Transcript 5 Ampere`s Law

Ampere’s Law
Andre Ampere
AP Physics C
Mrs. Coyle
Remember:
Biot-Savart Law: Field produced by current
carrying wires
   ds  rˆ
dB   0  I
2
 4  r
– Distance a from long straight wire
– Centre of a wire loop radius R
– Centre of a Solenoid with N turns
0 I
B
2a
B
B
0 I
2R
 0 NI
2R
Remember: There are two ways to find
the electric field around a charged object.
•
Coulomb’s Law (Superposition)
•
kdq
E 2
r
Gauss’s Law
• This is used for high symmetry cases.


surface
EdA cos  
qin
0
There are two ways to calculate
magnetic field.
• Biot-Savart Law
• Ampere’s Law
  0  ds  rˆ
dB    I
2
4

  r
 B  ds  μ
o
I
– Used for high symmetry cases.
Ampere’s Law
• For small length elements
ds on a closed path (not
necessarily circular)
 B  ds  μ
o
I
• I is enclosed current
passing through any
surface bounded by the
closed path.
• Note: dot product
• Use where there is high
symmetry
B
ds

I
Sign Convention for the Current in Ampere’s Law
I
I
negative I
positive I
B
B
ds
r
ds
r
The current I passing through a loop is positive if the
direction of B from the right hand rule is the same as the
direction of the integration (ds).
Field Outside a Long Straight Wire at a
distance r from the center, r > R
• The current is
uniformly distributed
through the cross
section of the wire
 B  ds  B( 2πr )  μ
o
μo I
B
2πr
I
Field Inside a Long Straight Wire at a
distance r from the center, r< R
• Inside the wire, the
current considered
is inside the
amperian circle
πr 
A enclosed by r 


r2
 B  ds = μ 0 Iencl = μ 0 I  A enclosed by R  = μ 0 I  πR 2  = μ 0 I R 2
2
r2
B 2 r  μ 0 I 2  B  μ I
0
R
r
Note the linear
2 R 2
relationship of B with r
Field Due to a Long Straight Wire
• The field is proportional
to r inside the wire
• The field varies as 1/r
outside the wire
• Both equations are
equal at r = R
Magnetic Field of a Toroid
• Find the field at a
point at distance r
from the center of
the toroid
• The toroid has N
turns of wire
 B  ds  B( 2πr )  μ N I
o
μo N I
B
2πr
Magnetic Field of an Thin Infinite Sheet
• Rectangular amperian
surface
• The w sides of the
rectangle do not contribute
to the field
• The two ℓ sides (parallel to
the surface) contribute to
the field
• Js =I/l is the linear current
density along the z
direction
• The current is in the y
direction
 B  ds  μ
o
Js
B  μo
2
I  μo Js
Magnetic Field of a Solenoid
• The field lines in the
interior are
– approximately parallel to
each other
– uniformly distributed
– close together
• The field is strong and
almost uniform in the
interior
Magnetic Field of a Tightly Wound Solenoid
• The field distribution
is similar to that of a
bar magnet
• As the length of the
solenoid increases
– the interior field
becomes more
uniform
– the exterior field
becomes weaker
Ideal Solenoid
– The turns are closely
spaced
– The length is much
greater than the
radius of the turns
Magnetic Field Inside a Long Solenoid
 B  ds =  B  ds   B  ds   B  ds   B  ds
1
 B  ds =
2

B
3
0

4
0

-The total current through the
rectangular path equals the
current through each turn
multiplied by the number of
turns
B = μ0 N I
B  μo
N
I  μo n I
0 = μ 0 Ienclosed
Note
• The magnetic field inside a long solenoid
does not depend on the position inside the
solenoid (if end effects are neglected).
Magnetic Field
– At a distance a from long straight wire B   0 I
2a
– At the centre of a wire loop radius R
B
– At the centre of a solenoid with N turns
-In the interior of a toroid
0 I
2R
B
0 NI
B
2 R
 0 NI
2R