Extra Office Hours and Review
Extra office hours next week
‹Monday
10:30 AM – 12:00 PM Avery (Rm. 2029)
‹Tuesday 1:55 PM – 2:45 PM
Takano (Rm. 2356)
‹Thursday 1:55 PM – 2:45 PM Takano (Rm. 2356)
Review for Exam 2
‹ Tuesday
6:15 PM – 8:05 PM
‹ Place to be announced on the course homepage
PHY2049: Chapter 29
1
Law of Magnetism
ÎUnlike
‹Piece
the law of electrostatics, comes in two pieces
1: Effect of B field on moving charge
r
r
F = qv × B
‹Piece
(Chapt. 28)
2: Strength and direction of B produced by current
Equivalent!
Biot-Savart Law
Ampere’s Law
Proof of equivalence not in the book (Requires
vector calculus and relies on the absence of
magnetic monopoles)
PHY2049: Chapter 29
2
Biot-Savart Law
ÎDeduced
from many experiments on B field produced by
currents. (Can be derived from Ampere’s law too.)
‹ Magnitude
dB =
µ0 i ds sin θ
4π
r2
Like Coulomb’s law
dB=0 ahead of ds and behind it. dB
maximum on plane perpendicular to ds.
‹ Direction:
RHR #2
‹ Vector notation
r µ0 i dsr × rˆ
dB =
4π r 2
‹ Applications
Reproduces formula for B around long, current-carrying wire
„ B by current loop (on axis)
„ In general cases, numerically integrate to find B
„
PHY2049: Chapter 29
3
B Field on Axis of Circular Current Loop
ÎRadius
R and current i: find B field at
center of loop
µ 0i
B=
From B-S law by integration
2R
‹ Direction:
RHR #3 (see picture)
B=
µ0
i R2
2 R2 + z2
(
)
32
ÎIf
N turns close together
N µ 0i
B=
2R
ÎB field on axis, including center
µ0
i R2
B=
2 R2 + z2
(
‹ z=0:
)
32
From B-S law by integration
agrees
µ0 i R 2
B=
‹ z>>R:
2 z3
Like E field around electric dipole!
PHY2049: Chapter 29
4
Current Loop Example
Îi
= 500 A, r = 5 cm, N=20. Field at center?
B=N
µ 0i
2r
=
( 20 ) ( 4π ×10−7 ) 500
2 × 0.05
= 1.26TT
500 A will melt even thick copper wire. Must
cool with large flow of cooling water.
PHY2049: Chapter 29
5
Checkpoint 3
ÎFigure
shows circular loops, centered on vertical
axes perpendicular to them, carrying identical
currents. They have only two sizes: small and large.
Which arrangement gives the highest field at the
dot?
(a)
(b)
(c)
PHY2049: Chapter 29
(d)
6
Field at Center of Partial Loop
ÎDirection
of B?
ÎSuppose
partial loop covers angle φ
‹ Calculate
B field from proportion of full circle
µ 0i  φ 
B=


2 R  2π 
ÎUse
example where φ = π (half circles)
‹ Define
direction into page as positive
µ 0i  π
B=

2 R1  2π
 µ 0i  π 
−


2
2
π
R


2
µ 0i  1
1 
B=
 − 
4  R1 R2 
PHY2049: Chapter 29
7
Partial Loops (cont.)
ÎNote
on problems when you have to evaluate a B field at
a point from several partial loops
‹Only loop parts contribute, proportional to angle
(previous slide)
‹Straight sections aimed at point contribute exactly 0
‹Be careful about signs. In (b), fields partially cancel,
whereas they add in (a) and (c).
PHY2049: Chapter 29
8
Magnetic Dipole
ÎBlah
‹Read
the book
PHY2049: Chapter 29
9
Long Solenoid
ÎRegard
ÎField
it as closely stacked circular loops
outside
‹No radial component
‹Draw rectangular Amperian
loop.
z component, if exists, must be
independent of r. Must be zero.
ÎField
r
inside
‹Draw
z
another rectangular loop.
∫ B ⋅ ds = Bh + 0 + 0 + 0
ienc = i (nh)
n: turns/length
PHY2049: Chapter 29
10
(continued)
B = µ0in
‹ Uniform.
Does not depend on r (distance from axis)!
‹ This makes a solenoid useful as an electromagnet.
PHY2049: Chapter 29
11
Solenoid Example
Îi
= 200 A, B=10 T. Required n?
B
10
=
= 4000 turns/m
n=
−7
µ0i 4π × 10 (200)
ÎWhat
‹(a)
is the unit?
A/m
‹(b) A/m2
‹(c) electrons/m3
‹(d) turns
‹(e) turns/m
‹ 4000 turns/m. 1 turn every 0.25 mm!
‹ Copper wire will melt. 0.25 mm diameter
superconducting wire cannot carry 200 A. Make a
superconducting magnet with a multilayer winding.
PHY2049: Chapter 29
12
FAQ on Magnetism
ÎAccording
to the law of magnetism, a current produces a
magnetic field, which exerts a force on a moving charge.
In the phenomenon of two bar magnets attracting each
other, I see no current in magnet 1 and no moving charge
in magnet 2, and vise versa. Doesn’t this example show
that the theory of magnetism is incomplete?
‹A: bar magnets consist of magnetic ions, in which
orbital motion of electrons do not cancel out. Orbital
motion of electrons (i.e. current) in magnet 1 produces
a magnetic field. Electrons orbiting in the magnetic ions
of magnet 2 receives force from this field. The force
between two magnets can be derived from the law of
magnetism, although the calculation is lengthy and you
first need to derive the formula for the force exerted on
a magnetic dipole by non-uniform magnetic field.
PHY2049: Chapter 29
13