Download Lesson 19 (a) Biot-Savart Law Magnetic fields are produced by

Lesson 19
(a) Biot-Savart Law
Magnetic fields are produced by electric currents just as electric fields are produced
by electric charges. For a small current of length d carrying current I , the magnetic
field it creates at a point P is given by
dB =
m0 Id ´ r̂
4p r 2
Here, the vector d is in the direction of the current, r
is the distance between the current element and the
point P, r̂ is the unit vector pointing from the current
element to P, and
m0 = 4p ´10-7 T × m / A
The diagrams below show various perspectives that illustrate the relations between
the directions of the current and the magnetic field it creates. The field lines are
circles. If you grab the current element with your right hand, with the thumb aligned
with the direction of the current, your fingers would indicate the direction of the
circulating magnetic field.
Example: A small current element of length 2.0mm carries a current of 5.0A in the
direction of the positive x-axis. Find the magnetic field it creates at the points
P1 : ( 2, 0, 0) , P2 : ( 0, 0, 4) , P3 : (3, 4, 0)
Solution: I = 5.0A d = 2 ´10-3 iˆ(m)
ˆ dB µ iˆ ´ iˆ = 0
For P , r̂ = i,
For
1
P2 , r = 4m, r̂ = k̂, dB =10-7 ´ 5.0 ´
For
2 ´10-3 iˆ ´ k̂
= -0.0625 ĵ ( nT )
42
(
)
2 ´10-3 iˆ ´ 3iˆ + 4 ĵ
3iˆ + 4 ĵ
-7
P3, r = 5m, r̂ =
, dB =10 ´ 5.0 ´
= 0.032 ĵ ( nT )
5
53
1
(b) Magnetic Field of a Circular Current Loop
The diagram shows the magnetic field created by a current element in a circular
loop of radius R at the center of the loop.
Elements of the same length d everywhere on the loop
creates magnetic field of the same strength
dB =
m0 I d
4p R 2
and in the same direction. The magnetic field due the current loop at the center
therefore has the magnitude
B=
m0 I 2p R m0 I
=
4p R 2
2R
This should be multiplied by 𝑁 if the coil has 𝑁 turns.
If you make a fist with your right hand with thumb stuck out and imagine the fingers
to represent the direction of the current, the thumb would point n the direction of
the magnetic field.
The magnetic field at any point on the axis of the circular loop can also be found. Let
the loop be placed on the x-y plane with its center at the origin. The axis of the loop
coincides with the z-axis. For any current element of length d in the loop, the
distance from a chosen point P on the axis is z 2 + R 2 where z is the z-coordinate of
the P. The magnitude of the magnetic field at the point P due to the current element
is
mI d
dB = 0 2
4p z + R 2
The direction of the field is as indicated, making an angle q with the axis where
cosq =
R
z2 + R2
When the contributions from all
elements are summed, because of
symmetry, we expect the resulting
magnetic field to have only z
component. Since both dB and cosq
2
are the same for elements of the same length d , this magnetic field is
m 0 I 2p R
m 0 2p R 2 I
Bz =
cosq =
4p z 2 + R 2
4p ( z 2 + R 2 ) 3 2
The magnetic field away from the axis is much harder to calculate. Near the loop, the
field is dominated by contributions from the current elements close by. The field
lines are rings around the wire. Overall, the field lines of the current loops resemble
a fountain.
For z >> R , and introducing the magnetic moment m = p R2 I ,
the approximation
Bz =
m0 2m
4p z 3 2
holds. This is to be compared with the electric field due to an
electric dipole, with dipole moment p :
Ez = k
2p
z3 2
Indeed, the magnetic field of the current loop at
large distances can be understood as that of a
small bar magnet aligned with the magnetic
moment vector. The direction of the magnetic
moment vector is the same as the direction
from the S to the N poles of the bar magnet.
Example: A 20-turn coil of radius 20.0cm carries a current of 5.0A and a concentric
10-turn coil of radius 0.30 cm caries a current of 0.40A. The planes of the two coils
are perpendicular to each other. Find the magnitude of the torque exerted by the
large coil on the small coil.
Solution: Because of the small size of the small coil, the magnetic field it experiences
from the large coil is approximately uniform. The value is
𝜇0 𝑁𝐼 4𝜋 × 10−7 × 20 × 5.0
𝐵=
=
= 3.14 × 10−4 𝑇
2𝑅
2 × 0.2
The magnetic moment of the small coil is
𝜇 = 𝑁𝐼𝐴 = 10 × 0.4 × 𝜋 × 0.0032 = 7.16 × 10−5 𝐴 ∙ 𝑚2
3
The direction of the magnetic moment vector is perpendicular to the direction of the
magnetic field.
The torque on the small coil is
𝜏 = 𝜇𝐵𝑠𝑖𝑛𝜃 = 𝜇𝐵 = 2.24 × 10−8 𝑁 ∙ 𝑚
(3) Solenoid
A solenoid can be considered as many circular loops of currents stacked together.
The magnetic field on the axis of the solenoid can be found by adding contributions
from these loops. Choose the axis to be the z-axis, and denote the coordinates of the
two ends by z1 and z2 , and coordinate of the point P where the field is to be
calculated by z . Let R be the radius of the solenoid, L its length, and N the number
of turns. Divide the solenoid into elemental rings. The center of a typical elemental
ring has z-coordinate z¢ and thickness dz¢. In terms of the turn density
n=
N
L
the current in the elemental ring is
di = Indz¢
The ring creates the magnetic field
dBz =
m0
2p R 2 Indz¢
4p é z - z¢ 2 + R 2 ù3 2
( )
ë
û
Using the substitution
z - z¢ = Rtanq dz¢ = -Rsec2 q dq
the field due to the whole solenoid is
Bz =
z2
ò
z1
q
m0
2p R 2 nIdz¢
m0
m nI
=
2
p
nI
cosq dq = 0 (sin q1 - sinq 2 )
(
)
ò
3
2
4p é z¢ - z 2 + R 2 ù
4p
2
q
(
)
2
ë
Using the relation
û
1
sin q =
z - z¢
( z - z¢)
4
2
+ R2
and choosing the origin of the z-axis so that z1 = - L 2 z2 = L 2 , we find
ì
Bz =
í
2 ï
î
m0 nI ï
ü
ï
ý
2
2
2
2
( z + L 2) + R
( z - L 2) + R ïþ
z+L 2
z-L 2
A plot of this magnetic field as a function of z shows that it is quite constant inside
the solenoid and falls off rapidly at the edges.
A useful limit of this formula is when the length is very long and the point P is inside
the solenoid. Taking L ® ¥, we find
Bz = m0 nI
The magnetic field of a long solenoid on the outside is
similar to that of a bar magnet. The magnitude is much
higher inside, and is also uniform. A good approximation is
to consider the magnetic field to be all confined to the
interior. In this respect, it is similar to a parallel plate
capacitor in that the electric field is uniform and confined to
the space between the plates.
Example: A 25.0-meter long copper wire, 2.00mm in diameter including insulation,
is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid
of diameter 3.0cm.. What is the magnetic field at the center when the current is
15.0A?
Solution:
Radius of solenoid 𝑅 = 0.015𝑚
Circumference of a turn = 𝜋 × 0.03 = 0.094𝑚
25
Number of turns 𝑁 = 0.094 = 266
Length of solenoid 𝐿 = 266 × 0.002 = 0.53𝑚
266
Turn density 𝑛 =
= 502𝑚−1
0.53
Field at the center 𝐵 =
𝜇0 𝑛𝐼
2
[
𝐿
√(𝐿 ⁄2)2 +𝑅2
5
] = 0.998 × 𝜇0 𝑛𝐼 = 9.46 × 10−3 𝑇