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6.5: Magnetic Field Near a Long, Straight, Current-carrying Conductor

[ "article:topic", "authorname:tatumj" ]

Consider a point $$P$$ at a distance $$a$$ from a conductor carrying a current $$I$$ (figure VI.4).

$$\text{FIGURE VI.4}$$

The contribution to the magnetic field at $$P$$ from the elemental length $$dx$$ is

$dB = \frac{\mu}{4\pi}\cdot \frac{I\,dx \cos \theta }{r^2}.\label{6.5.1}$

(Look at the way I have drawn $$\theta$$ if you are worried about the cosine.)

Here I have omitted the subscript zero on the permeability to allow for the possibility that the wire is immersed in a medium in which the permeability is not the same as that of a vacuum. (The permeability of liquid oxygen, for example, is slightly greater than that of free space.) The direction of the field at $$P$$ is into the plane of the “paper” (or of your computer screen).

We need to express this in terms of one variable, and we’ll choose $$\theta$$. We can see that $$r=a\sec \theta$$ and $$x=a\tan \theta$$ so that $$dx=a\sec^2 \theta \, d\theta$$. Thus Equation \ref{6.5.1} becomes

$dB=\frac{\mu I}{4\pi a}\sin \theta \, d\theta .$

Upon integrating this from $$-\pi/2 \text{ to }+ \pi/2$$ (or from $$0\text{ to }\pi/2$$ and then double it), we find that the field at $$P$$ is

$B=\frac{\mu I}{2\pi a}.$

Note the $$2\pi$$ in this problem with cylindrical symmetry.