Physics 1502: Lecture 17 Today’s Agenda Announcements: –Midterm 1 distributed today Homework 05...
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Transcript of Physics 1502: Lecture 17 Today’s Agenda Announcements: –Midterm 1 distributed today Homework 05...
Physics 1502: Lecture 17Today’s Agenda
• Announcements:– Midterm 1 distributed today
• Homework 05 due FridayHomework 05 due Friday
• Magnetism
Trajectory in Constant B Field
FFv
R
x x x x x x
x x x x x x
x x x x x xv B
q
x x x x x x
x x x x x x
x x x x x x
• Suppose charge q enters B field with velocity v as shown below. (vB) What will be the path q follows?
• Force is always to velocity and B. What is path? – Path will be circle. F will be the centripetal force needed to
keep the charge in its circular orbit. Calculate R:
Radius of Circular Orbit
• Lorentz force:
• centripetal acc:
• Newton's 2nd Law:
x x x x x x
x x x x x x
x x x x x xv
F
B
qFv
R
x x x x x x
x x x x x x
x x x x x x
This is an important result, with useful experimental consequences !
Ratio of charge to mass for an electron e-
3) Calculate B … next week; for now consider it a measurement
4) Rearrange in terms of measured values, V, R and B
1) Turn on electron ‘gun’
V
‘gun’
2) Turn on magnetic field B
R
&
Lawrence's Insight"R cancels R"
• We just derived the radius of curvature of the trajectory of a charged particle in a constant magnetic field.
• E.O. Lawrence realized in 1929 an important feature of this equation which became the basis for his invention of the cyclotron.
• R does indeed cancel R in above eqn. So What??– The angular velocity is independent of R!!
– Therefore the time for one revolution is independent of the particle's energy!
– We can write for the period, T=2/ or T = 2m/qB– This is the basis for building a cyclotron.
• Rewrite in terms of angular velocity !
The Hall Effect
cd
l
a
c
B
BI
I-
vd F
Hall voltage generatedacross the conductor
qEH
Force balance
Hd qEBqv =
Hd EBv =
BdvdEV dHH ==
Using the relation between drift velocity and current we can write:
tcoefficien Hall- /1 , nqRl
IBR
nql
IB
nqA
IBdBdvV H
HdH =====
The Laws of Biot-Savart & Ampere
x
Rr
P
Idx
dl
I
Calculation of Electric Field
• Two ways to calculate the Electric Field:
• Coulomb's Law:
• Gauss' Law
• What are the analogous equations for the Magnetic Field?
"Brute force"
"High symmetry"
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"
Biot-Savart Law…bits and pieces
I
dl
dBX
r
So, the magnetic field “circulates” around the wire
B in units of Tesla (T)
A
0= 4X 10-7 T m /A
Magnetic Field of Straight Wire
• Calculate field at point P using Biot-Savart Law:
• Rewrite in terms of R,:
x
Rr
P
IdxWhich way is B?
Magnetic Field of Straight Wire
x
Rr
P
Idx
1
Lecture 17, ACT 1• I have two wires, labeled 1 and 2, carrying equal
current, into the page. We know that wire 1 produces a magnetic field, and that wire 2 has moving charges. What is the force on wire 2 from wire 1 ?
(a) Force to the right (b) Force to the left (c) Force = 0
Wire 1
IX
Wire 2
IX
B
F
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
Circular Loop
x
•
z
R
R
• Circular loop of radius R carries current i. Calculate B along the axis of the loop:
• Magnitude of dB from element dl:
r dB
r
z
dB
• What is the direction of the field?• Symmetry B in z-direction.
Circular Loop
• Note the form the field takes for z>>R:
• Expressed in terms of the magnetic moment:
note the typical dipole field behavior!
x
•
z
R
R
r
r
dB
dB
z
Circular Loop
0 3
x =
y =
R
B
z
z
00
1
z3
Lecture 17, ACT 2• 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 17, ACT 3• 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 = 0 I
2 ra2
• Inside the wire: (r < a)
• Outside the wire: (r>a)
B = 0 I
2 r
0 4
x =
y =
r
B
a
Lecture 17, ACT 4• 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 17, ACT 4• 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 + oM = o (H +M) = o (H + H) = o (1+) H
» where H is magnetic field strength is magnetic susceptibility
• Alternatively, total magnetic field B can be expressed as:– B = m H
» where m is magnetic permeability» m = o (1 + )
• All the matter can be classified in terms of their response to applied magnetic field:
– Paramagnets m > o
– Diamagnets m < o
– Ferromagnets m >>> o