View
217
Download
0
Embed Size (px)
Citation preview
Physics 1502: Lecture 18Today’s Agenda
• Announcements:– Midterm 1 distributed available
• Homework 05 due FridayHomework 05 due Friday
• Magnetism
Calculation of Magnetic Field
• Two ways to calculate the Magnetic Field:
• Biot-Savart Law:
• Ampere's Law
• These are the analogous equations for the Magnetic Field!
"Brute force" I
"High symmetry"
Magnetic Fields
x
Rr
P
Idx
• Infinite line
• Circular loop
x
•
z
R
R r dB
r
z
dB
Force between two conductors
• Force on wire 2 due to B at wire 1:
• Total force between wires 1 and 2:
• Force on wire 2 due to B at wire 1:
• 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 (c) Bz(A) > 0
x
o x
o
z
I I
A B
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.
(a) Bz(B) < 0 (b) Bz(B) = 0 (c) Bz(B) > 0
– What is the magnetic field Bz(B) at point B, just to the right of the right loop?
x
o x
o
z
I I
A B
Magnetic Field of Straight Wire• Calculate field at distance R
from wire using Ampere's Law:
• Ampere's Law simplifies the calculation thanks to symmetry of the current! ( axial/cylindrical )
dl RI
• Choose loop to be circle of radius R
centered on the wire in a plane to wire. – Why?
» Magnitude of B is constant (fct of R only)» Direction of B is parallel to the path.
– Current enclosed by path = I
– Evaluate line integral in Ampere’s Law:
– Apply Ampere’s Law:
• 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.
– Current enclosed by path = J x Area of Loop
B Field inside a Long Wire ?
RI
Radius a
– Why?» Left Hand Side is same as before.
– Apply Ampere’s Law:
Review: B Field of aLong Wire
B = μ0 I
2 π ra2
• Inside the wire: (r < a)
• Outside the wire: (r>a)
B = μ0 I
2 πr
0 4
x =
y =
r
B
a
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. – What is the magnetic field Bx(a) at
point a, just outside the cylinder as shown?
2A
(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0
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
(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0
– What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?
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.
• 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.
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.
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.
xxx xx
•• • ••• The fields are in the same direction in the
region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).
xxx xx
•• • ••• Draw square path of side w:
(n: number ofturns per unitlength)
Toroid• Toroid defined by N total turns
with current i.
• B=0 outside toroid! (Consider integrating B on circle outside toroid)
• To find B inside, consider circle of radius r, centered at the center of the toroid.
x
x
x
x
x
x
x
x
x x
x
x x
x
x
x
•
•
•
•
• •
•
••
•
• •
•
•
••
r
B
Apply Ampere’s Law:
Magnetic FluxDefine the flux of the magnetic field through a surface (closed or open) from:
Gauss’s Law in Magnetism
dS
B B
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 + μoM = μo (H +M) = μo (H + H) = μo (1+) H
» where H is magnetic field strength is magnetic susceptibility
• Alternatively, total magnetic field B can be expressed as:– B = μm H
» where μm is magnetic permeability» μm = μo (1 + )
• All the matter can be classified in terms of their response to applied magnetic field:
– Paramagnets μm > μo
– Diamagnets μm < μo
– Ferromagnets μm >>> μo
Faraday's Law
dS
B Bv
BN S
v
BS N
Induction Effects
v
vS N
vN S
N S
S N
• Bar magnet moves through coil
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
Faraday's Law• Define the flux of the magnetic field through a surface
(closed or open) from:
• 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).
dS
B B
Faraday’’s law for many loops
• Circuit consists of N loops:
all same area
B 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.
• 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..
v
BS N
v
BN S
Lecture 18, ACT 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?
(c) no induced current(a) ccw (b) cw
4A
X X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X
v
x
y
Lecture 18, ACT 4
•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 induced current in the loop?4B
(a) ccw (b) cw (c) no induced current
v
I
x
yi
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!!
x x x x x x
x x x x x x
x x x x x x
x x x x x x
vw
x
I
– What is the magnitude?
» Magnetic Flux:
» Faraday’s Law:
B Ex x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
r
E
E
E
E
B
• 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 =
• 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.
• This work can also be calculated from
B E• Putting these 2 eqns together:
• Therefore, Faraday's law can be rewritten in terms of the fields as:
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
r
E
E
E
E
B
• 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• 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?
5A
t
Bz
t1
X X X X X X X X X X XX X X X X X X X
X X X X X X X X Xx
y
X X X X X X X X XX X X X X X X X X X X X X X X
X X X X
R
(a) < 0( E ccw)
(b) = 0 (c) > 0( E cw)
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 (c) E2R = 4ER
t
Bz
t1
X X X X X X X X X X XX X X X X X X X
X X X X X X X X Xx
y
X X X X X X X X XX X X X X X X X X X X X X X X
X X X X
R
ExampleAn 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 V 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
(t
) =
?
(t)
= ?
F(t
) =
?
Schematic Diagram of an AC Generator
d
dtB
d (cos( t))dt
sin( t))
Schematic Diagram of an DC Generator