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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. 
 
 
 
Find the magnetic field at P. B
 
Solution: 
 
According to the Biot-Savart Law, the magnitude of the magnetic field due to a 
differential current-carrying element Id s is given by 
 
 0 02 2
ˆ
4 4 4
dI I r ddB d
r r
µ µ µθ 0I
r
θ
π π
×
= = =
s r
π
 (1.1) 
 
For the outer arc, we have 
 
 0
04 4outer
0I IB d
b b
θµ µ θθ
π π
= =∫ (1.2) 
 
 
The direction of is determined by the cross productouterB ˆd ×s r which points into the 
page. 
 
Similarly, for the inner arc, we have 
 
 0
04 4inner
0I IB d
a a
θµ µ θθ
π π
= =∫ (1.3) 
 1
 
 
For , points out of the page. innerB ˆd ×s r
 
Therefore, the total magnitude of magnetic field is 
 
 0 1 1 (out of page)
4inner outer
I
a b
µ θ
π
 = = − 
 
B B + 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 
 (0 1cos cos4
IB
r
)2
µ θ θ
π
= − (2.1)
 
where 1 and 2θ θ 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= − +∞
1cos 1
to 
. The angles which parameterize this segment give ( ,a d− + ) θ = ( 1 0θ = ) and 
2/d d a= + 22cosθ . Therefore, 
 ( )0 01 1 2 2 2cos cos 14 4
I I dB
a a d a
µ µθ θ
π π
 
= − = −
+ 
 (2.2) 
 
The direction of is out of page, or1B ˆ+k . 
 
(ii) Next ,we consider the segment which extends from ( , ) ( , )x y a d= − + to ( . 
Again, the (cosine of the) angles are given by 
, )a d+ +
 1 2cos
a
a d
θ =
+ 2
 (2.3) 
 ( )2 1 1 2cos cos cos
a
a d
θ π θ θ= + = − = −
+ 2
 (2.4) 
 
This leads to 
 02 2 2 2 2 24 2
I a aB
d a d a d d a d
µ µ
π π
 
= + = 
+ + 
0
2
Ia
+
 (2.5) 
 
The direction of is into the page, or 2B ˆ−k . 
 
(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: 
 
 3 1B B= (2.6) 
 
The direction of is again out of page, or 3B ˆ+k . 
 
The total magnitude of the magnetic field is 
 
 
( )
1 2 3 1 2
0
2 2 2 2
2 2 2 20
2 2
2
ˆ1
2 2
ˆ
2
I d
a a d d a d
I d a d d a
ad a d
µ
π π
µ
π
= + + = +
 
= − − 
+ + 
= + −
+
B B B B B B
k
k
0 ˆIaµ
−
k (2.7) 
 
 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 1 1cosθ = ( 1 0θ = ) and cos 2 20 ( / 2)θ θ π= = . Therefore, its 
contribution to the magnetic field at P is 
 
 ( )0 01 1 2cos cos (1 0)4 4
0
4
I IB
r r
I
r
µ µθ θ µ
π π π
= − = − = (3.1) 
 
The direction of is out of page, or 1B ˆ+k . 
 
(ii) For the semi-circular arc of radius r, we make use of the Biot-Savart law: 
 
 0 2
ˆ
4
I d
r
µ
π
×
= ∫
s rB (3.2) 
 
and obtain 
 
 02 204 4
0I IrdB
r r
πµ µθ
π
= =∫ (3.3) 
 
The direction of is out of page, or2B ˆ+k . 
 
(iii) The third segment of the wire runs from ( , ) (0, )x y r= + to ( , )r−∞ + . One may readily 
show that it gives the same contribution as the first one: 
 4
 
 03 1 4
IB B
r
µ
π
= = (3.4) 
 
The direction of is again out of page, or 3B ˆ+k . 
 
The total magnitude of the magnetic field is 
 
 
0 0
1 2 3 1 2
0
ˆ ˆ2
2 4
ˆ(2 )
4
I I
r r
I
r
µ µ
π
µ π
π
= + + = + = +
= +
B B B B B B k k
k
 (3.5) 
 
Notice that the contribution from the two semi-infinite wires is equal to that due to an 
infinite wire: 
 
 01 3 1 ˆ2 2
I
r
µ
π
+ = =B B B k (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: 
 
 0 01 2 22 2
I IB
r a z
µ µ
π π
= =
+
 (4.1) 
 
Since the current is flowing in the –x direction, the magnetic field points in the direction 
of the cross product 
 
 1ˆ ˆ ˆ ˆˆ ˆˆ( ) ( ) (cos sin ) sin cos θ θ θ− × = − × + = −i r i θj k j k (4.2)
 
Thus, we have 
 
 ( )01 2 2 ˆ ˆsin cos 2
I
a z
µ θ θ
π
=
+
B −j k (4.3)
 
For the right wire, the magnetic field strength is the same as the left one: 1 2B B= . 
However, its direction is given by 
 
 2ˆ ˆ ˆ ˆˆˆ( ) ( ) ( cos sin ) sin cos ˆθ θ θ− × = − × − + =i r i θj k j+ k (4.4) 
 
Adding up the contributions from both wires, the z components cancel (as required by 
symmetry), and we arrive at 
 
 01 2 2 22 2
sin ˆ
( )
I
a za z
0 ˆIzµ θ µ
ππ
+ = =
++
B = B B j j (4.5) 
 
To locate the maximum of B, we set /dB dz 0= and find 
 6
 
 
( )
2 2
0 0
22 2 2 2 2 2 2
1 2 0
( )
I IdB z a z
dz a z a z a z
µ µ
π π
  −
= − = + +  +
2
= (4.6) 
 
which gives 
 z a= (4.7) 
 
Thus, at z=a, the magnetic field strength is a maximum, with a magnitude 
 
 0max 2
IB
a
µ
π
= (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: 
 
 0 encd Iµ⋅ =∫ B s (5.2) 
 
where the enclosed current Ienc is given by 
 
 ( )( )2encI d br rdπ= ⋅ = r∫ ∫J A (5.3) 
 
(a) For r , the enclosed current is R<
 
 
3
2
0
22
3
r
enc
brI br dr ππ= =∫ (5.4) 
 7
 
Applying Ampere’s law, the magnetic field at P1 is given by 
 
 ( )
3
0
1
22
3
brB r µ ππ = (5.5) 
or 
 201 3
bB rµ= (5.6) 
 
The direction of the magnetic field B1 is tangential to the Amperian loop which encloses 
the current. 
 
(b) For , the enclosed current is r R>
 
 
3
2
0
22
3
R
enc
bRI br dr ππ= =∫ (5.7) 
 
which yields 
 
 ( )
3
0
2
22
3
bRB r µ ππ = (5.8) 
 
Thus, the magnetic field at a point P2 outside the conductor is 
 
 
3
0
2 3
b RB
r
µ
= (5.9) 
 
A plot of B as a function of r is depicted below: 
 
 
 
 
 8
 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 and are both zero at a point midway between the coils. /dB dx 2 /d B dx2
 
Solution: 
 
We first consider the case where there is only one coil. Since each length element ds is 
perpendicular to the vector r and each length elements around the loop are at the same 
distance from the x-axis, hence at point P, we have 
ˆ
 
 
 0 02 2
ˆ
4 4
dI I dsdB
r x
µ µ
π π
×
= =
+
s r
2R
 (6.1) 
 
 
 
By symmetry, the magnetic field mustpoint along the x-axis: 
 
 0 02 2 2 2 3/
cos cos
4 4 (x
I Ids ds RdB dB
x R x R 2)
µ µθθ
π π
= = =
+ +
 (6.2) 
 9
 
where 2/ 2cos R x Rθ = + . Thus, the magnetic field at P is 
 
 ( )
2
0 0
2 2 3/ 2 2 2 3/ 2 2 2 3/24 ( ) 4 ( ) 2( )x
IR IR IRB ds R
x R x R x R
µ µ π
π π
= = =
+ +∫
0
2
µ
+
 (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: 
 
 
1 2
2
0
2 2 3/ 2 2 2 3/ 2
1 1
2 ( ) [( ) ]x x
IRB B B
x R x R R
µ  
= + = + + − + 
 (6.4) 
 
(b) Differentiating B with respect to x, we obtain 
 
 
( )
2
0
5/ 22 2 5/ 2 2 2
3 3( )
2 2( )
IRdB x x R
dx x R x R R
µ
 
− = − − +   − +  
 (6.5) 
 
At the midpoint where / 2x R= , the derivative vanishes: 
 
 
/ 2
0
x R
dB
dx =
= (6.6) 
 
Similarly, we can also show that 
 
22 2
0
2 2 2 5/ 2 2 2 7 / 2 2 2 5/ 2 2 2 7 / 2
3 1 5 1 5( )
2 ( ) ( ) [( ) ] [( ) ]
IRd B x x R
dx x R x R x R R x R R
µ  − −
= − + − + + − + − + 
2
 (6.7) 
 
which yields 
 
2
2
/ 2
0
x R
d B
dx
=
= (6.8) 
 
The fact that the first two derivatives vanish at / 2x R= indicates that the magnetic field 
is fairly uniform there.
 10
 
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 
 
 drdI I
w
= 
 

 (7.1) 
 
Using the Ampere’s law, we see that its contribution to the magnetic field at P is given by 
 
 0 0(2 ) ( )encdB r I dIπ µ µ= = (7.2) 
 
or 
 0 0
2 2
dI I drdB
r r w
µ µ
π π
= = 
 

 (7.3) 
 
Integrating over the expression, we obtain 
 
 0 0 ln
2 2
b w
b
I Idr b wB
w r w b
µ µ
π π
+ +  = =  
  ∫



 (7.4) 
 
Using the right-hand rule, the direction of the magnetic field can be shown to point in the 
+z direction, or 
 0 ˆln 1 
2
I w
w b
µ
π
 = + 
 
B k (7.5) 
 
 11
 
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 of the xy 
plane is given by 
, 0x y >
 
 0
2 2 2 2
1 1
4z
I x yB
x y y x y x x y
µ
π
 
= + + +
 + + 


 (8.1) 
 
Solution: 
 
Let P(x,y) be a point in the first quadrant at a distance from a point (0, y’) on the y-
axis and distance r from (x’, 0) on the x-axis. 
1r
2
 
 
 
Using the Biot-Savart law, the magnetic field at P is given by 
 
 0 0 01 1 2 22 2
1 2 
ˆ ˆˆ
4 4 4wire y wire x
I I Iddd
r r 2
d
r
µ µ µ
π π π
× ××
= = = +∫ ∫ ∫ ∫
s r s rs rB B (8.2) 
 
Let’s analyze each segment separately. 
 
(i) Along the y axis, consider a differential element 1 ˆ'd dy= −s j which is located at a 
distance 1 ˆ ( ˆ')x y y= + −r i j from P. This yields 
 
 1 1 ˆ ˆ ˆ ˆ( ' ) [ ( ') ] 'd dy x y y x d× = − × + − =s r yj i j k (8.3) 
 
(ii) Similarly, along the x-axis, we have 2 ˆ'd dx=s i and 2 ˆ( ') ˆx x y= − +r i j which gives 
 
 12
 2 2 ˆ'd y d× =s r kx (8.4) 
 
Thus, we see that the magnetic field at P points in the z direction. Using the above results 
and 2 21 ( 'r x y y= + − ) and ( )
2 2
2 0r x x= − + y , we obtain 
 
 0 02 2 3/ 2 20 0
'
4 [ ( ') ] 4 [ ( ') ]z
I Ix dy y dxB
x y y y x x 2 3/ 2
'µ µ
π π
∞ ∞
= +
+ − + −∫ ∫ (8.5) 
 
The integrals can be readily evaluated using 
 
 2 2 3/ 2 2 20
1
[ ( ) ]
b dx a
b a x b b a b
∞
= +
+ − +
∫ (8.6) 
 
The final result for the magnetic field is given by 
 
 0
2 2 2 2
1 1 ˆ
4
I y x
x yx x y y x y
µ
π
 
= + + + 
 + +  
B k (8.7) 
 
 13
	8.02 Spring 2003
	IX. Source of Magnetic Fields – Worked Examples
	
	Example 4: Two infinitely long wires