[1]. A low pass filter has an inductor, a resistor and a capacitor in series with an ac emf E = Vin cos ωt.
The output is the voltage, Vout is that across the capacitor. (a) Draw the circuit indicating the voltage
which is defined as Vout . (b) Derive an expression for Vout /Vin . (c) Find the limiting expressions for ω
large and small. Are these results consistent with the fact that this is called a low pass filter? Why?
(a) Draw figure
(b) The impedance of this is
Z=
p
R2 + (1/ωC)2
and so the current in the circuit is
i=
Vin cos ωt
Vin cos ωt
=p
.
Z
R2 + (1/ωC)2
The voltage across the capacitor is then, (using the same definition as for Vin )
Vout cos ωt = iXC =
so that
i
Vin
=p
cos ωt
2
ωC
R + (1/ωC)2 ωC
Vout
1
.
=p
Vin
R2 + (1/ωC)2 ωC
(c) If ω is large the Z ≈= R and
Vout
1
.
=
Vin
RωC
which is small when R > (1/ωC) which is what large ω means.
When ω is small Z ≈ (1/ωC) and
Vout
= 1.
Vin
So yes, these results are consistent with the name since the output is small of ω > 1/RC and is equal
to the input when the inverse is true.
1
~ = Bx î + By ĵ + Bz k̂, at the centre of a semi[2]. (a) By direct integration, calculate the magnetic field, B
circular segment carrying a current I in the clockwise sense. The radius of the segment is a, i.e., has
coordinates x, y such that x2 + y 2 = a2 . The semi-circle is such that y > 0, i.e., is centered at the origin
ad lies above the x-axis.
y
a
x
~ is
Consider small d~` in the clockwise direction. The contribution to B
~ =I
dB
µ0 d~` × r̂
.
4π r2
~ is into of the paper, i.e., in the negative z direction, so
The direction of dB
~ = −I
dB
µ0 d`
k̂.
4π a2
independent of which d~` is involved. Thus
Z
Z
~ = dB
~ = −I µ0 ( d`)k̂ = I µ0 (πa)k̂ = − Iµ0 k̂
B
4πa2
4πa2
4a
(b) A particle with charge q moves with velocity ~v = vx î + vy ĵ + vz k̂ in the resulting magnetic field.
Calculate the force F~ (magnitude and direction) as the particle passes through the origin.
Have
~ = q~v × B
~ = −q Iµ0 (vx î + vy ĵ + vz k̂) × k̂ = −q Iµ0 (vy î − vx ĵ)
F~ = q~v × B
4a
4a
(c) Determine the rate at which the magnetic field is doing work on the particle. (Hint: P = F~ · ~v .)
Iµ0
Iµ0
P = F~ · ~v = −q
(vy î − vx ĵ) · (vx î + vy ĵ + vz k̂) = −q
(vy vx − vx vy ) = 0
4a
4a
as it should be.
2
[3]. (a) An electromagnetic wave has an electric field given by
~
E(y,
t) = −(5.10 × 106 V/m)k̂ sin[(2.25 × 1012 rad/s)t − ky]
(a) In which direction is the wave moving?
Given sin[(2.25 × 1012 rad/s)t − ky] the wave is traveling in the positive y direction.
~
(b) Re-write E(y,
t) filling in the missing parameter k.
Have ω = ck or k = ω/c = (2.25 × 1012 rad/s)/(3.00 × 108 m/s) = 7.5 × 103 m−1 .
~
E(y,
t) = −(5.10 × 106 V/m)k̂ sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y]
(c) What is the wave length, λ, of the wave?
λ=
2π
2π
=
(7.5 × 103 m−1 ) = 0.84 × 10−3 m = 8.4 × 10−4 m
k
k
~
~ as part (b).
(d) Write the vector equation for B(y,
t), i.e., write the equivalent to the above for E
~ ×B
~ is in the direction of propagation which must be in the y direction. Have
~ = 1E
The vector S
µ0
~ must be in the −î direction, i.e.,
k̂ × î = ĵ so B
(5.10 × 106 V/m)
~
B(y,
t) = −
î sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y]
(3.00 × 108 m/s)
or
~
B(y,
t) == −(1.7 × 10−2 T)î sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y]
3
[4] A parallel plate capacitor has horizontal plates of area A and the distance between the plates is d.
(a) Find the capacitance C in terms of A, d and fundamental constants. Ignore edge effects.
The electric field close to a metallic surface is E = σ/0 , while σ = Q/A and V = Ed so that
V = Ed =
so that
C=
Qd
σ
d=
0
0 A
Q
0 A
=
.
V
d
(b) If there is a potential difference V across the plates what is the magnitude of the electric field E
between the capacitor plates? Again ignore edge effects.
As noted above
E=
V
.
d
(c) Ignoring edge effects, what is the energy density uE due to this electric field and what is the total
energy UC stored. Does this agree with 12 CV 2 using C from part (a)?
Have uE = (1/2)E 2 so the energy stored in the space between the plates is U = AduE = (1/2)0 E 2 Ad =
(1/2)) E 2 (A/d)V 2 which is to be compared with U = (/12)CV 2 with th result
C=
0 A
,
d
as in part (a).
A cork ball of mass M and charge q is placed between the plates and remains at rest.
(d) Find the voltage V across the capacitor in terms of M , q and g. Your solution must include a free
body diagram.
(Draw a diagramme.) The electric force is qE = q(V /d) while the gravitational force is M g and for
a = 0, this implies
V
q = Mg
d
or
M gd
V =
.
q
(e) If the space is filled with a dielectric with a dielectric constant κ what is the new value of V ?
If the potential difference was V , when the dielectric is included it becomes V /κ so
V =
M gdκ
.
q
4
[5] (a) Find the emf’s E1 and (b) E2 in the circuit shown in the figure. (c) Find the potential difference
between the points a and b.
20 V
1.00Ω
1.00A
+
4.00Ω
6.00Ω
1.00Ω
X1
b
a
2.00A 1.00Ω +
+
2.00Ω
X2
For E1 , note the potential difference between a and b is obtained from the top leg of the diagramme,
i.e., is 20 − 7.00Ω!.00A = 13V. The current in the middle brach is 1.00A in the direction b → a so 13 =
E1 − 5.00Ω1.00A or
E1 = 13 + 5 = 18.0V.
Then 13 = E2 + 3.00Ω2.00A so that
E2 = 13 − 6 = 7V.
It has already been determine that
Vab = 13V.
5
[6] The current in the long straight wire AB shown in the figure is upwards and increasing at a steady rate
di/dt.
~ at a distance
(a) At an instant when the current is i, what are the magnitude and direction of the field B
r to the right of the wire?
Consider the wire in isolation and use Ampere’s law,
I
~ · d~` = 2πrB = µ0 i that isB = µ0 i
B
2πr
and is into the paper.
(b) What is the flux dΦB through the narrow shaded strip?
dΦB = BLdr =
µ0 i
Ldr
2πr
(c) What is the total flux ΦB through the loop?
By integration
Z
ΦB =
µ0 i
dΦB =
L
2π
b
Z
a
dr
µ0 iL b
=
ln
r
2π
a
(d) What is the emf E induced in the loop?
Farday’s law states
µ0 L
dΦB
=−
E =−
dt
2π
b
ln
a
di
dt
(e) Evaluate the numerical value of E if a = 12.0cm, b = 36.0cm, L = 24.0cm and di/dt = 9.60A/s.
Have the
|E| =
µ0 × 0.24
2π
ln
36
12
6
9.6 = 5.06 × 10−7 V