Unformatted text preview: Physics 227: Lecture 16
Ampere’s Law • Lecture 15 review: •
• Magnetic ﬁeld magnitudes for charged
particle or current. Ratio of magnetic to electric force for two
charged particles. •
• Long straight wire: B = μ0I/2πr FM
2
= µ0 0 v
FE Parallel wires: F/L = μ0I2/2πr. µ0 q × r
vˆ
r
B () =
4π r 2
vˆ
µ0 Id × r
lˆ
µ0 q × r
() = nAdl
=
dB r
2
4π r
4π r 2
Monday, October 31, 2011 Physics 227:
Exam 2 Information • Note: exam 2: 16 questions covering chapters 25  28 •
• •
•
Monday, October 31, 2011 Thursday, Nov 17, 2011, 9:40 PM  11:00 PM
Room assignments: • AI Arc 103
• JM SEC 111
• NR PLH
• SZ Beck Auditorium, Livingston Campus!!! (NOT Hill 114) Anyone with a conﬂict should contact Prof. Cizewski ASAP
Bring pencils, 1 formula sheet w/ anything you want, NO
calculators or other electronics needed or allowed! Ampere’s Law
vs. Ampere’s Law involves a line integral over a closed path of the
magnetic ﬁeld. It relates the integral to the current through
the surface deﬁned by the closed path.
Like with Gauss’s Law in electrostatics, we will use it in
situations with a symmetry that allows us to calculate the ﬁeld. Monday, October 31, 2011 •
•
• Ampere’s Law Let’s do a simple case we know ﬁrst  a long straight wire
w ith magnetic ﬁeld B = μ0I/2πr. B is constant for ﬁxed r.
Thus the path is a circle centered on the wire, in a plane
perpendicular to it. B is tangential to the circle.
We ﬁnd the integral is ∫B.dl = BC = (μ0I/2πr)(2πr) = μ0I. •
•
• Monday, October 31, 2011 If we reverse the direction of
the current, the B ﬁeld
changes direction, and the sign
of the integral changes.
If we change the direction we
go around the circle, B.dl and
the integral change sign.
Thus, we ﬁnd the integral is
±μ0I, depending on how we
choose directions. iClicker: Ampere’s Law
with no Current Enclosed What is the magnitude of
∫B.dl around the quartercircle path shown?
A. 0.
B. μ0I.
C. μ0I (r1/r2).
D. (π/2) μ0I (r1/r2)2. E. (π/2) μ0I (r2r1).
Numerical integral done in
lecture. Also, since no current
is inside the loop, the integral
has to be 0. Monday, October 31, 2011 What if There are Several
Currents?
Add them up.
Note the direction to use  if the
thumb points in the direction of the
positive current, the direction
around is the way the RH ﬁngers
curl. In this case, the current is >0
and the integral will be +μ0I.
Positive current out of clock face:
integrate around CCW.
Positive current into clock face,
integrate around CW. Monday, October 31, 2011 A Note on Conservative Forces
The electric force is conservative. When a charge moves along
some path and returns to a previous position, the energy is the
same. The total work done by the electric force vanishes (=0).
Mathematically: W= l
q E · d → l
E · d = 0 The magnetic force is always perpendicular to the direction of
motion, so necessarily: W=
Fm · d = 0
l While there is often a symmetry between how we handle electric
and magnetic ﬁelds, the line integrals of the ﬁelds over a closed
loop are unrelated, with very different meanings.
Monday, October 31, 2011 iClicker: Ampere’s Law for
Uniform Current Density What if we have uniform
current density inside a wire?
How will B vary with r?
A. B is 0 inside the wire.
B. B is constant inside the
w ire.
C. B ≈ r.
D . B ≈ r 2. E. B ≈ 1/r. Answer derived on next slide. Monday, October 31, 2011 Use Ampere’s Law to Determine the Field of a
Uniform Current Density Wire •
•
•
•
• Current density in the
w ire is J = I/πR2.
For a circle of radius r:
∫B.dl = B 2πr =
μ0I = μ0 Jπr2 ➪
B = μ0Jr/2 = μ0Ir/2πR2.
The ﬁeld inside the
w ire varies with r.
Outside the wire, we
again have B = μ0I/2πr.
Note the units: B = μ0
x current / length Monday, October 31, 2011 Field of a Uniform Current Density Wire Monday, October 31, 2011 What is the Magnetic Field of a Solenoid?
Solenoid: a group of parallel, coaxial coils, with a current that ﬂows
through all of them. You can see from the drawing that as you get more coils, the ﬁeld
inside the solenoid grows, while the ﬁeld outside decreases.
In the limit of an inﬁnitely long solenoid, the ﬁeld inside is constant
while the ﬁeld outside vanishes.
Apply Ampere’s Law as shown above to the right...
Monday, October 31, 2011 Is it ``Obvious’’ that the Field Outside an
Infinitely Long Solenoid Vanishes?
The ``upper’’ wires lead to a ﬁeld in
the plane of the board that is
to the right above the wires
to the left below the wires
The ``lower’’ wires lead to a ﬁeld in
the plane of the board that is
to the left above the wires
to the right below the wires
The contributions cancel above and
below the solenoid, but add inside it.
Because the ﬁeld vanishes outside,
the ﬁeld inside cannot diverge  it
must remain parallel and constant.
Monday, October 31, 2011 iClicker: What is the Magnetic Field...?
... inside an inﬁnitely long solenoid?
Use n loops/m, current I A, radius
r m, an Amperian loop on length L,
and ∫B.dl = μ0I.
A. It depends on
where points
a&b are inside
the solenoid.
B. B = μ0nI.
C. B = μ0nIL.
D. B = μ0nI/L. E. B = μ0nI/L2.
Monday, October 31, 2011 Previous slides indicated the ﬁeld is constant. It
cannot depend on the length of the loop L that
you draw on the inﬁnite solenoid. Also recall
that the units of B are μ0 x current / length.
Since n has units of 1/length, only answer B has
the right units. What is the Magnetic Field of a ``Short’’
Solenoid? The ﬁeld is not constant, so an integration is needed, rather than
Ampere’s Law. Monday, October 31, 2011 What is the Magnetic Field for a toroid? The ﬁeld is constant. Use Ampere’s Law with n loops/m, current I A,
radius r m: ∫B.dl = μ0I ➪ B 2πr = μ0 In2πri ➪ B = μ0nIri/r. We can
use N = 2πrin to obtain B = μ0NI/2πr.
In the limit that ri ≈ ro, we also have ri ≈ r, and B = μ0nI, the same
as for an inﬁnitely long solenoid.
Monday, October 31, 2011 Thank you.
See you Thursday. Monday, October 31, 2011 ...
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This document was uploaded on 11/03/2011 for the course PHSYICS 227 at Rutgers.
 Fall '11
 StaceyJacos

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