Physics 102
Conference 6
Magnetic Field
Conference 6
Physics 102
General Physics II
Monday, March 3rd, 2014
6.1
Quiz
Problem 6.1
Think about the magnetic field associated with an infinite, current carrying wire.
Using this mental image, sketch the magnetic field lines associated with a loop
– you will find the field along the central axis in the next problem, but you can
get a sense of the field everywhere.
The progression from an infinite line of current to a loop is “shown” in Figure 6.1.
Figure 6.1: An infinite line has magnetic field that circulates around it – so does
a loop of wire.
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6.1. QUIZ
Conference 6
Problem 6.2
A circle of radius R has steady current I flowing around it. What is the magnetic
field (direction and magnitude) a height z above the center of the loop (the
Biot-Savart law, for the contribution of a segment of current to the magnetic
0 ])
field at r is: dB = 4µπ0 Id`×[r−r
).
|r−r0 |3
ẑ
r
r
r
ŷ
r
x̂
d
I
Referring to the figure:
ẑ
r
r
r
ŷ
r
x̂
d
I
(side view)
z
direction of magnetic field
R
the contribution from the r0 shown will have horizontal component cancelled by
the patch of current opposite it – only the vertical components will survive the
integration. So:
dBz =
µ0 I R dφ0
µ0 I R2 dφ0
sin
θ
=
,
4 π (R2 + z 2 )
4 π (R2 + z 2 )3/2
(6.1)
and then, integrating around the circle gives:
B=
µ0 I R 2
2 (R2 + z 2 )3/2
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ẑ.
(6.2)
6.1. QUIZ
Conference 6
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6.2. BIOT-SAVART LAW
6.2
Conference 6
Biot-Savart Law
Problem 6.3
Helmholtz coils are a pair of circular loops of radius R carrying steady current
I (in the same direction) – they are typically made using N turns of wire, so
that the net current carried by each loop is N I. The loops are arranged a
distance R apart. Find the magnetic field midway between the loops on the line
connecting their centers.
R
ẑ
NI
ŷ
R
x̂
Figure 6.2: A pair of circular loops carry current N I. Find the magnetic field
at the location shown.
Consider a centered pair of coils, so that the midpoint between the loops is at
y = 0. The magnetic field due to the loop on the left is:
µ0 I R 2
4 µ0 I
B` = − ŷ.
3/2 ŷ = − √
2
5 5R
2 R2 + R4
(6.3)
The magnetic field due to the loop on the right is:
µ0 I R 2
4 µ0 I
Br = − ŷ,
3/2 ŷ = − √
2
5 5R
2 R2 + R4
(6.4)
so the total field at the point of interest is (superposition):
8 µ0 I
ŷ.
B = B` + Br = − √
5 5R
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(6.5)
6.3. LORENTZ FORCE LAW
6.3
Conference 6
Lorentz Force Law
The role of the magnetic field in charge motion comes from the Lorentz force –
given a magnetic field B and a particle with charge q moving with velocity v,
the force on the particle, due to the field, is:
F = q v × B.
(6.6)
Problem 6.4
A charged particle (of mass m, charge q) enters a solenoid of radius R through
a hole along the x axis. It exits through a hole in the y axis.
a.
Assuming the velocity of the particle as it enters the solenoid is v = −v x̂,
find the magnitude and direction of the magnetic field that caused the motion
inside the solenoid.
R
ŷ
x̂
Figure 6.3: A top down view of the solenoid with the particle moving in along
the x̂ axis, and exiting along the ŷ axis.
Once inside the solenoid, the particle is in a region of uniform field. The direction
of the field must be ẑ (out of the page) in order to get a force, as the particle
enters, pointing to the right. We know that the moving charged particle will
travel along a circular arc, in this case, of radius R. Then the magnetic field is
providing the centripetal force, and we have:
mv
m v2
= q v B0 −→ B0 =
,
R
qR
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(6.7)
6.3. LORENTZ FORCE LAW
Conference 6
so the magnetic field inside the solenoid must be B = B0 ẑ =
mv
qR
ẑ.
b.
Assuming the solenoid in the previous problem is made up of wire carrying
current I, with n turns per unit length for the solenoid, find the current I that
must be flowing to induce the motion of the charged particle.
Using Ampere’s law, we know that the magnetic field associated with a solenoid
is:
B = µ0 I n ẑ,
(6.8)
and given our target magnitude, we must have:
µ0 I n =
mv
mv
−→ I =
.
qR
µ0 n q R
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(6.9)
6.3. LORENTZ FORCE LAW
Conference 6
Problem 6.5
A square loop of wire (mass m with side length `) carries uniform current I
flowing in the direction shown below. It is placed in a magnetic field, where its
magnetic dipole moment makes a small angle θ with respect to the magnetic
field. The moment of inertia of the loop (calculated about a rod through its
center) is:
m `2
I` =
,
(6.10)
6
use this, and the small angle approximation, to find the period of the oscillatory
motion of the loop.
I
B = B0 ŷ
ẑ
µ
ŷ
B = B0 ŷ
x̂
top down view
Figure 6.4: A square loop of wire has magnetic moment that makes an angle
of θ with respect to a constant magnetic field pointing to the right. For small
θ, find the period of the resulting oscillatory motion.
The torque on the loop is given by: τ = µ × B = I `2 B0 sin θ (−ẑ). Using the
small angle approximation (sin θ ≈ θ), together with I α = τ gives:
I`
so that:
d2 θ(t)
= −I `2 B0 θ(t),
dt2
(6.11)
d2 θ(t)
I ` 2 B0
6 I B0
=
−
θ(t) = −
θ(t).
2
2
dt
m ` /6
m
(6.12)
A typical solution to this ODE is:
r
θ(t) = θ0 cos
!
6 I B0
t ,
m
(6.13)
so the period is:
r
6 I B0
T = 2 π −→ T = 2 π
m
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r
m
.
6 I B0
(6.14)
6.3. LORENTZ FORCE LAW
Conference 6
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6.3. LORENTZ FORCE LAW
Conference 6
We saw that the Lorentz force law implied a force per unit length of:
F̃ =
µ0 I 2
2πd
(6.15)
for two parallel wires carrying the same current, and separated by a distance d.
We can use this expression to get a sense of the magnitude of the magnetic
force compared to the electric force.
Problem 6.6
Take two infinite line charges with uniform λ, and pull them “upwards” with
constant speed v. Assume the line charges are separated by a distance d.
v
v
d
Figure 6.5: Two infinite line charges pulled upward – these lines are separated
by a distance d.
a.
What is the electrostatic force per unit length acting between the wires?
(Think of the left-hand wire as the “source” wire, and the right-hand wire as
the “forced” wire).
The electrostatic field set up by the wire on the left has magnitude:
E=
λ
2 π 0 s
(6.16)
where s is the distance from the left-hand wire. The force exerted on a segment
of wire of length ` on the right is:
F =
λ2 `
2 π 0 d
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(6.17)
6.3. LORENTZ FORCE LAW
Conference 6
so the force per unit length, relevant for comparison is:
F̃E =
λ2
.
2 π 0 d
(6.18)
b.
What is the current I (for use in (6.15)) in this case? Write the magnitude
of the magnetic force in terms of λ and v.
The current associated with either wire is I = λ v, so that the magnetic force
per unit length is:
µ 0 λ2 v 2
F̃B =
(6.19)
2πd
c.
With what speed v must you pull the wires in order to get the electrostatic
force to have the same magnitude as the magnetostatic one? What does this
tell you about the relative size of the electrostatic and magnetostatic forces?
When the forces are equal:
F̃B
2
λ v2
µ0
2πd
we can solve for v:
r
v=
= F̃E
=
λ2
2 π 0 d
1
≈ 3 × 108 m/s.
µ0 0
The magnetic force is, numerically, much smaller than the electric one.
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(6.20)
(6.21)
6.4. FIRST ORDER ODES
6.4
Conference 6
First Order ODEs
Problem 6.7
Suppose you want to model a population – you denote the number of animals at
time t by N (t). It is reasonable to imagine that the rate of growth is proportional
to the population size, so that dNdt(t) ∝ N (t). Call the proportionality constant
α, then the differential equation describing the population is:
dN (t)
= α N (t)
dt
(6.22)
and we must be given the size of the population at time t = 0: N (0) = N0 .
a.
For ease of graphing, take N0 = 2 and α = 2 1/s. Then dN
dt |t=0 = 4, so
the slope of N (t) at t = 0 is four. Draw a line with that slope that starts at
t = 0 and goes to t = 1/2. What is the approximate value of N (1/2)? Using
that value, determine the slope of N (t) at t = 1/2 and draw a line with that
slope from t = 1/2 to t = 1, that will allow you to estimate the value of N (t)
at t = 1. Continue with t = 3/2, what sort of function are you getting?
This is a coarse approximation to the exponential, and it is off by quite a lot
because of our choice to take 1/2 s steps – but the curve is clearly growing
. . . quickly.
b.
One important feature of population models is the notion that there
is a natural maximum to the population, Nmax – modify the starting equation
dN
dN
dt = α N (t) so that if N (t) < Nmax you get dt > 0 and if N (t) > Nmax , you
have dN
dt < 0 (causing decay in the population). Take the simplest extension
– that will lead to an equation that is quadratic in N (t) on the right, and will
reduce to dNdt(t) = α N (t) for N (t) Nmax .
The simplest fix is to take
dN (t)
N (t)
=α 1−
N (t)
dt
Nmax
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(6.23)
6.4. FIRST ORDER ODES
Conference 6
32
24
16
8
1
2
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6.4. FIRST ORDER ODES
Conference 6
32
24
16
8
1
2
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