EXAMPLE 1 Calculating the Curl

Calculate the curl of \({\bf F} = \langle xy, e^x,y+z\rangle\).

Solution We compute the curl as a symbolic determinant: \begin{align*} {\bf curl}({\bf F}) &= \left| \begin{array}{ccc} {\bf i}& {\bf j} & {\bf k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ xy & e^x & y+z \\ \end{array} \right| \\ &= \left(\dfrac{\partial}{\partial y}(y+z) - \dfrac{\partial}{\partial{z}}e^x\right){\bf i} - \left(\dfrac{\partial}{\partial{x}}(y+z) - \dfrac{\partial}{\partial{z}} xy \right){\bf j} + \left(\dfrac{\partial}{\partial{x}} e^x - \dfrac{\partial}{\partial{y}} xy \right){\bf k}\\ &= {\bf i} +(e^x-x){\bf k} \end{align*}

EXAMPLE 2 Conservative Vector Fields Have Zero Curl

Verify:

\begin{equation} \boxed{ \textrm{If \({\bf F} = \nabla V\), then \({\bf curl}({\bf F})=0\). That is,\(\quad {\bf curl}(\nabla V)=0\).} }\tag{1} \end{equation}

Solution The curl of a vector field is zero if \[ \dfrac{\partial {F_3}}{\partial{y}} - \dfrac{\partial {F_2}}{\partial{z}} = 0,\qquad \dfrac{\partial {F_1}}{\partial{z}} - \dfrac{\partial {F_3}}{\partial{x}} = 0,\qquad \dfrac{\partial{F_2}}{\partial{x}} - \dfrac{\partial{F_1}}{\partial{y}} = 0 \]

But these equations are equivalent to the cross-partials condition that is satisfied by every conservative vector field \({\bf F} = \nabla V\).

THEOREM 1 Stokes’ Theorem

For surfaces \(S\) as described above, \begin{equation} \boxed{ \oint_{\partial S} {\bf F}\cdot d{\bf s} = \iint_{S\,} {\bf curl}({\bf F})\cdot d{\bf S}}\tag{2} \end{equation}

The integral on the left is defined relative to the boundary orientation of \(\partial S\). If \(S\) is closed (that is, \(\partial S\) is empty), then the surface integral on the right is zero.

Note

The curl measures the extent to which \({\bf F}\) fails to be conservative. If \({\bf F}\) is conservative, then \({\bf curl}({\bf F})={\bf 0}\) and Stokes’ Theorem merely confirms what we already know: The circulation of a conservative vector field around a closed path is zero.

Proof

Each side of Eq. (2) is equal to a sum over the components of \({\bf F}\): \begin{eqnarray*} \oint_C\,{\bf F}\cdot d{\bf s} &=&\oint_C\,F_1\,dx+F_2\,dy+F_3\,dz \\ \iint_{S\,} {\bf curl}({\bf F})\cdot d{\bf S} &=& \iint_{S\,} {\bf curl}(F_1{\bf i})\cdot d{\bf S} + \iint_{S\,} {\bf curl}(F_2{\bf j})\cdot d{\bf S} + \iint_{S\,} {\bf curl}(F_3{\bf k})\cdot d{\bf S} \end{eqnarray*}

The proof consists of showing that the \(F_1\)-, \(F_2\)-, and \(F_3\)-terms are separately equal.

Because a complete proof is quite technical, we will prove it under the simplifying assumption that \(S\) is the graph of a function \(z = f(x,y)\) lying over a domain \(D\) in the \(xy\)-plane. Furthermore, we will carry the details only for the \(F_1\)-terms. The calculation for \(F_2\)-components is similar, and we leave as an exercise the equality of the \(F_3\)-terms (Exercise 31). Thus, we shall prove that \begin{equation} \oint_C\,F_1\,dx = \iint_{S\,} {\bf curl}(F_1(x,y,z){\bf i})\cdot d{\bf S}\tag{3} \end{equation}

Orient \(S\) with upward-pointing normal as in Figure 17.33 and let \(C=\partial S\) be the boundary curve. Let \(C_0\) be the boundary of \(D\) in the \(xy\)-plane, and let \({\bf c}_0(t)= (x(t), y(t))\) (for \(a\le t\le b\)) be a counterclockwise parametrization of \(C_0\) as in Figure 17.33. The boundary curve \(C\) projects onto \(C_0\), so \(C\) has parametrization \begin{align*} {\bf c}(t) &= \bigl( x(t), y(t), f(x(t),y(t)) \bigr) \end{align*} and thus \begin{align*} \oint_C F_1(x,y,z) \, dx &= \int_a^b F_1\bigl(x(t),y(t),f(x(t),y(t))\bigr)\frac{{dx}}{dt}\,dt \end{align*}

The integral on the right is precisely the integral we obtain by integrating \(F_1\bigl(x,y,f(x,y)\bigr)\,dx\) over the curve \(C_0\) in the plane \({\bf R}^2\). In other words, \[ \oint_C F_1(x,y,z)\,dx = \int_{C_0} F_1\bigl(x,y,f(x,y)\bigr)\,dx \]

By Green’s Theorem applied to the integral on the right, \[ \oint_C F_1(x,y,z)\, dx = - \iint_{D\,}\dfrac{\partial}{\partial{y}} F_1(x,y,f(x,y)) \,dA \]

By the Chain Rule, \[ \dfrac{\partial}{\partial{y}} F_1\big(x,y,f(x,y)\big) = F_{1y}\big(x,y,f(x,y)\big)+F_{1z}\big(x,y,f(x,y)\big)f_y(x,y) \] so finally we obtain \begin{equation} \oint_C F_1\,dx = - \iint_{D\,} \Big(F_{1y}\big(x,y,f(x,y)\big)+F_{1z}\big(x,y,f(x,y)\big)f_y(x,y)\Big)\,dA\tag{4} \end{equation}

To finish the proof, we compute the surface integral of \({\bf curl}({\bf F}_1{\bf i})\) using the parametrization \(G(x,y)=(x,y,f(x,y))\) of \(S\): \begin{eqnarray} {\bf n} &=& \langle-f_x(x,y), -f_y(x,y), 1\rangle\qquad\textrm{(upward-pointing normal)}\notag\\ {\bf curl}(F_1{\bf i})\cdot{\bf n} &=& \langle 0, F_{1z}, -F_{1y}\rangle\cdot\langle-f_x(x,y), -f_y(x,y), 1\rangle\notag\\ &=& -F_{1z}\big(x,y,f(x,y)\big)f_y(x,y)-F_{1y}\big(x,y,f(x,y)\big)\notag\\ \iint_{S\,} {\bf curl}(F_1{\bf i})\cdot d{\bf S} &=& - \iint_{D\,}\notag \Big(F_{1z}(x,y,z)f_y(x,y)+F_{1y}\big(x,y,f(x,y)\big)\Big)\,dA\tag{5} \end{eqnarray}

The right-hand sides of Eq. (4) and Eq. (5) are equal. This proves Eq. (3).

EXAMPLE 3 Verifying Stokes’ Theorem

Verify Stokes’ Theorem for \[ {\bf F} = \langle -y, 2x, x+z\rangle \] and the upper hemisphere with outward-pointing normal vectors (Figure 17.34): \[ S=\{(x,y,z): x^2+y^2+z^2=1, z\ge 0\} \]

Solution We will show that both the line integral and the surface integral in Stokes’ Theorem are equal to \(3\pi\).

Figure 17.34: Upper hemisphere with oriented boundary.

Step 1. Compute the line integral around the boundary curve.

The boundary of \(S\) is the unit circle oriented in the counterclockwise direction with parametrization \({\bf c}(t)=(\cos t, \sin t, 0)\). Thus, \begin{align} {\bf c}'(t) &= \langle-\sin t, \cos t, 0\rangle\notag\\ {\bf F}({\bf c}(t)) &= \langle -\sin t, 2\cos t, \cos t\rangle\notag\\ {\bf F}({\bf c}(t))\cdot{\bf c}'(t) &=\langle -\sin t, 2\cos t, \cos t\rangle\cdot\langle-\sin t, \cos t, 0\rangle\notag\\ & = \sin^2t+2\cos^2t=1+\cos^2t\notag\\ \oint_{\partial S} {\bf F}\cdot d{\bf s} &= \int_0^{2\pi}(1+\cos^2t)\,dt = 2\pi +\pi= \boxed{3\pi}\tag{6} \end{align}

Step 2. Compute the curl.

\begin{eqnarray*} {\bf curl}({\bf F}) &=& \left| \begin{array}{ccc} {\bf i}& {\bf j} & {\bf k} \\ \displaystyle\dfrac{\partial}{\partial{x}} & \displaystyle\dfrac{\partial}{\partial{y}} & \displaystyle\dfrac{\partial}{\partial{z}} \\ -y & 2x & x+z \\ \end{array} \right| \\ &=& \left(\dfrac{\partial}{\partial{y}}(x+z) - \dfrac{\partial}{\partial{z}} 2x\right){\bf i} - \left(\dfrac{\partial}{\partial{x}}(x+z) - \dfrac{\partial}{\partial{z}}(-y)\right){\bf j} \\ &&{} + \left(\dfrac{\partial}{\partial{x}}\,2x - \dfrac{\partial}{\partial{y}}\, (-y)\right){\bf k} \\ &=& \langle 0, -1,3\rangle \end{eqnarray*}

Step 3. Compute the surface integral of the curl.

We parametrize the hemisphere using spherical coordinates: \[ G(\theta,\phi)=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi) \]

By Eq. (2) of Section 16.4, the outward-pointing normal vector is \[ {\bf n} = \sin\phi\langle \cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\rangle \]

Therefore, \begin{align*} {\bf curl}({\bf F})\cdot {\bf n} &= \sin\phi\langle 0, -1, 3\rangle\cdot \langle \cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\rangle\\ &= -\sin\theta\sin^2\phi + 3\cos\phi\sin\phi \end{align*}

The upper hemisphere \(S\) corresponds to \(0\le \phi\le\frac{\pi}2\), so \begin{align*} \iint_{S\,} {\bf curl}({\bf F})\cdot d{\bf S} &= \int_{\phi=0}^{\pi/2}\int_{\theta=0}^{2\pi} (-\sin\theta\sin^2\phi + 3\cos\phi\sin\phi)\,d\theta\,d\phi\\ &= 0+2\pi\int_{\phi=0}^{\pi/2} 3\cos\phi\sin\phi \,d\phi = 2\pi\left(\frac32\sin^2\phi\right)\bigg|_{\phi=0}^{\pi/2}\\ &= \boxed{3\pi} \end{align*}

EXAMPLE 4

Use Stokes’ Theorem to show that \(\displaystyle{\oint_C {\bf F}\cdot d{\bf s} = 0}\), where \[ {\bf F} = \langle \sin(x^2), e^{y^2}+ x^2, z^4+2x^2\rangle \] and \(C\) is the boundary of the triangle in Figure 17.35 with the indicated orientation.

Solution We apply Stokes’ Theorem \[ \oint_C {\bf F}\cdot d{\bf s} = \iint_{S\,} {\bf curl}({\bf F})\cdot \,d{\bf S} \] and show that the integral on the right is zero.

By the preceding remark, the first component \(\sin(x^2)\) does not contribute to the curl since it depends only on \(x\). Similarly, \(e^{y^2}\) and \(z^4\) drop out of the curl, and we have \begin{align*} {\bf curl}\Bigl(\langle \sin x^2, e^{y^2} + x^2, z^4+2x^2\rangle\Bigr) \quad& = \quad \overbrace{{\bf curl}\Bigl(\langle \sin x^2,e^{y^2},z^4 \rangle\Bigr)}^{\textrm{Automatically zero}} {}+ {\bf curl}\Bigl(\langle 0,x^2,2x^2 \rangle\Bigr) \\ &= \langle 0,-\dfrac{\partial}{\partial{x}} 2x^2,\dfrac{\partial}{\partial{x}} x^2 \rangle = \langle 0,-4x,2x \rangle \end{align*}

Now, it turns out (by the author’s design) that we can show the surface integral is zero without actually computing it. Referring to Figure 17.35, we see that \(C\) is the boundary of the triangular surface \(S\) contained in the plane \[ \frac{x}3+\frac{y}2+z=1 \]

Therefore, \({{\bf u} = \big\langle\frac13, \frac12,1\big\rangle}\) is a normal vector to this plane. But \({\bf u}\) and \({\bf curl}({\bf F})\) are orthogonal: \[ {\bf curl}({\bf F})\cdot{\bf u} = \langle 0, - 4x, 2x\rangle \cdot \langle \frac13, \frac12,1\rangle = -2x+2x = 0 \]

In other words, the normal component of \({\bf curl}({\bf F})\) along \(S\) is zero. Since the surface integral of a vector field is equal to the surface integral of the normal component, we conclude that \(\displaystyle{ \iint_{S\,} {\bf curl}({\bf F})\cdot \,d{\bf S} = 0}\).

CONCEPTUAL INSIGHT

Recall that if \({\bf F}\) is conservative—that is, \({\bf F} = \nabla V\)—then for any two paths \(C_1\) and \(C_2\) from \(P\) to \(Q\) (Figure 17.36), \[ \int_{C_1} {\bf F} \cdot d{\bf s} = \int_{C_2} {\bf F} \cdot d{\bf s} = V(Q)- V(P) \]

In other words, the line integral is path independent. In particular, \(\displaystyle{\oint_{C} {\bf F} \cdot d{\bf s}}\) is zero if \(C\) is closed (\(P=Q\)).

Analogous facts are true for surface integrals when \({\bf F}={\bf curl}({\bf A})\). The vector field \({\bf A}\) is called a vector potential for \({\bf F}\). Stokes’ Theorem tells us that for any two surfaces \(S_1\) and \(S_2\) with the same oriented boundary \(C\) (Figure 17.37), \[ \iint_{S_1\,}{\bf F}\cdot d{\bf S} = \iint_{S_2\,}{\bf F}\cdot d{\bf S} = \oint_{C}{\bf A}\cdot d{\bf s} \]

In other words, the surface integral of a vector field with vector potential \({\bf A}\) is surface independent, just as a vector field with a potential function \(V\) is path independent.

If the surface is closed, then the boundary curve is empty and the surface integral is zero: \[ \iint_{S\,}{\bf F}\cdot d{\bf S}=0\qquad\textrm{if}\quad \textrm{\({\bf F}={\bf curl}({\bf A})\) and \(S\) is closed} \]

Note

Vector potentials are not unique: If \({\bf F}={\bf curl}({\bf A})\), then \({\bf F}={\bf curl}({\bf A}+{\bf B})\) for any vector field \({\bf B}\) such that \({\bf curl}({\bf B})={\bf 0}\).

THEOREM 2 Surface Independence for Curl Vector Fields

If \({\bf F} = {\bf curl}({\bf A})\), then the flux of \({\bf F}\) through a surface \(S\) depends only on the oriented boundary \(\partial S\) and not on the surface itself: \begin{equation} \iint_{S\,} {\bf F} \cdot d{\bf S} = \oint_{\partial S} {\bf A} \cdot d{\bf s}\tag{7} \end{equation}

In particular, if \(S\) is closed (that is, \(\partial S\) is empty), then \(\displaystyle{ \iint_{S\,} {\bf F}\cdot d{\bf S} = 0}\).

EXAMPLE 5

Let \({\bf F} = {\bf curl}({\bf A})\), where \({\bf A} = \langle y+z,\sin(xy),e^{xyz}\rangle\). Find the flux of \({\bf F}\) through the surfaces \(S_1\) and \(S_2\) in Figure 17.38 whose common boundary \(C\) is the unit circle in the \(xz\)-plane.

Solution With \(C\) oriented in the direction of the arrow, \(S_1\) lies to the left, and by Eq. (7), \[ \iint_{S_1\,} {\bf F}\cdot d{\bf S} = \oint_C {\bf A}\cdot d{\bf s} \]

We shall compute the line integral on the right. The parametrization \({\bf c}(t)=(\cos t, 0, \sin t)\) traces \(C\) in the direction of the arrow because it begins at \({\bf c}(0)=(1,0,0)\) and moves in the direction of \({\bf c}\big(\frac{\pi}2\big)=(0,0,1)\). We have \begin{align*} {\bf A}({\bf c}(t))&= \langle 0 + \sin t, \sin(0), e^0\rangle = \langle \sin t, 0, 1\rangle \\ {\bf A}({\bf c}(t))\cdot {\bf c}'(t)&= \langle \sin t, 0, 1\rangle \cdot \langle -\sin t, 0, \cos t\rangle =-\sin^2t+\cos t\\ \oint_C {\bf A}\cdot d{\bf s} &= \int_0^{2\pi} (-\sin^2t+\cos t)\,dt =-\pi \end{align*}

We conclude that \(\displaystyle{\iint_{S_1\,} {\bf F}\cdot d{\bf S}=-\pi}\). On the other hand, \(S_2\) lies on the right as you traverse \(C\). Therefore \(S_2\) has oriented boundary \(-C\), and \[ \iint_{S_2\,} {\bf F}\cdot d{\bf S}= \oint_{-C} {\bf A}\cdot d{\bf s}=-\oint_C {\bf A}\cdot d{\bf s} = \pi \]

CONCEPTUAL INSIGHT Interpretation of the Curl

In Section 17.1, we showed that the quantity \(\displaystyle{\dfrac{\partial{F_2}}{\partial{x}}-\dfrac{\partial{F_1}}{\partial{y}}}\) in Green’s Theorem is the “circulation per unit of enclosed area.” A similar interpretation is valid in \({\bf R}^3\).

Consider a plane through a point \(P\) with unit normal vector \({\bf e}_{\bf n}\) and let \(D\) be a small domain containing \(P\) with boundary curve \(C\) (Figure 17.39). By Stokes’ Theorem, \begin{equation} \oint_{C}{\bf F}\cdot\,d{\bf s} \approx \iint_{D}({\bf curl}({\bf F})\cdot{\bf e}_{\bf n}) \, dS\tag{8} \end{equation}

The vector field \(\displaystyle{{\bf curl}({\bf F})}\) is continuous (its components are derivatives of the components of \({\bf F}\)), so its value does not change much on \(D\) if \(D\) is sufficiently small. To a first approximation, we can replace \({\bf curl}({\bf F})\) by the constant value \({\bf curl}({\bf F})(P)\), giving us the approximation \begin{align} \iint_{D}({\bf curl}({\bf F})\cdot{\bf e}_{\bf n}) \, dS &\approx \iint_{D}({\bf curl}({\bf F})(P)\cdot{\bf e}_{\bf n}) \, dS\notag\\ &\approx ({\bf curl}({\bf F})(P)\cdot{\bf e}_{\bf n}) \, \textrm{Area}(D)\tag{9} \end{align}

Furthermore, \(\displaystyle{{\bf curl}({\bf F})(P)\cdot{\bf e}_{\bf n} = \|{{\bf curl}({\bf F})(P)}\|\cos\theta}\), where \(\theta\) is the angle between \({\bf curl}({\bf F})\) and \({\bf e}_{\bf n}\). Together, Eq. (8) and Eq. (9) give us \begin{equation} \oint_{C}{\bf F}\cdot\,d{\bf s} \approx \|{{\bf curl}({\bf F})(P)}\|(\cos\theta) \, \textrm{Area}(D)\tag{10} \end{equation}

This is a remarkable result. It tells us that \({\bf curl}({\bf F})\) encodes the the circulation per unit of enclosed area in every plane through \(P\) in a simple way—namely, as the dot product \(\displaystyle{{\bf curl}({\bf F})(P)\cdot{\bf e}_{\bf n}}\). In particular, the circulation rate varies (to a first-order approximation) as the cosine of the angle \(\theta\) between \({\bf curl}({\bf F})(P)\) and \({\bf e}_{\bf n}\).

We can also argue (as in Section 17.1 for vector fields in the plane) that if \({\bf F}\) is the velocity field of a fluid, then a small paddle wheel with normal \({\bf e}_{\bf n}\) will rotate with an angular velocity of approximately \({\frac12{\bf curl}({\bf F})(P)\cdot{\bf e}_{\bf n}}\) (see Figure 17.40).

EXAMPLE 6 Vector Potential for a Solenoid

An electric current flowing through a solenoid (a tightly wound spiral of wire; see Figure 17.41) creates a magnetic field \({\bf B}\). If we assume that the solenoid is infinitely long, with radius \(R\) and the \(z\)-axis as central axis, then \[ {\bf B} = \begin{cases} {\bf 0} & \textrm{if \(r > R\)} \\ B{\bf k} &\textrm{if \(r<R\)} \end{cases} \]

Figure 17.41: 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.

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 \({\bf B}\) is \[ {\bf A} = \begin{cases} \dfrac12 R^2 B\langle -\dfrac{y}{r^2}, \dfrac{x}{r^2}, 0\rangle & \textrm{if \(r> R\)} \\ \dfrac12B\langle -y, x, 0\rangle &\textrm{if \(r<R\)} \end{cases} \]
• (b) Calculate the flux of \({\bf B}\) through the surface \(S\) (with upward-pointing normal) in Figure 17.41 whose boundary is a circle of radius \(r\) where \(r>R\).

Solution

• (a) For any functions \(f\) and \(g\), \[ {\bf curl}(\langle f, g, 0 \rangle) = \langle -g_z, f_z, g_x-f_y\rangle \] Applying this to \({\bf A}\) for \(r<R\), we obtain \[ {\bf curl}({\bf A}) = \frac1{2}B\langle 0,0, \dfrac{\partial{}}{\partial{x}} x - \dfrac{\partial{}}{\partial{y}}(-y )\rangle = \langle 0,0,B\rangle = B{\bf k} = {\bf B} \] We leave it as an exercise [Exercise 29] to show that \({\bf curl}({\bf A}) ={\bf B}={\bf 0}\) for \(r>R\).
• (b) The boundary circle of \(S\) with counterclockwise parametrization \({\bf c}(t)=(r\cos t,r\sin t, 0)\), so \begin{align*} {\bf c}'(t) &= \langle -r\sin t, r\cos t, 0\rangle\\ {\bf A}({\bf c}(t))&=\frac12R^2Br^{-1}\langle -\sin t, \cos t, 0\rangle \\ {\bf A}({\bf c}(t))\cdot {\bf c}'(t) &=\frac12R^2B \left((-\sin t)^2+\cos^2t \right) = \frac12R^2B \end{align*}

Figure 17.42: The magnitude \(\|{{\bf A}}\|\) of the vector potential as a function of distance \(r\) to the \(z\)-axis.

By Stokes’ Theorem, the flux of \({\bf B}\) through \(S\) is equal to \begin{align*} \iint_{S\,}{\bf B}\cdot d{\bf S} &= \oint_{\partial S}{\bf A}\cdot d{\bf s} = \int_0^{2\pi} {\bf A}({\bf c}(t))\cdot {\bf c}'(t)\,dt = \frac12R^2B \int_0^{2\pi} dt =\pi R^2B \end{align*}

CONCEPTUAL INSIGHT

There is an interesting difference between scalar and vector potentials. If \({\bf F}=\nabla V\), then the scalar potential \( V\) is constant in regions where the field \({\bf F}\) is zero (since a function with zero gradient is constant). This is not true for vector potentials. As we saw in Example 6, the magnetic field \({\bf B}\) produced by a solenoid is zero everywhere outside the solenoid, but the vector potential \({\bf A}\) is not constant outside the solenoid. In fact, \({\bf A}\) is proportional to \(\displaystyle{\langle -\dfrac{y}{r^2},\dfrac{x}{r^2},0\rangle}\). This is related to an intriguing phenomenon in physics called the Aharonov-Bohm (AB) effect, first proposed on theoretical grounds in the 1940s.

According to electromagnetic theory, a magnetic field \({\bf B}\) exerts a force on a moving electron, causing a deflection in the electron’s path. We do not expect any deflection when an electron moves past a solenoid because \({\bf B}\) is zero outside the solenoid (in practice, the field is not actually zero, but it is very small—we ignore this difficulty). However, according to quantum mechanics, electrons have both particle and wave properties. In a double-slit experiment, a stream of electrons passing through two small slits creates a wavelike interference pattern on a detection screen (Figure 17.43). The AB effect predicts that if we place a small solenoid between the slits as in the figure (the solenoid is so small that the electrons never pass through it), then the interference pattern will shift slightly. It is as if the electrons are “aware” of the magnetic field inside the solenoid, even though they never encounter the field directly.

The AB effect was hotly debated until it was confirmed definitively in 1985, in experiments carried out by a team of Japanese physicists led by Akira Tonomura. The AB effect appeared to contradict “classical” electromagnetic theory, according to which the trajectory of an electron is determined by \({\bf B}\) alone. There is no such contradiction in quantum mechanics, because the behavior of the electrons is governed not by \({\bf B}\) but by a “wave function” derived from the nonconstant vector potential \({\bf A}\).