Physics 9 Fall 2011 Homework 7 - Solutions Friday October 14, 2011

Physics 9 Fall 2011
Homework 7 - Solutions
Friday October 14, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Friday, October 21st. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. In a pre-quantum mechanical model of the hydrogen atom, an electron orbits a proton
at a radius of 5.29 × 10 −11 meters. According to this model, what is the magnitude
of the magnetic field at the proton due to the orbital motion of the electron? Neglect
any motion of the proton.
————————————————————————————————————
Solution
The electron moves in a circle around the proton and so it behaves like a ring of current,
generating a magnetic field. The magnetic field at the center of a ring of current I is,
as we’ve seen,
B = µ0I
2R .
The current of a moving charge is just the charge, divided by the time it takes to go
around the orbit, T = 2πR/v,
I = e
T
= ev
2πR .
Hence,
B = µ0ev
.
4πR2 Now, to get the velocity of the electron we use Coulomb’s law and equate it to the
centripetal force on the electron,
1 e
4πɛ0
2
mv2
e
= ⇒ v =
R2 R 2
4πɛ0mR ,
and so
Plugging in the numbers gives
which is a huge field!
B = µ0e
4πR 2
B = µ0e
4πR2
e2 4πɛ0mR .
e 2
4πɛ0mR
= 4π×10−7 (1.602×10 −19 )
4π(5.29×10 −11 ) 2
= 12.5 Teslas,
1
9×10 9 (1.602×10 −19 ) 2
(9.11×10 −31 )(5.29×10 −11 )

2. A particle of mass m and charge q enters a region where there is a uniform magnetic
field B parallel with the x axis. The initial velocity of the particle is v = v0xî + v0yˆj, so the particle moves in a helix.
(a) Show that the radius of the helix is r = mv0y
qB .
(b) Show that the particle takes a time ∆t = 2πm
qB
to complete each turn of the helix.
(c) What is the x component of the displacement of the particle during the time given
in part (b)?
————————————————————————————————————
Solution
(a) The magnetic force on a charge, q, moving at a speed v through a magnetic field
B is given by the Lorentz force law,
F = qv × B.
Now, the magnetic field is along the î axis, so we can write B = Bî, and so
F = q î + v0yˆj × Bî = −qv0yB ˆ k,
v0x
where we have recalled that î × î = 0, and ˆj × î = − ˆ k. Thus, there is a net force
on the charge in the negative z direction, pushing it downwards. This net force
causes it to move in a circle in the yz plane at a speed v0y that drifts along the
x direction at a speed v0x (i.e., it moves in a helix). The centripetal force is just
equal to the magnetic force,
Fmag = qv0yB = mv2 0y
r
Solving this expression for the radius gives
r = mv0y
qB .
= Fcent.
(b) The time to complete one turn of the helix is just the time it takes to go around
a circle of radius r, which is
∆t = 2πr
v0y
= 2πmvoy
v0yqB
= 2πm
qB .
(c) During the time, ∆t, it takes for the particle to make a circle it moves a distance
∆x = v0x∆t along the x direction. This distance is
∆x = v0x∆t = 2πmv0x
qB .
2

3. The closed loop shown in the figure to the right
carries a current of 8.0 A in the counterclockwise
direction. The radius of the outer arc is 0.60 m
and that of the inner arc is 0.40 m. Find the
magnetic field (magnitude and direction) at point
P .
————————————————————————————————————
Solution
By the right-hand rule for currents, the magnetic field from the top of the circuit points
out of the page, while that from the bottom of the circuit points into the page. Since
both branches carry the same current, and since the bottom branch is closer, it wins
out and the magnetic field points into the page.
The total magnetic field is just the sum of the two semicircles (the straight-line segments
contribute nothing, since they are inline with the point P). We can use the
Biot-Savart law to determine the magnetic field from each semicircle. From the top,
since the distance from the wire to the point is constant, r = R1, and the current is
always tangent to the point, the Biot-Savart law says
Btop = µ0I
4π
= µ0I
4πR 2 1
= µ0I
4πR 2 1
= µ0I
12R1 ,
ds׈r
r2
ds
where we have recalled that the distance around the semicircle is s = θR, and θ = π/3
for a 60◦ angle. The magnetic field from the bottom wire is done in exactly the same
way, but now the radius is only r = R2. So, Bbottom = − µ0I
, where the minus sign
12R2
accounts for the direction of the field from the bottom branch (into the page).
R1θ
Combining the two fields gives the total,
B = µ0I
1
−
12 R1
1
= −
R2
µ0I
(R1 − R2) .
12R1R2
4. Plugging in the numbers gives
B = − µ0I
(R1 − R2) = −
12R1R2
4π × 10−7 × 8
(.6 − .4) = −7.0 × 10
12(.6 × .4)
−7 T,
or about 0.7 µT, into the page.
3

5. A long cylindrical shell has an inner radius a and an outer radius b and carries current
I parallel to the central axis. Assume that within the material of the shell the current
density is uniformly distributed. Find an expression for the magnitude of the magnetic
field for
(a) 0 < r < a,
(b) a < r < b, and
(c) r > b.
————————————————————————————————————
Solution
We can determine the magnetic field using Ampere’s law, which says that
B · ds = µ0Ithrough,
where Ithrough is the current passing through the Amperian loop.
(a) Inside the inner radius, (i.e., for r < a), there is no current inside this region, and
so there is no magnetic field.
(b) Inside the shell, itself (where a < r < b), then the current is uniformly distributed,
such that the current density J is constant. Now, since the total current through
the full area A = π (b 2 − a 2 ) is I, then the current density is
J =
I
(b 2 − a 2 ) .
Now, if we take our Amperian loop to be a circle of radius r, where a < r < b,
then the current passing through this loops is
Ithrough = JAthrough =
I
(b2 − a2 2 2
r − a
)
.
Notice that if r = a, then the total current is zero, as we should expect. Furthermore,
if the loop encloses the entire shell (such that r = b), then the total current
is I. Thus, we have the correct Ithrough.
Now, the magnetic field is constant around our Amperian loop, so
2 2
B
r − a
· ds = B ds = (2πr) B = µ0I
,
and so the magnetic field is,
B = µ0I
2πr
r 2 − a 2
b 2 − a 2
.
b 2 − a 2
Notice that, if r = a, then B = 0, and so we get the correct continuity of the
magnetic field at the inner boundary.
4

(c) Finally, if we are outside the shell (when r > b), then an Amperian loop contains
the entire current, and so we just get back the field of a long straight current, or
B = µ0I
2πr .
Again, notice that this result agrees with our answer from part (b) when r = b,
again yielding the correct continuity of the field at the boundary.
5

6. A nonconducting disk has radius R, carries a uniform surface charge density σ, and
rotates with angular speed ω.
(a) Consider an annular strip that has a radius r, a width dr, and a charge dq. Show
that the current, dI, produced by this rotating strip is given by ωσrdr
(b) Use your result from part (a) to show that the magnetic field strength at the
center of the disk is given by the expression 1
2 µ0σωR.
(c) Use your result from part (a) to find and expression for the magnetic field strength
at a point on the central axis of the disk a distance z from its center.
————————————————————————————————————
Solution
(a) We can imagine that the strip has a radius r and thickness dr. If the surface
charge density is σ, then the ring contains a small charge dq which is the area
enclosed in the ring, times the surface charge density. Now, the ring has an area
dA = 2πrdr, since the length of a ring of radius r is 2πr. So, dq = 2πrσdr. The
current is dI = dq
, where dt is the time it takes for the ring to go around in an
dt
orbit. So, in terms of the orbital frequency dt = 1/f, and since 2πf = ω, we have
dI = dq
dt
= fdq = (2πf) rσdr = ωσrdr.
(b) Now, as we’ve seen, the magnetic field at a distance z away from a ring of radius
r and current dI is
dB = µ0
4π
2πr2dI Plugging in for dI gives
dB = µ0ωσ
2
(z2 + r2 . 3/2
)
r 3 dr
(z2 + r2 . 3/2
)
Now, integrating this result from r = 0 to r = R gives
B(z) = µ0ωσ
2
= µ0ωσ
2
= µ0ωσ
2
R
0
r3dr (z2 +r2 ) 3/2
r2 +2z2
√
r2 +z2 R
0
R2 +2z2 √
R2 +z2 − 2z
This is actually the magnetic field at a distance z away from the center, and so
we’ve done part (c), already. To find the field at the center, we simply set z = 0
in our above expression,
as claimed.
B(0) = µ0ωσ
2
.
2 2 R + 0
√ − 0 =
R2 + 02 1
2 µ0σωR,
6