MAGNETIC FIELD INTENSITY

1. Magnetism – Lecture 10
Magnetic Fields & Currents
SITARAM APPARI
BVCITS ELECTRICAL Department

2. November 7, 2007
Magnetic Field Review
 Magnets only come in pairs of N
and S poles (no monopoles).
 Magnetic field exerts a force on
moving charges (i.e. on currents).
 The force is perpendicular to both
and the direction of motion (i.e.
must use cross product).
 Because of this perpendicular
direction of force, a moving
charged particle in a uniform
magnetic field moves in a circle or
a spiral.
 Because a moving charge is a
current, we can write the force in
terms of current, but since current
is not a vector, it leads to a kind of
messy way of writing the equation:
B

v

N S N S N S

 BvqFB
BLiFB



3. November 7, 2007
N S
Magnetic Field Caused by Current
 As you may know, it is possible to
make a magnet by winding wire in
a coil and running a current
through the wire.
 From this and other experiments,
it can be seen that currents create
magnetic fields.
 In fact, that is the only way that
magnetic fields are created.
 If you zoom in to a permanent
magnet, you will find that it
contains a tremendous number of
atoms whose charges whiz around
to create a current.
 The strength of the magnetic field
created by a current depends on
the current, and falls off as 1/r2.
N S
N S
N S
N S
N S
N S
N S
N S
N S
Electromagnetic
crane

4. November 7, 2007
Biot-Savart Law
 The magnetic field due to an element
of current is
 The magnetic field wraps in circles
around a wire. The direction of the
magnetic field is easy to find using the
right-hand rule.
 Put the thumb of your right hand in
the direction of the current, and your
fingers curl in the direction of B.
3
0
2
0
4
ˆ
4 r
rsdi
r
rsdi
Bd
 







sdi

Bd

(out of
page)
Biot-Savart sounds like “Leo Bazaar”
lightofspeed
1
00
 c

0 = permeability constant
exactly m/AT104 7
 


5. November 7, 2007
1. Which drawing below shows the correct
direction of the magnetic field, B, at the point P?
A. I.
B. II.
C. III.
D. IV.
Direction of Magnetic Field
I II III IV V
i i ii
P P P P P
i
B B
B into
page
B into
page
B into
page

6. November 7, 2007
 Just add up all of the contributions ds to
the current, keeping track of distance r.
 Notice that . And r sin q = R,
So the integral becomes
 The integral is a little tricky, but is
B due to a Long Straight Wire
22
sRr 
3
0
4 r
rsdi
Bd
 



 
 

0 0 3
0 sin
2
2
r
dsri
dBB
q






0 2/322
0
)(2 sR
dsRi
B


R
i
sR
s
R
i
B




22
0
0
22
0









R
i
B


2
0
 B due to current in a long straight wire

7. November 7, 2007
 Just add up all of the contributions ds to the
current, but now distance r=R is constant,
and .
 Notice that . So the integral
becomes
 For a complete loop, f = 2, so
B at Center of a Circular Arc of Wire
fRdds 
3
0
4 r
rsdi
Bd
 



 
f f


0 02
0
4
ds
R
i
dBB
R
i
Rd
R
i
B

f
f

 f
44
0
02
0
 
R
i
B

f
4
0
 B due to current in circular arc
sdr


R
i
B
2
0
 B at center of a full circle

8. November 7, 2007
 How would you determine B in the center of
this loop of wire?
B for Lines and Arcs
70
°
90°
95
°
R
2R
3R
2.43 A
T7.812T
1.0
10812.7
222.1
3
571.1
2
833.1
3
658.1
)43.2(10
7
7












RRRR
B
T7.458T
1.0
10458.7
062.5
3
571.1
2
833.1
3
658.1
)43.2(10
7
7












RRRR
B
 Say R = 10 cm, i = 2.43 A. Since 95° = 1.658
radians, 90° = 1.571 radians, 70° = 1.222
radians, 105° = 1.833 radians, we have
?
R
i
B

f
4
0
 circular arc
(out of page)
(into page)
3
0
4 r
rsdi
Bd
 



0
4 3
0



r
rsdi
Bd




9. November 7, 2007
2. The three loops below have the same current.
Rank them in terms of magnitude of magnetic
field at the point shown, greatest first.
A. I, II, III.
B. II, I, III.
C. III, I, II.
D. III, II, I.
E. II, III, I.
Magnetic Field from Loops
I. II. III.

10. November 7, 2007
 Recall that a wire carrying a current in a
magnetic field feels a force.
 When there are two parallel wires carrying
current, the magnetic field from one causes a
force on the other.
 When the currents are parallel, the two wires are
pulled together.
 When the currents are anti-parallel, the two wires
are forced apart.
Force Between Two Parallel Currents
FF
 To calculate the force on b due to a, abba BLiF


d
ia
2
0



R
i
B


2
0

d
Lii
F ba
ba


2
0
 Force between two parallel currents
BLiFB



11. November 7, 2007
3. Which of the four situations below has the
greatest force to the right on the central
conductor?
A. I.
B. II.
C. III.
D. IV.
E. Cannot
determine.
Forces on Parallel Currents
I.
II.
III.
IV.
F greatest?

12. November 7, 2007
 Ampere’s Law for magnetic fields is analogous to
Gauss’ Law for electric fields.
 Draw an “amperian loop” around a system of
currents (like the two wires at right). The loop
can be any shape, but it must be closed.
 Add up the component of along the loop, for
each element of length ds around this closed loop.
 The value of this integral is proportional to the
current enclosed:
Ampere’s Law
i1 i2
B

  encisdB 0

Ampere’s Law

13. November 7, 2007
Magnetic Field Outside a Long
Straight Wire with Current
 We already used the Biot-Savart Law to show
that, for this case, .
 Let’s show it again, using Ampere’s Law:
 First, we are free to draw an Amperian loop of
any shape, but since we know that the
magnetic field goes in circles around a wire,
let’s choose a circular loop (of radius r).
 Then B and ds are parallel, and B is constant
on the loop, so
 And solving for B gives our earlier expression.
r
i
B


2
0

  encisdB 0

Ampere’s Law  encirBsdB 02 

r
i
B


2
0


14. November 7, 2007
Magnetic Field Inside a Long
Straight Wire with Current
 Now we can even calculate B inside the wire.
 Because the current is evenly distributed over
the cross-section of the wire, it must be
cylindrically symmetric.
 So we again draw a circular Amperian loop
around the axis, of radius r < R.
 The enclosed current is less than the total
current, because some is outside the
Amperian loop. The amount enclosed is
 so
2
2
R
r
iienc



inside a straight wire
2
2
002
R
r
iirBsdB enc  

r
R
i
B 





 2
0
2
 rR
~1/r
~r
B

15. November 7, 2007
4. Rank the paths according to the value of
taken in the directions shown, most positive
first.
A. I, II, III, IV, V.
B. II, III, IV, I, V.
C. III, V, IV, II, I.
D. IV, V, III, II, I.
E. I, II, III, V, IV.
Fun With Amperian Loops
I.
II.
III.
IV.
V.
  sdB


16. November 7, 2007
Solenoids
 We saw earlier that a complete loop of
wire has a magnetic field at its center:
 We can make the field stronger by
simply adding more loops. A many
turn coil of wire with current is called a
solenoid.
 We can use Ampere’s Law to calculate
B inside the solenoid.
R
i
B
2
0

 The field near the wires is still circular,
but farther away the fields blend into a
nearly constant field down the axis.

17. November 7, 2007
Solenoids
 The actual field looks more like this:
 Approximate that the field is constant inside
and zero outside (just like capacitor).
 Characterize the windings in terms of
number of turns per unit length, n. Each
turn carries current i, so total current over
length h is inh.
 Compare with electric field in a capacitor.
 Like a capacitor, the field is uniform inside
(except near the ends), but the direction
of the field is different.
inhiBhsdB enc 00  

only section that has non-zero
contribution
inB 0 ideal solenoid

18. November 7, 2007
Toroids
 Notice that the field of the solenoid sticks out
both ends, and spreads apart (weakens) at the
ends.
 We can wrap our coil around like a doughnut, so
that it has no ends. This is called a toroid.
 Now the field has no ends, but wraps uniformly
around in a circle.
 What is B inside? We draw an Amperian loop
parallel to the field, with radius r. If the coil has
a total of N turns, then the Amperian loop
encloses current Ni.
iNirBsdB enc 002  

r
iN
B
2
0


 inside toroid

19. November 7, 2007
Current-Carrying Coils
 Last week we learned that a current-carrying
coil of wire acts like a small magnet, and we
defined the “dipole moment” (a vector) as
 The direction is given by the right-hand rule.
Let your fingers curl around the loop in the
direction of i, and your thumb points in the
direction of B. Notice that the field lines of the
loop look just like they would if the loop were
replaced by a magnet.
 We are able to calculate the field in the center
of such a loop, but what about other places.
In general, it is hard to calculate in places
where the symmetry is broken.
 But what about along the z axis?
NiA

N is number of turns, A
is area of loop

20. November 7, 2007
B on Axis of Current-Carrying Coil
 What is B at a point P on the z axis of the
current loop?
 We use the Biot-Savart Law
to integrate around the current loop, noting
that the field is perpendicular to r.
 By symmetry, the perpendicular part of B is
going to cancel around the loop, and only the
parallel part will survive.
3
0
4 r
rsdi
Bd
 






 cos
4
cos 2
0
||
r
dsi
dBdB 
r
R
cos
22
zRr 
R
zR
dsi
2/322
0
)(4 



 
 ds
zR
iR
dBB 2/322
0
||
)(4

2/322
2
0
)(2
)(
zR
iR
zB



R
i
B
2
)0( 0


21. November 7, 2007
5. The magnetic field inside a Toroid is .
Using an Amperian loop, what is the expression
for the magnetic field outside?
A. Zero
B. The same, decreasing as 1/r.
C. The same, except decreasing as
1/r2.
D. The same, except increase as r.
E. Cannot determine.
B Outside a Toroid
r
iN
B
2
0




22. November 7, 2007
Summary
 Calculate the B field due to a current using Biot-Savart Law
 Permiability constant:
 B due to long straight wire: circular arc: complete loop:
 Force between two parallel currents
 Another way to calculate B is using Ampere’s Law (integrate B around
closed Amperian loops):
 B inside a long straight wire: a solenoid: a torus:
 B on axis of current-carrying coil:
3
0
4 r
rsdi
Bd
 



0 = permeability constant
exactly m/AT104 7
 

r
i
B


2
0

R
i
B

f
4
0

R
i
B
2
0

d
Lii
F ba
ba


2
0

  encisdB 0

r
R
i
B 





 2
0
2

inB 0
r
iN
B
2
0



2/322
2
0
)(2
)(
zR
iR
zB


