W9_parallel_resonance.eps

Week 7: Sources of the Magnetic Field 241

up liker^2 , we got a quantity that only depended on the charge (and Gauss’s Law for Electrostatics).
Here we can clearly do the same thing and write:

B× 2 πr=

μ 0 I

2 πr×^2 πr=μ^0 I (532)

So far, this is only suggestive. However, consider the geometry offigure 85. where the currentI

C

∆θ

r

∆s

∆r

B

I (in)

Figure 85: A circular path of radiusraround a long straight wire with a “notch” of angular width
∆θand radiusr+ ∆r.

is drawn directly into the page so we can concentrate on the plane in which the magnetic field lies.

TheB-field is of course constant in magnitude on the circle of radiusras before, and so we can
multiply it by the length of the circular arc at the same radius right up to the notch. However, our
path now stepsoutby ∆ralong the radius (and perpendicular to the field). Along the curved path
∆s, the field is somewhat weaker, but the path itself is somewhat longer.

In fact, we can cleverly add up the following:

B(2π−∆θ)r+B∆s = B(2π−∆θ)r+B(r+ ∆r)∆θ

=

μ 0 I

2 πr

(2π−∆θ)r+

μ 0 I

2 π(r+ ∆r)

(r+ ∆r)∆θ

=

μ 0 I

2 πr

2 πr−

μ 0 I

2 π

∆θ+

μ 0 I

2 π

∆θ

= μ 0 I (533)

and we see that deforming the circle with the notchdid not alter the value of the sum we got
from multiplying the field times the length of the curved pathC (circle with notch)alongthe
magnetic field, while ignoring the part ofCperpendicular to the magnetic field. Again, this should
be reminding you of what we did for Gauss’s Law only it is a bit simpler.

Well, we can add more notches; in fact, we can deform the curveCin, we can deform it out, we
can deform it so that it goes along the wire and no longer lies in a plane, and as long as we break
upCinto teensy segments that either lie perpendicular to the field or follow a curved arc tangent
to the field, we only get a contribution from the piece of lengthdstangent to the field, and that
contribution is always of the form:

Bds=

μ 0 I

2 πrrdθ=

μ 0 I

2 πdθ (534)

If we clean up the geometry of this, picking a path element alongCwith avectorlengthd~ℓand
selecting only the component parallel toB~at that point with the dot product, we get:
∮

C

B~·d~ℓ=

∮

C

μ 0 I

2 πr

θˆ·rdθθˆ=μ^0 I

2 π

∫ 2 π

0

dθ=μ 0 I (535)