Transcript Document

Ampere’s Law
Outline
• Introduce Ampere’s Law as an analogy to
Gauss’ Law.
• Define Ampere’s Law.
• Show how to use Ampere’s Law for cases
with symmetry.
Student Objectives
• Recognise Ampere’s Law to be analogous
to Gauss’ Law.
• Recognise similar concepts: (1) draw an
imaginary shape enclosing the current
carrying conductor, (2) current enclosed.
Ampere’s Law
• Gauss’ law allowed us to find the net
electric field due to any charge distribution
(with little effort) by applying symmetry.
• Similarly the net magnetic field can be
found with little effort if there is symmetry
using Ampere’s law.
Ampere’s Law
• Ampere’s law,

 
B.ds   0ienc
• Where the integral is a line integral.
• B.ds is integrated around a closed loop called
an Amperian loop.
• The current ienc is net current enclosed by the
loop.
Ampere’s Law
• ie,

 
B.ds   0
• ie ienc 
N
i
n 1
N
i
n 1
n
n
Ampere’s Law
• The figure shows the cross section of 3
arbitrary long straight wires with current as
shown.
i1
i3
i2
Ampere’s Law
• Two of the currents are enclosed by an
Amperian loop.
i1
i3
i2
Ampere’s Law
• An arbitrary direction for the integration is
chosen.
i1
Direction of
integration
i3
i2
Ampere’s Law
• The loop is broken into elements of length ds
(choose in the direction of the integration).
• Direction of B doesn’t need to be known
before the integration!


ds

B
i1
Direction of
integration
i3
i2
Ampere’s Law
• B can be in an arbitrary direction at some
angle to ds as shown (from the right hand
grip rule we know B must in the plane of
page).

ds
• We choose B to be in the

i1
direction as ds.

B
Direction of
integration
i3
i2
Ampere’s Law
• The right hand screw (grip) rule is used to
assign a direction to the enclosed currents.
• A current passing through the loop in the
same direction as the thumb are positive ( in
the opposite direction -ve).
Ampere’s Law
• Consider the integral,
N
 
B.ds   0
in  B cosds

 
n 1

ds


B
i1
Direction of
integration
i3
i2
Ampere’s Law
• Applying the screw rule,

B cosds   0 i1  i2 

ds


B
i1
Direction of
integration
i3
i2
Ampere’s Law
• Example. Find the magnetic field outside a
long straight wire with current.
I
r
Ampere’s Law
• We draw an Amperian loop and the
direction of integration.
Amperian Loop
Wire surface
Direction of
Integration

ds

B
 0
Ampere’s Law
 
B.ds   0
N
i


• Therefore, B cosds  B  ds  B2r 
• Recall,
n
n 1
 B2r     0 I
0 I
B 
2r
• The equation derived earlier.
Ampere’s Law
• The positive sign for the current
collaborates that the direction of B was
correct.
Ampere’s Law
• Example. Magnetic Field inside a Long
Straight wire with current.
Wire surface
r

ds
R

B
Amperian Loop
Ampere’s Law
 
• Ampere’s Law, B.ds   0



B cosds  B ds  B2r 
N
i
n 1
n
Ampere’s Law
 
• Ampere’s Law, B.ds   0



B cosds  B ds  B2r 
N
i
n
n 1
• The charge enclosed is proportional to the
area encircled by the loop,
r
 2i
R
2
ienc
Ampere’s Law
• The current enclosed is positive from the
right hand rule.
2
r
 B2r    0i 2
R
 0i
B
r
2
2R