Physics 102
Conference 8
Magnetic Flux
Conference 8
Physics 102
General Physics II
Monday, March 24th, 2014
8.1
Quiz
Problem 8.1
Suppose we want to set up an EMF of 12 Volts in a circular loop of wire (radius
R = 1 m) by changing a magnetic field that bathes the wire (the magnetic field
points into the page with magnitude B(t) as in Figure 8.1). Assuming we want
the twelve volt EMF over a time period of 10 s, what linear function of time
should we use to modify the magnetic field if we want current to flow clockwise
through the loop?
B = −B(t) x̂
I
ẑ
R
ŷ
x̂
Figure 8.1: A magnetic field goes through a circular loop of wire. The magnetic
field’s magnitude changes linearly in time, so there is a time-varying flux through
the loop, and hence an EMF.
If we set the magnetic field to have generic linear time dependence, B(t) =
B0 − α t, then the flux through the circular loop of wire is:
Φ(t) = (B0 − α t) π R2
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(8.1)
8.1. QUIZ
Conference 8
and the EMF generated by the temporal dependence of the field is:
E =−
dΦ(t)
= π R2 α.
dt
(8.2)
If we want this to equal a particular value, then:
π R2 α = V0 −→ α =
V0
π R2
(8.3)
and our magnetic field is:
B(t) = B0 −
V0
t.
π R2
(8.4)
To ensure that we can run the circuit for T = 10 s, we must start with B0 =
V0
T (at least), so:
π R2
V0
B(t) =
(T − t) .
(8.5)
π R2
If we insert the target quantities:
B(t) =
12 V
(10 s − t) .
π (1 m)2
(8.6)
Note that we took B(t) decreasing to get current flowing clockwise (Lenz’s
law).
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8.1. QUIZ
Conference 8
Problem 8.2
A rectangle has side lengths that sum to 2 L – the rectangle has equal side
length 21 L at t = 0. We pull apart the rectangle with constant speed until the
rectangle has two sides of length L/4 and two with 3 L/4 at time T . What
current runs through the loop if it has net resistance R and sits in a constant
magnetic field pointing into the page with magnitude B0 ?
B0 into page
B0 into page
L/2
L/4
3 L/4
L/2
Call the horizontal length of the rectangle `(t), and the height h(t) – we know
that `(t) + h(t) = L so that h(t) = L − `(t). We are also told that the
horizontal length increases with constant speed, call it v: `(t) = L/2 + v t,
L
and `(T ) = 3 L/4, so that L/2 + vT = 3 L/4 and
v = 4 T . The flux is
L
L
Φ(t) = B0 `(t) h(t) = B0 L/2 + 4 T t L/2 − 4 T t . Then:
dΦ(t)
E =−
= 2 B0
dt
L
4T
2
t,
(8.7)
B0
E
L 2
and I = R
so that I = 2 R
t – the flux was calculated using an oriented
4T
boundary circulating in the clockwise direction, and the positive current here
tells us that I goes in the same direction. Can also get from Lenz’s law – the
area in the field is decreasing, so to oppose the loss of flux into the page, the
wire loop “wants” to generate a magnetic field pointing into the page, hence
current is clockwise.
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8.2. FARADAY’S LAW
8.2
Conference 8
Faraday’s Law
Problem 8.3
A pair of conducting rails are separated by a distance d, with a uniform magnetic
field B = −B0 x̂ between them. A conducting bar (of mass M ) is free to slide
along the rails – if the bar travels to the right with initial speed v0 , find v(t),
the speed of the bar as a function of time. What is the power loss of kinetic
energy of the bar as a function of time? (Assume that a resistance R acts in
the circuit, as shown below)
B = −B0 x̂
d
R
y(t)
ẑ
ŷ
x̂
Figure 8.2: A pair of conducting rails separated by a distance d. A bar completes
the circuit, and the entire configuration is in a constant magnetic field.
The movement of the bar changes the size of the circuit, and hence the magnetic
flux through it – the flux is given by:
Φ(t) = −B0 d y(t)
(8.8)
where y(t) is the location of the bar at time t. The EMF generated is:
E =−
dΦ(t)
= B0 d v(t),
dt
(8.9)
E
B0 d v(t)
=
,
R
R
(8.10)
so the current in the circuit is:
I(t) =
where the sign is positive, indicating that current flows in the counterclockwise
direction. Now, the bar carries this current, and is therefore a wire with current
I moving in a magnetic field, so there is a force on the bar given by:
F = d I × B = −d I(t) B0 ŷ.
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(8.11)
8.2. FARADAY’S LAW
Conference 8
From Newton’s second law, we have:
M
dv(t)
B0 d v(t)
= −d B0
,
dt
R
(8.12)
or
B 2 d2
dv(t)
= − 0 v(t).
dt
MR
The solution to this differential equation is a decaying exponential:
v(t) = A e−
2 d2
B0
MR
t
(8.13)
(8.14)
for arbitrary A – since we are given the initial speed, we have:
v(t) = v0 e−
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2 d2
B0
MR
t
.
(8.15)
8.2. FARADAY’S LAW
Conference 8
Problem 8.4
A time-varying magnetic field (with magnitude B(t) = α eβ t ) points into the
page. With what velocity v(t) must you pull a loop of wire to the right such
that no current flows through the resistor? Assume at t = 0, half of the loop is
in the magnetic field region.
w
B(t) = ↵ e
t
into page
v(t) =?
R
`
x(t)
The total flux is: Φ(t) = B(t) ` (w − x(t)), and then the EMF is:
dΦ(t)
dx
= −Ḃ(t) ` (w − x(t)) + B(t) `
dt
dt
dx
= −α β eβ t ` (w − x) + α eβ t ` .
dt
E =−
We want E = 0 to keep current from flowing. Then: β (w − x) =
z ≡ w − x, then
dz
= −β z −→ z(t) = z0 e−β t ,
dt
(8.16)
dx
dt ,
and take
(8.17)
and x = w − z = w − z0 e−β t with x(0) = w/2 giving z0 = w/2 so x(t) =
w 1 − 21 e−β t and
dx
w
= β e−β t .
(8.18)
v(t) =
dt
2
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8.2. FARADAY’S LAW
Conference 8
The new component of Maxwell’s equations is the sourcing of an electric field
by a time-varying magnetic field:
I
dΦ(t)
E · d` = −
.
(8.19)
dt
This equation has interesting physical content (the change in flux through an
arbitrary surface is related to the integral of E around the boundary of that
surface), and its form is similar to another one of Maxwell’s equations:
I
B · d` = µ0 Ienc .
(8.20)
In the next problem, you will find the electric field generated by a time-varying
magnetic field, use your method from Ampere’s law to help solve (8.19) by
analogy with your solution to (8.20) for an infinite current carrying wire. You
can check your result using Faraday’s law of induction.
Problem 8.5
An infinite solenoid of radius R with n turns per unit length has slowly, linearly
increasing, current I(t) = α t in each turn. Find the electric field outside of the
solenoid. Notice that while there is no magnetic field outside the solenoid (as
far as we have been concerned), you could detect the presence of a time-varying
magnetic field inside the solenoid by watching a charged particle outside the
solenoid.
The magnetic field inside the solenoid is B = µ0 n I(t) ẑ. Thinking about the
relevant equation for E, we have:
Z
I
d
E · d` = −
B · da .
(8.21)
dt
The geometry of this setup is identical to an infinite wire sourcing a magnetic
field – so we use the same Amperian loop and ansatz. The electric field encircles
the solenoid (outside), and has magnitude that depends only on r, the distance
from the solenoid’s center. Consider a circular loop of radius r centered at
the center of the solenoid. Then the left-hand side of (8.21) reads (using our
assumptions about the form of E):
I
E · d` = E(r) 2 π r,
(8.22)
and the right-hand side is:
−
dΦ(t)
d
dI(t)
=−
B(t) π R2 = −π R2 µ0 n
.
dt
dt
dt
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(8.23)
8.2. FARADAY’S LAW
Conference 8
We can solve (8.21) for E(r):
E(r) 2 π r = −π R2 µ0 n
dI(t)
µ0 n α R 2
−→ E(r) = −
.
dt
2r
(8.24)
The electric field circulates in a direction opposite the current flow. The integral of this electric field, around a closed circular loop of radius r is E =
−µ0 n α π R2 , which is precisely what we would have gotten from Faraday’s
law.
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