Calculus 2nd Edition Rogawski

SECTION 18.2 Stokes’ Theorem 1029
z
z
I
B
B
S
A
y
r
y
FIGURE 11 The magnetic field of a long
solenoid is nearly uniform inside and weak
outside. In practice, we treat the solenoid
as “infinitely long” if it is very long in
comparison with its radius.
x
I
R
x
where r = (x 2 + y 2 ) 1/2 and B is a constant that depends on the current strength and the
spacing of the turns of wire.
(a) Show that a vector potential for B is
⎧
1
⎪⎨
2 R2 B
〈− y r
A =
2 , x 〉
r 2 , 0 if r>R
1 ⎪⎩ B ⟨−y,x,0⟩ if rR.
Solution
The vector potential A is continuous but
not differentiable on the cylinder r = R,
that is, on the solenoid itself (Figure 12).
The magnetic field B = curl(A) has a
jump discontinuity where r = R. We take
for granted the fact that Stokes’ Theorem
remains valid in this setting.
1
BR
2
A
R
FIGURE 12 The magnitude ∥A∥ of the
vector potential as a function of distance r
to the z-axis.
r
(a) For any functions f and g,
Applying this to A for rR.
(b) The boundary circle of S with counterclockwise parametrization c(t) =
(r cos t,r sin t,0), so
c ′ (t) = ⟨−r sin t,r cos t,0⟩
A(c(t)) = 1 2 R2 Br −1 ⟨− sin t,cos t,0⟩
A(c(t)) · c ′ (t) = 1 (
)
2 R2 B (− sin t) 2 + cos 2 t = 1 2 R2 B
By Stokes’ Theorem, the flux of B through S is equal to
∫∫ ∮ ∫ 2π
B · dS = A · ds =
S
∂S
0
A(c(t)) · c ′ (t) dt = 1 ∫ 2π
2 R2 B dt = πR 2 B
0