### 1-up - MIT

```MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
November 9, 2014
Prof. Alan Guth
PROBLEM SET 7 SOLUTIONS
PROBLEM 1: J.J. THOMPSON AND THE CHARGE TO MASS RATIO
OF THE ELECTRON (10 points)
Griffiths Problem 5.3 (p. 216).
~ and the electric field E
~
(a) The beam is not deflected by the uniform magnetic field B
when the forces exerted by these fields cancel each other:
~ + (~v × B)
~ =0.
F~ = q E
(1.1)
~ B
~ and the velocity ~v of the particles in the beam are perpendicular
Since the fields E,
to each other, both fields will exert a force along the same direction. This allows
us to write Eq. (1.1) in terms of the magnitudes of the fields E and B and the
magnitude of the velocity v:
E = vB
=⇒
v=
E
.
B
(1.2)
(b) When the electric field is turned off, each particle will experience only the force from
~ This force will always be orthogonal to its velocity and thus
the magnetic field B.
will cause it to move along a circular orbit with the radius R. In this case, the force
exerted by the magnetic field must supply the centripetal force for the orbit of radius
R, which gives us
v2
qvB = m
R
=⇒
q
v
=
=
m
BR
E
.
B2R
(1.3)
PROBLEM 2: EXAMPLES OF THE USE OF THE BIOT-SAVART LAW
(15 points)
Griffiths, Problem 5.8 (p. 228).
(a) We can find the magnetic field in the center of a loop
by summing the contributions of each of the four straight
sides of the square loop. The figure on the right illustrates one convenient way to parametrize the calculation
for a single side: we consider a straight piece of wire
located at a distance s from P, the center of the loop,
and break it into infinitesimal segments of wire, each lo-
8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014
p. 2
cated in the angular sector (θ; θ + dθ) with respect to P. Then the horizontal
coordinate of each infinitesimal segment is `0 = s tan θ along the wire, and the length
of the segment is d`0 = s d(tan θ) = s dθ/ cos2 θ. The distance from d~` to P is
= s/ cos θ. The infinitesimal magnetic field at point P is pointed towards the
s dθ
cos2 θ
µ0 I
µ0 I |d~`0 × ˆ |
µ0 I
cos
θ
=
=
dθ cos θ.
dB =
2
2
2
4π cos θ
s
4πs
4π
(2.1)
Then the magnetic field produced by a piece of wire at a distance s in the angular
sector (θ1 , θ2 ) with respect to P is equal
θ2
Z
B=
dB(θ) =
θ1
where s = R, θ2 = −θ1 =
~ tot
B
π
4
µ0 I
(sin θ2 − sin θ1 ) ,
4πs
(2.2)
for one side of the square. The total magnetic field is
µ0 I
=4
4πR
√
√ !
2
2
− −
zˆ =
2
2
√
2µ0 I
zˆ ,
πR
(2.3)
where the z axis is perpendicular to the square, upwards in Fig. 5.22 of the problem
set.
(b) For an n-sided polygon, the angle between two lines drawn from the center to the
two corners of one edge is 2π/n. Then we get θ2 = −θ1 = nπ . The total magnetic
field is
h
i
~ tot = n µ0 I sin π − sin − π zˆ =
B
4πR
n
n
nµ0 I
π
sin zˆ .
2πR
n
(2.4)
(c) When n → ∞,
µ0 I
sin x
µ0 I sin nπ
~
zˆ =
lim
=
Btot = lim
π
n→∞ 2R
2R x→0 x
n
µ0 I
zˆ .
2R
(2.5)
This result agrees with our previous result for the magnetic field at the center of a
circular current carrying wire.
8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014
p. 3
PROBLEM 3: MAGNETIC FIELD ON THE AXIS OF A TIGHTLY
WOUND SOLENOID (15 points)
(a) The wire forms a helix, but we can approximate each turn as a circle in the x-y
plane. Since the number of turns per unit length is n, the current of a ring of length
dz is given as dI = nIdz. Then the magnetic field at the origin produced by a ring
at height z is given by Griffiths’ Eq. (5.41) (which was also derived in lecture) as
dB =
µ0 nIdz
R2
,
2
(R2 + z 2 )3/2
(3.1)
~ = dB zˆ is the contribution from the ring at height z. The total magnetic
where dB
field at the origin is then
Z
~ = µ0 nI zˆ
B
2
z2
z1
R2
dz .
(R2 + z 2 )3/2
(3.2)
Using the change of variable z = R tan θ, with dz = R cosdθ2 θ , we find
Z
Using sin θ =
R2
dz =
(R2 + z 2 )3/2
√ z
z 2 +R2
z2 →∞
find
R2 cos3 θ Rdθ
=
R3
cos2 θ
Z
cos θdθ = sin θ .
(3.3)
and restoring the limits of integration,
~ = µ0 nI
B
2
z2
(b) Since lim p
Z
z22 + R2
= 1 and
"
z2
p
z22 + R2
lim
z1 →−∞
−p
z1
p
z1
z12 + R2
z12 + R2
#
zˆ .
(3.4)
= −1, for an infinite solenoid we
~ solenoid = µ0 nI zˆ .
B
(3.5)
8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014
p. 4
~ FIELD (10
PROBLEM 4: VECTOR POTENTIAL FOR A UNIFORM B
points)
Griffiths Problem 5.25 (p. 248).
~ ·A
~ = 0:
Let’s first check ∇
~ ·A
~ = −1∇
~ · (~r × B),
~
∇
2
1 ~ ~
~ × B)
~ =0,
=− B
· (∇ × ~r) − ~r · (∇
2
(4.1)
~ × ~r) = 0,
where the first term vanishes because the curl of a radial vector field is zero, (∇
~ is uniform. Then
and the second term vanishes because the magnetic field B
~ ×A
~ = −1∇
~ × (~r × B)
~
∇
2
1 ~ ~
~ B
~ + ~r(∇
~ · B)
~ − B(
~ ∇
~ · ~r) .
· ∇)~r − (~r · ∇)
= − (B
2
(4.2)
~ B
~ = 0,
The terms involving the partial derivative of the magnetic field give zero, (~r · ∇)
~ · B)
~ = 0. The first term in Eq. (4.2) is
and ~r(∇
~ · ∇)~
~ r=
(B
∂
∂
∂
+ By
+ Bz
(xˆ
ex + yˆ
ey + zˆ
ez )
Bx
∂x
∂y
∂z
~ .
= Bx eˆx + By eˆy + Bz eˆz = B
(4.3)
The last term in Eq (4.2) is,
∂x
∂y
∂z
~ ∇
~ · ~r) = B
~
~ .
B(
+
+
= 3B
∂x ∂y ∂z
(4.4)
~ ×A
~ = − 1 (B
~ − 3B)
~ =B
~ .
∇
2
(4.5)
Then we get
Is this result unique? In localized situations, where we can impose the boundary condition
~ → ~0 as |~r | → ∞, the conditions ∇
~ ×A
~=B
~ and ∇
~ ·A
~ have a unique solution. In
that |A|
~ field extending to infinity, no such boundary condition is
this case, however, with the B
~ is not unique. One possible change would be to add any constant vector.
possible, so A
~ = B0 zˆ, then A
~ = B0 xˆ
~ = −B0 yˆ
Bigger differences are possible, too. If B
y or A
x both
work.
8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014
p. 5
PROBLEM 5: A CHARGED PARTICLE IN THE FIELD OF A MAGNETIC
MONOPOLE (20 points)
Griffiths Problem 5.45 (p. 258).
(a) The force F~ on qe is,
~ = µ0 qe qm (~v × rˆ) .
F~ = qe (~v × B)
4π r2
(5.1)
Let the accelaration of qe be ~a, then
F~ = m~a
=⇒
~a =
µ0 qe qm
(~v × ~r) .
4π mr3
(5.2)
(b) Since the force on particle qe is found to be perpendicular to its velocity ~v , this force
will not perform work on the particle. Therefore the kinetic energy of particle does
not change and the speed v = |~v | is a constant of the motion.
d 1
dEkin
=
m~v · ~v = m~a · ~v = 0 .
dt
dt 2
(5.3)
(c) It is claimed that the vector quantity
~ ≡ m~r × ~v − µ0 qe qm rˆ
Q
4π
(5.4)
~ with respect to time, finding
is conserved. To verify this, we differentiate Q
~
dQ
µ0 qe qm d ~r
= m(~v × ~v ) + m(~r × ~a) −
,
dt
4π dt r
µ0 qe qm ~v
µ0 qe qm ~r dr
=0+
~r × (~v × ~r) −
− 2
.
4πr3
4π
r
r dt
(5.5)
The time derivative or r, dr
ˆ, dr
v · rˆ = 1r ~v · ~r.
dt is the velocity component along r
dt = ~
The triple product rule gives ~r × (~v ×~r) = r2~v −~r(~r ·~v ). Then putting these relations
into Eq. (5.5), we get
~
~v
dQ
µ0 qe qm 1 2
~r (~v · ~r)
=
r ~v − ~r(~r · ~v ) − + 2
=0.
dt
4π
r3
r
r
r
(5.6)
8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014
p. 6
~ points along z-axis, so Q
~ · rˆ = Q zˆ · rˆ = Q cos θ.
(d) In the new coordinate system Q
Then, from Eq. (5.4), we find
µ0 qe qm
~ · rˆ = m(~r × ~v ) · rˆ − µ0 qe qm (ˆ
Q
r · rˆ) =
,
4π
4π
so
Q cos θ = −
µ0 qe qm
.
4π
(5.7)
(5.8)
~ is a constant of the motion. Then Eq. (5.8) implies that
In part (c) we found that Q
the motion is confined to a fixed value of the polar angle θ, hence to a cone. The
polar angle θ is
cos θ = −
µ0 qe qm
.
4πQ
(5.9)
PROBLEM 6: THE MAGNETIC FIELD OF A SPINNING, UNIFORMLY
CHARGED SPHERE (25 points)
Griffiths Problem 5.60 (p. 264).
This problem has been held over to Problem Set 8.
PROBLEM 7: THE COMPLETE MAGNETIC FIELD OF A MAGNETIC
DIPOLE: (10 points)
This problem has been held over to Problem Set 8.
```