The vector nature of a long current carrying wire.
Magnetic fields have different properties than Newtonian gravity and static electric fields. Since there are no magnetic monopoles magnetic fields
circle around the current. For a long current carrying wire the magnetic field
circles around the wire. We will find out how to find the components of a
current carrying wire for any point in space. This is a very useful problem
since there are very few example that we can find every point in space so
simply. The main problem is finding the tangent vector to a circle. We do
not need calculus, just geometry! Figure 1 shows the tangent vectors on a
circle.
~ @
I
@
B
¾
@
¡
¡
µ
¡
¡
¡
@
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¡
¡
¡
ª
@
@
@
@
R
I
¶³
@
¡
¡
@
µ´
µ
¡
¡
?
¡
@
I
@
@
¡
?
@
-
@
-
6
I
¶³
z
µ´
@
~
B
¾
@
R
@
(a) Current going out of the page
¡
¡
¡
¡
ª
(b) Current going into of the page
Figure 1: Magnetic field of a long current carrying wire
Now we can see that the geometry is just our usually polar coordinate θ̂.
Looking at just one vector arbitrarily we find Figure 2:
1
¾
φ­
H
Y
HH θ 6
¤¡
yb
¨¥
H ­
θb H
6
­
Á
­£ ¢
­φ
­
­
­
­
­
­
­©
θ
­ ª
- x
b
Figure 2: Geometry of the unit vector theta
From Figure 2 the unit vectors components are observed to be:
θ̂x = − sin(θ)
θ̂y = cos(θ)
It is good to remember that our angle θ is defined from the horizontal
axis. If we define it from the vertical, sine goes to cosine and vice versa. Now
we can write down the x and y components of the long current carrying wire’s
~ and the magnitude by
magnetic field! Note that the vector is denoted by B
B. The magnitude B is given by Ampere’s Law or:
B=
2
µ0 I
2πr
The magnitude and the vector nature that we constructed are all you
need for wire problems. Finally the x and y components are just the magnitude of the vector times the components of the unit vector θ̂:
~ x = B θ̂x = B(− sin(θ)) = −B sin(θ) = − µ0 I sin(θ)
B
2πr
~ y = B θ̂y = B(cos(θ)) = B cos(θ) = µ0 I cos(θ)
B
2πr
We can write and vector in the x, y plane by using the x and y unit
vectors: x̂ = (1, 0) and ŷ = (0, 1) respectively.
~ = Bx x̂ + By ŷ = − µ0 I sin(θ)x̂ + µ0 I cos(θ)ŷ
B
2πr
2πr
µ0 I
µ0 I
µ0 I
θ̂ = B θ̂
(− sin(θ)x̂ + cos(θ)ŷ) =
(θ̂x + θ̂y ) =
2πr
2πr
2πr
This is a nice way to write it. Let us review and test our definition.
First we defined our positive to be out of the page or Figure 1 (a). This is
Counter Clockwise. The Clockwise direction has an extra negative sign
or −θ̂, −θ̂x = sin(θ) and −θ̂y = − cos(θ). An interesting example of our
vector is to test it on the axes or θ = 0◦ , 90◦ , 180◦ and 270◦ . The x and y
components of the unit vector theta are listed:
0◦
θ̂x = − sin(0◦ ) = 0
θ̂y = cos(0◦ ) = 1
3
90◦
θ̂x = − sin(90◦ ) = −1
θ̂y = cos(90◦ ) = 0
180◦
θ̂x = − sin(180◦ ) = 0
θ̂y = cos(180◦ ) = −1
270◦
θ̂x = − sin(270◦ ) = 1
θ̂y = cos(270◦ ) = 0
We see that these give us the unit directions for Figure 1 (a). All we need
0I
to do now is multiply by the magnitude B = µ2πr
for the magnetic field or
0◦
~ x = B θ̂x = −B sin(0◦ ) = 0
B
~ y = B θ̂y = B cos(0◦ ) = B = µ0 I
B
2πr
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90◦
~ x = B θ̂x = −B sin(90◦ ) = −B = − µ0 I
B
2πr
~ y = B θ̂y = B cos(90◦ ) = 0
B
180◦
~ x = B θ̂x = −B sin(180◦ ) = 0
B
~ y = B θ̂y = B cos(180◦ ) = −B = − µ0 I
B
2πr
270◦
~ x = B θ̂x = −B sin(270◦ ) = B = µ0 I
B
2πr
~ y = B θ̂y = B cos(270◦ ) = 0
B
Compare these answers with Figure 1 (a) and confirm that (b) is the
negative of all the components.
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Exercises:
1. A good exercise is to work out the components of a long current carrying wire’s magnetic field for all the integer multiples of 30◦ , 45◦ and 60◦
less than 360◦ . Just like the lat discussion.
2. Two wire both have currents coming out of the page. They are positioned on corners of an equilateral triangle with length 1 m. The two have
the same current 1 A. What is the magnitude and direction of the magnetic
field at the third corner?
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