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Problem 3
Figure 3 shows a diagram of a reflection grating spectrometer: Light from an argon lamp
propagates through air (refractive index n = 1) via Mirror 1 to a diffraction grating, which splits
the light into its constituent wavelengths. The diffracted light is further reflected at the grating
and onto a second Mirror 2, from which it finally impinges on a detector. The diffraction grating
has N = 500 narrow slits evenly spaced across a distance of Δx = 1.0 mm.
Figure 3
A) Mirror 1 is made out of a material with refractive index n = 1.75 which transmits a small
amount of the incoming light rather than reflecting it. The incidence angle of light from the
normal to the mirror is θ = 18⁰. What is the angle between the reflected ray and the ray that is
refracted within the mirror (before it exits into air again)?
B) The light from the argon lamp contains two closely spaced emission lines at wavelengths
λ1 = 670.95 nm and λ2 = 671.85 nm. What is the angular separation between these
wavelengths diffracted by the grating in the first order (m = 1)? Assume the grating slits have
negligible width.
C) Can you resolve these two lines with this grating for any order m? Explain your answer.
D) Green light from the grating with wavelength λ = 530 nm impinges on Mirror 2 with intensity
I = 0.21 W/m2. The mirror has an area of 0.04 m2 and is fully covered by this light. What is the
force exerted by this light on the mirror?
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Solution:
A) Snell’s law: n1sinθ = n2sinθ2 → angle of refraction θ2 = asin(1.0*sin(18⁰)/1.75) = 10.17⁰. The
relevant angle becomes 180⁰ – θ + θ2 = 180⁰ - 18⁰ - 10.17⁰ = 151.83⁰ (= 2.650 rad)
B) d = Δx/N = 0.001 m/500 = 2 x 10-6 m, and d*sin θm = mλ → θ1 = asin(λ /2x10-6 m) = 19.6014⁰
and 19.6288⁰ → difference is 0.0274⁰ (= 4.78 x 10-4 rad).
C) Wavelength difference of the two lines: Δλ′ = 671.85 nm – 670.95 nm = 0.90 nm. Resolving
power of grating (Eq. 33-27 in Tipler & Mosca): R = λ/|Δλ| = mN → spectral resolution at λ ≈
671 nm is Δλ = λ/(mN) ≈ 671 nm/(m*500) ≈ 1.34 nm for m = 1 and 0.67 nm for m = 2. Hence,
the lines can be resolved for m = 2, but not for m = 1 (Δλ too large), nor for m = 3 (because
θ3 > 90⁰).
D) Radiation pressure Pr = I/c (fully absorbed light) or Pr = 2I/c (fully reflected light). Since it is
not specified in the text whether light is assumed to be fully reflected, both answers are
deemed correct. So F = Pr*Amirror = I*Amirror/c = 0.21 W/m2 * 0.04 m2 / 3 x 108 m/s = 2.8 x 10-11
N or 2 x 2.8 x 10-11 N = 5.6 x 10-11 N.
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