Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1502: Lecture 18
Today’s Agenda
• Announcements:
– Midterm 1 distributed available
• Homework 05 due Friday
• Magnetism
Calculation of Magnetic Field
• Two ways to calculate the Magnetic Field:
• Biot-Savart Law:

I
"Brute force"
• Ampere's Law
"High symmetry"
• These are the analogous equations for the Magnetic Field!
Magnetic Fields
• Infinite line
P
q
r
R
q
• Circular loop
x
I
dx
•
Rq
r
q
z
R
x
dB
z
r
dB
Force between two conductors
• Force on wire 2 due to B at wire 1:
• Force on wire 2 due to B at wire 1:
• Total force between wires 1 and 2:
• Direction:
attractive for I1, I2 same direction
repulsive for I1, I2 opposite direction
Lecture 18, ACT 1
• Equal currents I flow in identical
circular loops as shown in the
diagram. The loop on the right (left)
carries current in the ccw (cw)
direction as seen looking along the
+z direction.
– What is the magnetic field Bz(A)
at point A, the midpoint between
the two loops?
(a) Bz(A) < 0
(b) Bz(A) = 0
I
o
I
x
B
A
x
(c) Bz(A) > 0
o
z
Lecture 18, ACT 1
• Equal currents I flow in identical
circular loops as shown in the
diagram. The loop on the right
(left) carries current in the ccw
(cw) direction as seen looking
along the +z direction.
I
o
I
x
B
A
x
o
– What is the magnetic field Bz(B) at point B, just to the right of
the right loop?
(a) Bz(B) < 0
(b) Bz(B) = 0
(c) Bz(B) > 0
z
Magnetic Field of  Straight Wire
• Calculate field at distance R
from wire using Ampere's
Law:
dl
• Choose loop to be circle of radius R
centered on the wire in a plane ^ to wire.
I R
– Why?
» Magnitude of B is constant (fct of R only)
» Direction of B is parallel to the path.
–
Evaluate line integral in Ampere’s Law:
–
Current enclosed by path = I
–
Apply Ampere’s Law:

• Ampere's Law simplifies the calculation thanks to symmetry of
the current! ( axial/cylindrical )
B Field inside a Long Wire ?
• What is the B field at a distance R,
with R<a (a: radius of wire)?
• Choose loop to be circle of radius R,
whose edges are inside the wire.
– Why?
» Left Hand Side is same as before.
– Current enclosed by path = J x Area of Loop
– Apply Ampere’s Law:

Radius a
I
R
Review: B Field of a
Long Wire
• Inside the wire: (r < a)
y=
a
b1(x);b2(x)
1
0 I r
B=
2 a2
B
• Outside the wire: (r>a)
0 I
B=
2 r
0
0
4
r
x=
x
Lecture 18, ACT 3
• A current I flows in an infinite straight
wire in the +z direction as shown. A
2A
concentric infinite cylinder of radius R
carries current I in the -z direction.
– What is the magnetic field Bx(a) at
point a, just outside the cylinder as
shown?
(a) Bx(a) < 0
(b) Bx(a) = 0
y
x
x
a
x
b
x
x 2I
I
x
(c) Bx(a) > 0
x
x
x
Lecture 18, ACT 3
• A current I flows in an infinite straight wire
in the +z direction as shown. A concentric
infinite cylinder of radius R carries current
I in the -z direction.
2B
– What is the magnetic field Bx(b) at
point b, just inside the cylinder as
shown?
(a) Bx(b) < 0
(b) Bx(b) = 0
y
x
x
a
x
b
x
x 2I
I
x
x
x
x
(c) Bx(b) > 0
B Field of a
Solenoid
• A constant magnetic field can (in principle) be produced by an
 sheet of current. In practice, however, a constant magnetic
field is often produced by a solenoid.
L
• A solenoid is defined by a current I flowing
through a wire which is wrapped n turns per
unit length on a cylinder of radius a and
length L.
• If a << L, the B field is to first order contained within the
solenoid, in the axial direction, and of constant magnitude.
In this limit, we can calculate the field using Ampere's Law.
a
B Field of a
 Solenoid
• To calculate the B field of the  solenoid using Ampere's Law,
we need to justify the claim that the B field is 0 outside the
solenoid.
• To do this, view the  solenoid from the
side as 2  current sheets.
• The fields are in the same direction in the
region between the sheets (inside the
solenoid) and cancel outside the sheets
(outside the solenoid).
• Draw square path of side w:
xxxxx
• •• • •
xxxxx
• •• • •

(n: number of
turns per unit
length)
Toroid
• Toroid defined by N total turns with
current i.
•
x
x
x
x
xx
•
• B=0 outside toroid! (Consider
integrating B on circle outside toroid) •
Apply Ampere’s Law:

•
•
xx x
•
• To find B inside, consider circle of radius
r, centered at the center of the toroid.
•
•
•
•
x
•
x
x
r xx
xx
• B•
•
•
•
Magnetic Flux
Define the flux of the magnetic field
through a surface (closed or open) from:
dS
B
B
Gauss’s Law in Magnetism
Magnetism in Matter
• When a substance is placed in an external magnetic field Bo,
the total magnetic field B is a combination of Bo and field due to
magnetic moments (Magnetization; M):
–
B = Bo + oM = o (H +M) = o (H + c H) = o (1+c) H
» where H is magnetic field strength
 c is magnetic susceptibility
• Alternatively, total magnetic field B can be expressed as:
– B = m H
» where m is magnetic permeability
» m = o (1 + c )
• All the matter can be classified in terms of their response to
applied magnetic field:
– Paramagnets
– Diamagnets
– Ferromagnets
m > o
m < o
m >>> o
Faraday's Law
dS
B
N
S
v
B
S
N
v
B
B
Induction Effects
• Bar magnet moves through coil
S
N
N
S
N
S
 Current induced in coil
• Change pole that enters
 Induced current changes sign
• Bar magnet stationary inside coil
 No current induced in coil
• Coil moves past fixed bar magnet
 Current induced in coil
v
S
N
v
v
Faraday's Law
• Define the flux of the magnetic field through a surface
(closed or open) from:
dS
B
B
• Faraday's Law:
The emf induced around a closed circuit is determined by
the time rate of change of the magnetic flux through that
circuit.
The minus sign indicates direction of induced current
(given by Lenz's Law).
Faraday’’s law for many loops
• Circuit consists of N loops:
all same area
FB magn. flux through one loop
loops in “series”
emfs add!
• Lenz's Law:
Lenz's Law
The induced current will appear in such a direction that it
opposes the change in flux that produced it.
B
B
S
N
v
N
S
v
• Conservation of energy considerations:
Claim: Direction of induced current must be so as to
oppose the change; otherwise conservation of energy
would be violated.
» Why???
• If current reinforced the change, then the change
would get bigger and that would in turn induce a
larger current which would increase the change,
etc..
•
Lecture 18, ACTy 4
A conducting rectangular loop moves with constant
velocity v in the +x direction through a region of
constant magnetic field B in the -z direction as shown.
– What is the direction of the induced current in the
loop?
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
4A
(a) ccw
(b) cw
(c) no induced current
Lecture 18, ACT 4
y
•A conducting rectangular loop
moves with constant velocity v in the
-y direction away from a wire with a
constant current I as shown.
• What is the direction of the
4B
induced current in the loop?
(a) ccw
(b) cw
I
i
v
(c) no induced current
x
Calculation
• Suppose we pull with velocity v a
coil of resistance R through a
region of constant magnetic field
B.
– What will be the induced current?
» What direction?
• Lenz’ Law  clockwise!!
xxxxxx
xxxxxx
xxxxxx
xxxxxx
x
– What is the magnitude?
» Magnetic Flux:
» Faraday’s Law:
\

I
w
v
DB  E
• Faraday's law  a changing B induces
an emf which can produce a current in
a loop.
• In order for charges to move (i.e., the
current) there must be an electric field.
 \ we can state Faraday's law more
generally in terms of the E field which
is produced by a changing B field.
x x xEx x x x x x x
E
xxxxxxxxxx
r
xxxxxxxxxx
B
xxxxxxxxxx
E
x x x x x x x xEx x
• Suppose B is increasing into the screen as shown above. An E
field is induced in the direction shown. To move a charge q
around the circle would require an amount of work =
• This work can also be calculated from
DB  E
• Putting these 2 eqns together:

• Therefore, Faraday's law can be
rewritten in terms of the fields as:
x x xEx x x x x x x
E
xxxxxxxxxx
r
xxxxxxxxxx
B
xxxxxxxxxx
E
x x x x x x x xEx x
• Note that
for E fields generated by charges at
rest (electrostatics) since this would correspond to the
potential difference between a point and itself. Consequently,
there can be no "potential function" corresponding to these
induced E fields.
Lecture 18, ACT 5
•
5A
The magnetic field in a region of space of
radius 2R is aligned with the z-direction and
changes in time as shown in the plot.
– What is sign of the induced emf in a ring of
radius R at time t=t1?
(a) e < 0
( E ccw)
(b) e = 0
(c) e > 0
( E cw)
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
t1
x
t
Lecture 18, ACT 5
5B
– What is the relation between
the magnitudes of the induced
electric fields ER at radius R and
E2R at radius 2R ?
(a) E2R = ER (b) E2R = 2ER
y
XXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
R
XXXXXXXXX
XXXXXXXX
XXXXXXX
XXXX
Bz
x
(c) E2R = 4ER
t1
t
Example
An instrument based on induced emf has been used to measure projectile
speeds up to 6 km/s. A small magnet is imbedded in the projectile, as
shown in Figure below. The projectile passes through two coils separated
by a distance d. As the projectile passes through each coil a pulse of emf
is induced in the coil. The time interval between pulses can be measured
accurately with an oscilloscope, and thus the speed can be determined.
(a) Sketch a graph of DV versus t for the arrangement shown. Consider a
current that flows counterclockwise as viewed from the starting point of
the projectile as positive. On your graph, indicate which pulse is from coil
1 and which is from coil 2.
(b) If the pulse separation is 2.40 ms and d = 1.50 m, what is the projectile
speed ?
A Loop Moving Through a Magnetic Field
Schematic Diagram of an AC Generator
dF B
d (cos( wt))
= - NAB
e= -N
dt
dt
= - NAB w sin( wt))
Schematic Diagram of an DC Generator