Ch. 30
•Magnetic fields go around the wire – they are
perpendicular to the direction of current
•Magnetic fields are perpendicular to the
separation between the wire and the point where
you measure it - Sounds like a cross product!
The BiotSavartI Law
dB =
r
2
ds  rˆ
r
I
ds
0 I
B
4
ds  rˆ
0  4 107 T  m/A
 r 2 •Permeability of free space
0 I
B
4
ds  r
 r3
•The Amp is defined to work out this way
Sample Problem
A loop of wire consists of two quarter
circles of radii R and 2R, both centered at
a point P, and connected with wires going
radially from one to the other. If a
current I flows in the loop, what is the
magnetic field at the point P?
0 I
B
4
ds
ds  rˆ
 r2
ds
r̂
r̂
r̂ ds
•Do one side at a time
P
r̂ds
•First do one of the straight segments
•ds and r-hat are parallel ds  rˆ  0
R
2R
•No contribution to the integral
0 I ˆ
•Other straight segment is the same
B
k
ˆ
•Now do inner quarter loop (assume x-y) ds  rˆ  kds
16 R
Binner
0 I ˆ ds
0 I ˆ 2 R  0 I ˆ

k

k 2
k
2

8R  I
4
R
4 R
4
B outer   0 kˆ
•Outer loop opposite direction, similar
16 R
I
Magnetic Field from a Finite Wire
r   xˆi  aˆj
•Magnetic field from a finite straight wire:
•Let a be the distance the point is from the wire
•Let x be the horizontal separation
P
1
-x1
a
r
O
x
cos 1 
cos  2 
ds 2
I
r
x2  a2
ds  ˆidx
ˆidx   xˆi  aˆj
  
0 I x
B
32


x
2
2
4
x  a 
0 Ia ˆ x
dx

k
x
2
2 32
4
x  a 
2

1
x2
x1
x12  a 2
x2
x22  a 2
Warning: My  differs
from that of the book
2
1
0 Ikˆ 
x2



4 a  x22  a 2



x12  a 2 
x1
0 I
B
 cos 1  cos  2 
4 a
Magnetic Field from a Wire
0 I
•Magnitude is found from the formula
B
 cos 1  cos  2 
4 a
•Direction is found from the right hand rule
•Place thumb in direction of current flow
•Fingers curl in direction of B-field
•Infinite wire:
0 I
1
B
•Angles are simple
1   2  0
cos 1  cos  2  1
2 a
Two wires have the same current I flowing
through them. If we want the magnetic field
between them to be large and up, we should
have the current in the upper one flow _____
and the lower one ______
A) Left, Right
B) Left, Left
D) Right, Left
E) Right, Right
2
I
a
CT-1-A battery establishes a steady current around the circuit below. A compass needle is placed in the plane
of the circuit successively just to the above points P, Q, and R (where the letters are) in the plane of the circuit.
The relative deflection of the needle, in de-scending order, is
A.
B.
C.
D.
E.
P, Q, R.
Q, R, P.
R, Q, P.
P, R, Q.
Q, P, R
Ans E
Sample problem
A regular hexagon whose center is a
distance a = 1 cm from the nearest side has
current I = 4.00 A flowing around it. The
current flows N = 500 times around. What
is the total magnetic field at the center?
I
60
a
60 60
•Draw in the two directions from the center to the corners of one segment
•Top angle is one-sixth of a circle, or 60 degrees
•Total angles in circle is 180, so other two angles are 60 each
•Use formula to get magnetic field – right hand rule says up.
0 I
0 I
B
 cos 60  cos 60  
4 a
4 a
•Multiply by all six side, and then by 500 cycles
Btot
6 0 IN


4 a
6  4  10 7 T  m/A   4 A  500
4  0.01 m 
 0.120 T
I
Right Hand Rule for Loops
•If you curl your fingers in the direction the
current flows, thumb points in direction of
B-field inside the loop
•Works for solenoids too (later)
B
Curl fingers around I and
Thumb point to Mag
dipole moment.
Also gives B inside a
loop
Warmup 14
Force Between Parallel Wires
I1
I2
F
d
L
•One wire – infinite – creates a magnetic field
•Other wire – finite or infinite – feels the force
F2  I 2L  B
 0 I1
B
2 d
F 0 I1 I 2

L
2 d
•Attractive if current is parallel, repulsive if anti-parallel
CT-2 Two parallel wires are connected to two separate DC power supplies so that the direction of their currents
are initially parallel. When the leads to one of the wires are reversed
I1
+
-
I2
+
-
A. The wires move closer to each other
B. The wires mover apart
C. The wires do not move.
Ans B
CT-3- Consider two parallel wires carrying currents I1 and I2 respectively. The wires are a small distance a
apart. Which of the following (is) are true:
A. If I1 = 2I2 and the directions of the currents are in the same direction, then the attractive force on the wire
carrying I2 is 2 times that on the wire carrying I1.
B. If I1 = 2I2 and the directions of the currents are in the same direction, then the attractive force on the wire
carrying I1 is 2 times that on the wire carrying I2.
C. If the magnitudes of the currents are the same but their directions are opposite to each other the magnetic
field at a distance r > a is twice what it would be if only one wire were present.
D. If the magnitudes of the currents are the same but their directions are opposite to each other the magnetic
field at a distance r > a is zero or close to zero.
E. Two of the above
F. None of the above [Don’t click]
Ans D
Warmup 14
Ampere’s Law (original recipe)
•Suppose we have a wire coming out of the plane
•Let’s integrate the magnetic field around a closed path
•There’s a new symbol for such an integral
•Circle means “over a closed loop”
•The magnetic field is parallel to direction of integration
0 I
 B  ds  B  ds  2 r 2 r 0 I
ds
ds cos
•What if we pick a different path?
 0 I 
 B  ds   Bds cos     2 r  rd
ds cos   rd
B  ds   I

0 I
B
2 a
d
r
I
0
•We have demonstrated this is true no matter what path you take
•We don’t even need a straight infinite wire
•All that matters is that current passes through the closed Ampere
loop
Understanding Ampere’s Law
•If multiple currents flow through, add up all that are inside the loop
•Use right-hand rule to determine if they count as + or –
B  ds  0 I
•Curl fingers in direction of Ampere loop
•If thumb points in direction of current, plus, otherwise minus
•The wire can be bent, the loop can be any shape, even non-planar

There are currents going in and out of the
screen as sketched at right. What is the
ingtegral of the magnetic field around the path
sketched in purple?
A) 0(11 A) B) 0(-11 A) C) 0(3 A)
D) 0(-3 A) E) None of the above
•Right hand rule causes thumb to point down
•Downward currents count as +, upwards as –
 B  ds    4 A  2 A  5 A 
0
1A
2A
4A
5A
7A
JIT Quick Quiz 30.3
Rank the magnitudes of
·d for the closed paths in the figure, from greatest to least.
Ans c>a>d>b
JIT Quick Quiz 30.4
Rank the magnitudes of
·d for the closed paths in the figure, from greatest to least.
Ans a=c=d>b
Using Ampere’s Law
•Ampere’s Law can be used – rarely – to calculate magnetic fields
•Need lots of symmetry – usually cylindrical
A wire of radius a has total current I distributed uniformly across its cross-sectional area.
Find the magnetic field everywhere.
I
I
End-on view
•Draw an Ampere loop outside the wire – it contains all the current
•Magnetic field is parallel to the direction of this loop, and constant
around it
0 I  B  ds  B ds  2 rB
•Use Ampere’s Law
•But we used a loop outside the wire, so we only have it for B   0 I
2 r
r>a


Using Ampere’s Law (2)
•Now do it inside the wire
•Ampere loop inside the wire does not contain all the current
•The fraction is proportional to the area I r
 r2
I r  Ir
0 I r 
2
a
2
I

a
2
 B  ds  2 rB
0 I r
0 Ir
B

2 r
2 a 2
 0 I 2 r
B
2

Ir
2

a
 0
r  a End-on view
ra
a
Warmup 15
Solenoids
•Consider a planar loop of wire – any
shape – with a current I going around it
•Now, stack many, many such loops
•Treat spacing as very closely spaced
•Assume stack is tall compared to
size of loop
•Outside, most of the field cancels for
ideal solenoid (long, thin, tightly packed)
Boutside neglible (ideal solenoid)
Field Inside a Solenoid
•It remains only to calculate the magnetic field inside
•We use Ampere’s law
•Recall, neglible B-field outside
0 I tot 
 B  ds
 Bin L
I tot  NI
•There may be many (N) current loops
within this Ampere loop
•Let n = N/L be loops per unit length
Bin 
0 NI
L
Bin  0 nI
L
•Works for any shape solenoid, not just cylindrical
•For finite length solenoids, there are “end effects”
Magnetic Flux
•Magnetic flux is defined exactly the same way
for magnetism as it was for electricity
 B   B  nˆ dA
A cylindrical solenoid of radius 10 cm has
length 50 cm and has 1000 turns of wire going
around it. What is the magnetic field inside
it, and the magnetic flux through it, when a
current of 2.00 A is passing through the wire?
Bin  0 nI  4 107 T  m/A
Bin  0.00503 T
1000  2 A 
0.5 m
 B  BA  B R   0.00503 T    0.1 m 
2
 B  1.579  104 T  m 2
A Tesla meter2 is also
called a Weber (Wb)
2
Gauss’s Law for Magnetism
•Magnetic field lines always go in circles – there are no magnetic
monopole sources
•For closed surfaces, any flux in must go out somewhere else
B 
 B  nˆ dA  0
A regular tetrahedron (four sides, all congruent) has a cylindrical magnet placed in the middle of the bottom face. There
is a total of 0.012 Tm2 of magnetic flux entering the bottom
face. What is the total flux from one of the three top faces?
A) 0.006 T m2
B) 0.004 T m2
C) 0.003 T m2
D) 0.012 T m2
E) None of the above
•Flux in bottom must equal total flux out other three sides
•Other three sides must have equal flux, by symmetry
1   2   3  0.012 T  m 2 31  0.012 T  m 2
Sign of
entering
flux is
negative
CT-4 - A sphere of radius R is placed near a long, straight wire that carries a steady current I. The
magnetic field generated by the current is B. The total magnetic flux passing through the sphere is
A. o I.
B. o I/(4 R 2 ).
C. 4 R 2/ o I.
D. zero.
E. need more information
Ans D
Ex – Serway 30.47 A cube of edge length = 2.50 cm is positioned as shown below. A uniform magnetic field
given by B = (5.00 i + 4.00 j + 3.00 k) T exists throughout the region. (a) Calculate the flux through the shaded
region. (b) What is the total flux through the six faces?
Solve on
Board
Warmup 15
http://www.youtube.com/watch?v=A1vyB-O5i6E