Final exam solutions
PROBLEM 1
1 Q
a. Potential created by single charge Q is given by V = 4π
, where r is the distance from the charge to the
0 r
point of interest. The principle of superposition works for potential too, so the total potential at point P is
(−2q)
1 q
1
√2 ) 1 q .
VP = 2 · 4π
+ 4π
◦ = (1 −
0 d
0 d cos 30
3 2π0 d
b. By symmetry electric field can be only up or down.
Let’s project all fields to the axis looking up: Ez =
√
(−2q)
q
q 4
3
1
1
1
◦
cos
30
−
=
(
−
).
−2 4π
2
4π0 (d cos 30◦ )2
2π0 d2 3
2
0 d
c. By doing that we change the potential energy of the charges from zero to the present potential energy. The
current potential energy of the charges equals to the potential energy due to interaction of all pairs of charges
2
2
1 q·(−2q)
7 q
1 q
+ 2 · 4π
= − 4π
. Work needed is
(#3 is the charge in the middle): U = U12 + U13 + U23 = 4π
d/2
0 d
0
0 d
2
7 q
the same as the change in the potential energy when bringing in the charges from inifinity: W = − 4π
.
0 d
PROBLEM 2
a. In the equilibrium all charges can be located only at the surfaces of the conductor. In this case we have inner
and outer cylindrical surfaces. Let’s use a Gaussian law to determine whether we have any charge on the inner
surface. Gaussian surface will be a cylinder with radius r such that r1 < r < r2 . The electric field inside the
conductor is zero, so charge inside this surface is zero too. This tells us that there is no charge on the inner
surface, all charge is on the outer surface of the cylinder. Alternatively: equal charges repel each other so they
want to put the largest possible distance between them.
b. Gaussian surface will be a cylinder with radius r such that r < r1 . There is no charge inside, so the electric
field should be zero too.
c. Electric field is zero inside the conductor.
d. Gaussian surface will be a cylinder with radius r > r2 , so the total charge inside will Qin = Q · L, where
L is the length of the Gaussian surface. Due to the symmetry the electric field is radial from the center
axis of the cylinder, and perpendicular to the Gaussian surface at all points. The Gaussian integral is then
Q
ΦE = EL2πr = Qin /0 , so the electric field outside of the cylinder is E = 2πr
.
0
e. Charge density on the outer surface of the cylinder is σ =
QL
2πr2 L
=
Q
2πr2 .
f. Let’s have a small Gaussian box intersecting the outer surface of the cylinder: two of the sides will be parallel
to the surface of the cylinder (we can assume they are parallel to the surface of the cylinder due to the fact that
the box is small), and the other four sides will be perpendicular to the surface of the cylinder. The total flux
of the electric field will be due to the side of the box outside of the cylinder that is parallel to the surface. Our
2
box will be a cube with a side d. The flux is then ΦE = E · d2 = Qinside
= σd
0
0 , so electric field close to the
charged surface of the conductor is E = σ0 . In our case E = 2πrQ2 0 using e) for σ.
g. Answer in f) is the same as answer in d) if r = r2 .
PROBLEM 3
~ = −∇V
~ = ( 2ax 2 , 2ay 2 , 0) = ( 2 a 2 )~r with ~r = (x, y, 0) a vector perpendicular to the z-axis.
a. E
x +y
x +y
x +y
b. E = √
a
.
x2 +y 2
c. E ∝ 1r , where r is the distance from z axis. The direction of the electric field is radial away from z axis.
2
PROBLEM 4
a. Force acting on the electron is F = eQ
r 2 , and directed to the left, so the acceleration is not constant. We can
only say it is a linear, decelerating motion.
b. Using the conservation of energy Ui + Ki = Uf + Kf , we get
q
vf = v 2 − 2|e|Q
ml .
|e|Q
l
+
mv 2
2
=
|e|Q
l/2
+
mvf2
2 ,
so the final velocity is
PROBLEM 5
a. Let’s combine resistors 3 and 4, R34 = R3 + R4 , then we can combine resistors R2 and R34 , they are in parallel:
·R34
R234 = RR22+R
= 50·140
190 ≈ 36.84 Ω. The total resistance is R = R1 + R234 ≈ 86.84 Ω. The current through
34
10
resistor R1 is then I1 = VR ≈ 86.84
≈ 0.115 A. Now consider the junction between R1 , R2 , and R34 . The Junction
Theorem (charge conservation) says that I1 = I2 + I34 . Using the LoopTheorem on the loop with R2 , R3 , and
R4 in it shows that I2 R2 = I34 R34 . We have two equations with two unknowns I2 and I34 . These can be solved
2
34
and I34 = I1 R2 R
to get I2 = I1 R2R+R
+R34 . I34 = I3 = I4 , so the currents are I2 = 0.085 A, I3 = I4 = 0.030 A.
34
PROBLEM 6
~E
~ = ρ/0 , so the Gauss law for the fictional magnetic
a. Analogous equation for the electric field looks like ∇
field will be ΦB = Minside · µ0 . The spherical Gaussian surface of radius r around the single charge will yield
mµ0
~
B · 4πr2 = mµ0 , so magnetic field will be B = 4πr
2 . For reasons of symmetry B is directed radially outward.
PROBLEM 7
a. Only the torque is not zero for such construction in the uniform magnetic field. To prove that the force is zero,
~ 1 and dl
~ 2 of the loop that are at opposite sides of the loop. We have
consider two equal length short sections dl
~
~
~
~ 1 we have that dF~2 = −dF~1 and the two forces sum
~
~
~
~
dF1 = I dl1 × B and dF2 = I dl2 × B. Because dl2 = −dl
to zero. This is true for any pair of opposite sections so the total force on the loop is zero as well. Another
~ × B,
~ so the total force is
way to see this is to again look at Rthe force on the small part of the loop: dF~ = I dl
~
~
~
the integral around the loop: F = I dl × B. The distributive properties of the cross product and the fact that
R
~ × B.
~ But this is a
magnetic field is uniform allow integral to be moved inside the cross product: F~ = I( dl)
~
closed loop, so integral of the vector dl is zero. Hence the total force is zero. A third slightly “illegal” way to
see that the force is zero is to consider the magnetic dipole as a dipole consisting of a positive and a negative
magnetic charge at a certain distance from each other. The magnetic field will produce forces on them that are
equal and opposite so they cancel. Compare with an electric dipole in a uniform electric field.
~ so minimum energy corresponds to the situation
b. Potential energy of the magnetic dipole is given by U = −~
µ · B,
when magnetic moment µ
~ and magnetic field are aligned. Magnetic moment direction is given by right hand
rule applied to the current and is perpendicular to the loop. So we can say that the direction of the current
should be perpendicular to the magnetic field, going around it and its direction should be given by the right
~
hand rule with fingers along the current and the thumb along B.
PROBLEM 8
a. Voltage across the capacitor V = Ed, where E is the electric field inside the capacitor and d is Rthe distance
between plates. The electric field is, then, E = V /d and the energy stored in the capacitor is WE = 20 E 2 dV =
0 2
−9
J. Another way to calculate WE
2 E Ad, where A is the cross-sectional area of the capacitor. WE ≈ 442 · 10
1
2
is to use WE = 2 CV with C = 0 A/d.
b. Force on the any small part of the sphere is directed along the electric field, so the net force on the sphere will
be F = Qsphere · E = ρ 43 πr3 V /d ≈ 10.5 · 10−12 N.
3
c. The sphere is not moving, so the force is zero.
PROBLEM 9
a. Let’s calculate what force exerts the upper wire on the wire below. The magnetic field due to upper wire is
µ0 I
, so the force (per unit
directed to the right at the position of the second wire and its magnitude is B = 2πL
~ the direction of the force is down, from the first wire, and the magnitude is F = IB = µ0 I 2 .
length) is F~ = I~2 × B,
2πL
The force on the first wire is opposite in the direction — it will point up, so the wires repel each other.
b. The magnitude of the magnetic fields due to the two wires will be equal: B1 = B2 =
2π
√
µ0 I
(L/2)2 +x2
and the
direction will be to the up-right and down-right. The only component that will survive after summation is
µ0 IL
horizontal component to the right: Bx = B1x + B2x = 2B1 cos α = 2B1 √ L/22 2 = 2π((L/2)
2 +x2 ) .
(L/2) +x
PROBLEM 10
a. The magnetic field inside the solenoid can be found by considering Ampere’s rectangular loop with one side
parallel
to the solenoid axis and lying inside the solenoid and the opposite side lying outside the solenoid.
R
~ = BL, where L is the length of the loop and the length of the solenoid. So, BL = µ0 Ienc = µ0 IN ,
~ dl
B
because the
R wire will N times cross the surface of this loop. B = µ0 IN/L then and the stored magnetic energy
is WB = 2µ1 0 B 2 dV = 2µ1 0 B 2 · LπD2 /4 ≈ 158 · 10−7 J.
R
~ = − dΦB , let’s choose the loop of radius r = 5 mm, oriented perpendicular to the magnetic field and
~ dl
b. E
dt
~ is constant along the loop. So,
centered on the axis of the solenoid. Because of symmetry the magitude of E
2
E2πr = | dB
dt |πr and magnitude of the electric field is E =
dB
| dt |r
2
= 150 · 10−6 V/m.
c. It’s three time higher, E = 450 · 10−6 V/m.
PROBLEM 11
a. Steady State means that all time derivatives are zero. Thus there is no induced voltage across the inductance.
The current is E/R.
di
b. When we flip the switch, the battery is not in the circuit anymore. The Loop Theorem gives 0 = L dt
+Q
C.
dQ
Also i(t) = + dt (think carefully about the sign: positive current increases the charge on the capacitor where
2
Q
the current enters it), so L ddtQ
2 + C = 0. This is a second order differential equation so it needs two initial
√
conditions. The solution is Q = A cos ωt + B sin ωt, where ω = 1/ LC as can be seen by substitution of the
solution in the differential equation. At t = 0 the charge of the capacitor was zero (first initial condition), so
A = 0. At t = 0 the current i(0) was equal to E/R (second initial condition), so i(0) = dQ
dt |t=0 = Bω = E/R.
√
E LC
E
Thus B = R . The current at t > 0 is then I = Bω cos ωt = R cos ωt. The current for t < 0 and t > 0 is
continuous at t = 0 as it should be (no jumps). When the cosine is positive, the current flows clockwise, and
when the cosine is negative it flows counterclockwise. It’s oscillating.
√
c. The current is a constant E/R for t < 0 and for t > 0 is oscillating with frequency f = ω/(2π) = LC/(2π)
and amplitude E/R.
PROBLEM 12
a. The electron will exit perpendicular to the other side if the radius of the circular motion while in the magnetic
mv
field region is d so that the electron follows 1/4 of a circle. Thus d = |e|B
and so v = |e|Bd
m .
b. Magnetic field does not do work because the force is always perpendicular to ~v , the direction of motion, so the
2
force does not change the kinetic energy and the magnitude of the velocity is constant. Kf = mv
2 .
4
PROBLEM 13
E0
a. L dI
dt = E = E0 sin ωt, so integrating that, I = − Lω cos ωt + A, where A is constant. Assuming that at t = 0
E0
(1 − cos ωt). If you left out the integration constant A you will lose no points in
there was no current, I = Lω
E0
this case because the initial condition was not specified. In that case you obtain I = − Lω
cos ωt.
b. Energy stored in the inductor is U =
E02
2Lω 2
1
2
2 LI
=
E02
2Lω 2 (1
− cos ωt)2 . The alternative answer is U =
1
2
2 LI
=
2
cos ωt.
c. Average energy is Uave =
1
T
R
U dt integrated over one period T = 2π/ω. Expanding (1 − cos ωt)2 , we note that
averages: < 1 >= 1, < 2 cos ωt >= 0, < cos2 ωt >=
answer is < U >= Uave =
E02
4Lω 2 .
1
2,
so < U >= Uave =
E02 3
2Lω 2 2
=
3E02
4Lω 2 .
The alternative
E2
2
0
d. sin (ωt − π/2) = − cos ωt, so U = 2Lω
2 (1 + sin (ωt − π/2)) . To comment on the phase difference we need to
E0
convert current to sine: I = const + sin (ωt − π/2) Lω , so the current lags the voltage by π/2 in the inductor.
The alternative solution gives the same lag using again − cos ωt = sin (ωt − π/2) in the alternative expression
for I.
PROBLEM 14
~ r
~ = µ0 I dl×~
~
a. We will use Biot-Savart law to calculate the magnetic field at point P. dB
4π r 3 . The direction of dl is
down, the ~r lies in the plane of the sheet, so the result will be perpendicular to the board, specifically out of
the sheet plane. Then we can just sum up the magnitudes of the elementary contributions. Let’s introduce y –
~ Also, ~r will be a vector that
distance from the top point of the wire to the small segment
in consideration dl.
p
2
2
~ and ~r. This
~
~
~
connects dl to point P directed from dl to P. dl × ~r = dl y + D sin θ with θ the angle between dl
angle is exterior to the triangle definedpby the segment of conductor and P but because sin (π − θ) = sin θ we
R 0I
D dy
µ0 I
l
have from the triangle that sin θ = D/ y 2 + D2 . So B = µ4π
= 4πD
.
(y 2 +D 2 )3/2
(l2 +D 2 )1/2
PROBLEM 15
~ =∇
~ ×A
~ = (∂y Az , −∂x Az , 0) =
a. Magnetic field is B
b. B = (x2 + y 2 )1/2 Ri2 e−
x2 +y 2
R2
−
( −iy
R2 e
x2 +y 2
R2
,
ix −
R2 e
x2 +y 2
R2
−
, 0) = i
e
x2 +y 2
R2
R2
(−y, +x, 0).
.
~ = (x, y, z) · (−y, x, 0) i2 e−
c. ~r · B
R
x2 +y 2
R2
= 0.
d. Magnetic field is perpendicular to the radius-vector pointing from zero.
Extra credit: magnetic field is perpendicular not only to radius-vector pointing from the origin, but also perpendicular to the vector perpendicular to the z-axis that connects the current point to z axis. Magnetic field
~ does not depend upon the azimuthal
is cylindrical symmetric around the z-axis because the magnitude of B
angle φ (remember cylindrical coordinates). So the magnetic field resembles the magnetic field due to the wire
carrying current aligned along z axis but it has the wrong dependence upon distance from the wire.