MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2003
IX. Source of Magnetic Fields – Worked Examples
Example 1: Current-carrying arc
Consider the current-carrying loop formed of radial lines and segments of circles whose
centers are at point P as shown below.
G
Find the magnetic field B at P.
Solution:
According to the Biot-Savart Law, the magnitude of the magnetic field due to a
G
differential current-carrying element Id s is given by
G
µ I d s × rˆ µ0 I r dθ µ0 I
dB = 0
dθ
=
=
r2
4π
4π r 2
4π r
(1.1)
For the outer arc, we have
Bouter =
µ0 I θ
µ Iθ
dθ = 0
∫
4π b 0
4π b
(1.2)
G
G
The direction of B outer is determined by the cross product d s × rˆ which points into the
page.
Similarly, for the inner arc, we have
Binner =
µ0 I θ
µ Iθ
dθ = 0
∫
0
4π a
4π a
(1.3)
1
G
G
For Binner , d s × rˆ points out of the page.
Therefore, the total magnitude of magnetic field is
G G
G
µ Iθ  1 1 
B = Binner + B outer = 0  −  (out of page)
4π  a b 
(1.4)
Example 2: Rectangular current loop
Determine the magnetic field (in terms of I, a and b) at the origin due to the current loop
shown below.
Solution:
For a finite wire carrying a current I, its contribution to the
magnetic field at a point P is given by
B=
µ0 I
( cos θ1 − cos θ 2 )
4π r
(2.1)
where θ1 and θ 2 are the angles which parameterize the
length of the wire.
To obtain the magnetic field at O, we make use of the above formula. The cobtributions
can be divided into 3 parts:
2
(i) Consider the left segment of the wire which extends from ( x, y ) = (− a, +∞) to
(− a, + d ) . The angles which parameterize this segment give cos θ1 = 1 ( θ1 = 0 ) and
cos θ 2 = d / d 2 + a 2 . Therefore,
B1 =

µ0 I
µI
d
( cosθ1 − cosθ 2 ) = 0 1 − 2 2 
4π a
4π a 
d +a 
(2.2)
G
The direction of B1 is out of page, or +kˆ .
(ii) Next ,we consider the segment which extends from ( x, y ) = (− a, + d ) to (+ a, + d ) .
Again, the (cosine of the) angles are given by
a
cos θ1 =
(2.3)
a2 + d 2
a
cos θ 2 = cos (π + θ1 ) = − cos θ1 = −
(2.4)
2
a + d2
This leads to
B2 =

µ0 I 
µ 0 Ia
a
a
+
 2
=
2
2
2
4π d  a + d
a + d  2π d a 2 + d 2
(2.5)
G
The direction of B 2 is into the page, or −kˆ .
(iii) The third segment of the wire runs from ( x, y ) = (+ a, + d ) to (+ a, +∞) . One may
readily show that it gives the same contribution as the first one:
B3 = B1
(2.6)
G
The direction of B3 is again out of page, or +kˆ .
The total magnitude of the magnetic field is
G G G
G
G G
B = B1 + B 2 + B3 = 2B1 + B 2
=
=
µ0 I 
d
1 − 2
2π a 
a + d2
µ0 I
2π ad a + d
2
2
(d
ˆ
µ0 Ia
kˆ
k −
2
2
2π d a + d

(2.7)
)
a 2 + d 2 − d 2 − a 2 kˆ
3
Example 3: Hairpin
An infinitely long current-carrying wire is bent into a hairpin-like shape shown in the
figure below.
Find the magnetic field at the point P which lies at the center of the half-circle.
Solution:
Again we break the wire into three parts: two semi-infinite plus a semi-circular segments.
(i) Let P be located at the origin on the xy plane. The first semi-infinite segment then
extends from ( x, y ) = (−∞, − r ) to (0, − r ) . The two angles which parameterize this
segment are characterized by cos θ1 = 1 ( θ1 = 0 ) and cos θ 2 = 0 (θ 2 = π / 2) . Therefore, its
contribution to the magnetic field at P is
B1 =
µ0 I
µI
µI
( cos θ1 − cos θ 2 ) = 0 (1 − 0) = 0
4π r
4π r
4π r
(3.1)
G
The direction of B1 is out of page, or +kˆ .
(ii) For the semi-circular arc of radius r, we make use of the Biot-Savart law:
G µ0 I d Gs × rˆ
B=
4π ∫ r 2
(3.2)
and obtain
B2 =
µ0 I
4π
∫
π
0
rdθ µ0 I
=
r2
4r
(3.3)
G
The direction of B 2 is out of page, or +kˆ .
(iii) The third segment of the wire runs from ( x, y ) = (0, + r ) to (−∞, + r ) . One may readily
show that it gives the same contribution as the first one:
4
B3 = B1 =
µ0 I
4π r
(3.4)
G
The direction of B3 is again out of page, or +kˆ .
The total magnitude of the magnetic field is
G G G
G
G G
µI
µI
B = B1 + B 2 + B3 = 2B1 + B 2 = 0 kˆ + 0 kˆ
2π r
4r
µI
= 0 (2 + π )kˆ
4π r
(3.5)
Notice that the contribution from the two semi-infinite wires is equal to that due to an
infinite wire:
G G
G
µI
B1 + B3 = 2B1 = 0 kˆ
2π r
(3.6)
The result can be readily obtained by using Ampere’s law.
Example 4: Two infinitely long wires
Consider two infinitely long wires carrying currents are in the negative x direction.
(a) Plot the magnetic field pattern in the yz plane.
(b) Find the distance d along the z axis where the magnetic field is a maximum.
Solution:
(a) The magnetic field lines are shown in the figure below. Notice that the directions of
both currents are into the page.
5
(b) The magnetic field at (0, 0, z) due to the left wire is, using Ampere’s law:
B1 =
µ0 I
µ0 I
=
2π r 2π a 2 + z 2
(4.1)
Since the current is flowing in the –x direction, the magnetic field points in the direction
of the cross product
(−ˆi ) × rˆ1 = (−ˆi ) × (cos θ ˆj + sin θ kˆ ) = sin θ ˆj − cos θ kˆ (4.2)
Thus, we have
G
B1 =
µ0 I
2π a + z
2
2
(sin θ ˆj − cosθ kˆ )
(4.3)
For the right wire, the magnetic field strength is the same as the left one: B1 = B2 .
However, its direction is given by
(−ˆi ) × rˆ2 = (−ˆi ) × (− cos θ ˆj + sin θ kˆ ) = sin θ ˆj + cos θ kˆ
(4.4)
Adding up the contributions from both wires, the z components cancel (as required by
symmetry), and we arrive at
G G G
µ I sin θ ˆ
µ0 Iz ˆ
B = B1 + B 2 = 0
j=
j
2
2
π (a 2 + z 2 )
π a +z
(4.5)
To locate the maximum of B, we set dB / dz = 0 and find
6
 µ0 I a 2 − z 2
dB µ0 I  1
2z2
=
−
=0

=
π  a 2 + z 2 (a 2 + z 2 ) 2  π ( a 2 + z 2 )2
dz
(4.6)
z=a
(4.7)
which gives
Thus, at z=a, the magnetic field strength is a maximum, with a magnitude
Bmax =
µ0 I
2π a
(4.8)
Example 5: Non-uniform current density
Consider an infinitely long, cylindrical conductor of radius R carrying a current I with a
non-uniform current density
J = br
(5.1)
where b is a constant. Find the magnetic field everywhere.
Solution:
The problem can be solved by using the Ampere’s law:
G G
B
v∫ ⋅ d s = µ0 I enc
(5.2)
where the enclosed current Ienc is given by
G G
I enc = ∫ J ⋅ dA = ∫ ( br )( 2π rdr )
(5.3)
(a) For r < R , the enclosed current is
r
I enc = ∫ 2π br 2 dr =
0
2π br 3
3
(5.4)
7
Applying Ampere’s law, the magnetic field at P1 is given by
B1 ( 2π r ) =
2 µ0π br 3
3
(5.5)
or
B1 =
bµ 0 2
r
3
(5.6)
G
The direction of the magnetic field B1 is tangential to the Amperian loop which encloses
the current.
(b) For r > R , the enclosed current is
R
I enc = ∫ 2π br 2 dr =
0
2π bR 3
3
(5.7)
which yields
B2 ( 2π r ) =
2µ0π bR 3
3
(5.8)
Thus, the magnetic field at a point P2 outside the conductor is
B2 =
bµ 0 R 3
3r
(5.9)
A plot of B as a function of r is depicted below:
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Example 6: Helmholtz coils
The Helmholtz coils consist of two circular coils of radius R, each perpendicular to the
symmetric axis, with their centers being separated by a distance R. There is a steady
current I flowing in the same direction around each coil.
(a) Find the magnetic field on the axis at a distance x from the center of one coil.
(b) Verify that dB / dx and d 2 B / dx 2 are both zero at a point midway between the coils.
Solution:
G
We first consider the case where there is only one coil. Since each length element d s is
perpendicular to the vector rˆ and each length elements around the loop are at the same
distance from the x-axis, hence at point P, we have
G
µ I d s × rˆ µ0 I ds
=
dB = 0
4π r 2
4π x 2 + R 2
(6.1)
By symmetry, the magnetic field must point along the x-axis:
dBx = dB cos θ =
µ0 I ds cos θ µ0 I
ds R
=
2
2
2
4π x + R
4π ( x + R 2 )3/ 2
(6.2)
9
where cos θ = R / x 2 + R 2 . Thus, the magnetic field at P is
Bx =
µ0 IR
4π ( x 2 + R 2 )3/ 2
µ0 IR
v∫ ds = 4π ( x
2
+ R 2 )3/ 2
( 2π R ) =
µ0 IR 2
2( x 2 + R 2 )3/ 2
(6.3)
Next, by applying the superposition principle, the magnetic field at P (a point at a
distance x away from one center and R− x from the other) due to the two coils can be
obtained as:
B = Bx1 + Bx2 =
µ0 IR 2 
2

1
1
 ( x 2 + R 2 )3/ 2 + [( x − R) 2 + R 2 ]3/ 2 


(6.4)
(b) Differentiating B with respect to x, we obtain


dB µ0 IR 2 
3x
3( x − R)

=
−
−

5
/
2
2
2 5/ 2
dx
2  2( x + R )
( x − R ) 2 + R 2  

 

(6.5)
At the midpoint where x = R / 2 , the derivative vanishes:
dB
dx
=0
(6.6)
x= R / 2
Similarly, we can also show that
d 2 B −3µ0 IR 2
=
dx 2
2


1
5x2
1
5( x − R) 2
(6.7)
−
+
−
 2
2 5/ 2
2
2 7/2
2
2 5/ 2
2
2 7/2 
(x + R )
[( x − R) + R ]
[( x − R) + R ] 
(x + R )
which yields
d 2B
dx 2
=0
(6.8)
x= R / 2
The fact that the first two derivatives vanish at x = R / 2 indicates that the magnetic field
is fairly uniform there.
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Example 7: Thin strip of metal
Consider an infinitely long, thin strip of metal of width w lying on the xy plane. The strip
carries a current I along the +x direction, as shown in the figure below.
Find the magnetic field at point P which is in the plane of the strip and at a distance b
away from it.
Solution:
Consider a thin strip of width dr parallel to the direction of the current and at a distance r
away from P. The amount of current carried by this differential element is
 dr 
dI = I  
w
(7.1)
Using the Ampere’s law, we see that its contribution to the magnetic field at P is given by
dB(2π r ) = µ0 I enc = µ0 (dI )
(7.2)
µ 0 dI µ 0  I dr 
=


2π r 2π r  w 
(7.3)
or
dB =
Integrating over the expression, we obtain
B=∫
b+ w
b
µ 0 I  dr  µ 0 I  b + w 
ln 
 =

2π w  r  2π w  b 
(7.4)
Using the right-hand rule, the direction of the magnetic field can be shown to point in the
+z direction, or
G µ I  w
B = 0 ln  1 +  kˆ
(7.5)
2π w  b 
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Example 8: Two semi-infinite wires
A wire carrying current I runs down the y axis to the origin, thence out to infinity along
the positive x axis. Show that the magnetic field in the quadrant with x, y > 0 of the xy
plane is given by
Bz =
µ0 I  1 1
x
y
 + +
+
2
2
2

4π  x y y x + y
x x + y2




(8.1)
Solution:
Let P(x,y) be a point in the first quadrant at a distance r1 from a point (0, y’) on the yaxis and distance r2 from (x’, 0) on the x-axis.
Using the Biot-Savart law, the magnetic field at P is given by
G
G µ 0 I d Gs × rˆ µ 0 I
B = ∫ dB =
=
4π ∫ r 2
4π
G
d s1 × rˆ1 µ 0 I
∫ r12 + 4π
wire y
G
d s2 × rˆ2
∫ r22
wire x
(8.2)
Let’s analyze each segment separately.
G
(i) Along the y axis, consider a differential element d s1 = −dy ' ˆj which is located at a
G
distance r = xˆi + ( y − y ')ˆj from P. This yields
1
G G
d s1 × r1 = (− dy ' ˆj) × [ xˆi + ( y − y ')ˆj] = x dy ' kˆ
(8.3)
G
G
(ii) Similarly, along the x-axis, we have d s2 = dx ' ˆi and r2 = ( x − x ')ˆi + yˆj which gives
12
G G
d s2 × r2 = y dx ' kˆ
(8.4)
Thus, we see that the magnetic field at P points in the z direction. Using the above results
( x − x0 )
and r1 = x 2 + ( y − y ') 2 and r2 =
Bz =
µ0 I
4π
∫
∞
2
+ y 2 , we obtain
µI
x dy '
+ 0
2 3/ 2
[ x + ( y − y ') ]
4π
2
0
∫
∞
0
y dx '
[ y + ( x − x ') 2 ]3/ 2
2
(8.5)
The integrals can be readily evaluated using
∫
∞
0
b dx
1
a
= +
2 3/ 2
2
[b + (a − x) ]
b b a + b2
2
(8.6)
The final result for the magnetic field is given by
G µ I
B= 0
4π
1

y
x
1
+ +
 +
 kˆ
2
2
2
2
x
y

x x +y
y x + y 
(8.7)
13