Just as ∇ · B = 0 allows us to express B as the curl of a vector potential (B = ∇ × A), so ∇ · A = 0 permits us to write A itself as the curl of a “higher” potential: A = ∇ × W. (And this hierarchy can be extended ad infinitum.)

(a) Find the general formula for W (as an integral over B), which holds when B → 0 at ∞.

(b) Determine W for the case of a uniform magnetic field B. [Hint: see Prob. 5.25.]

(c) Find W inside and outside an infinite solenoid. [Hint: see Ex. 5.12.]

Question

Just as ∇ · B = 0 allows us to express B as the curl of a vector potential (B = ∇ × A), so ∇ · A = 0 permits us to write A itself as the curl of a “higher” potential: A = ∇ × W. (And this hierarchy can be extended ad infinitum.)

(a) Find the general formula for W (as an integral over B), which holds when B → 0 at ∞.

(b) Determine W for the case of a uniform magnetic field B. [Hint: see Prob. 5.25.]

(c) Find W inside and outside an infinite solenoid. [Hint: see Ex. 5.12.]

Accepted Answer

[latex] ext { (a) } abla \cdot B =0, abla imes B =\mu_{0} J , ext { and } abla \cdot A =0, abla imes A = B \Rightarrow A =\frac{\mu_{0}}{4 \pi} \int \frac{ J }{ᴫ} d au^{\prime}[/latex] , so

[latex] abla \cdot A =0, abla imes A = B , ext { and } abla \cdot W =0 ext { (we'll choose it so), } abla imes W = A \Rightarrow W =\frac{1}{4 \pi} \int \frac{ B }{ᴫ} d au^{\prime} .[/latex]

(b) W will be proportional to B and to two factors of r (since differentiating twice must recover B), so I’ll try something of the form [latex]W =\alpha r ( r \cdot B )+\beta r^{2} B[/latex] , and see if I can pick the constants [latex] \alpha ext { and } \beta [/latex] in such a way that [latex] abla \cdot W =0 ext { and } abla imes W = A[/latex] .

[latex] abla \cdot W =\alpha[( r \cdot B )( abla \cdot r )+ r \cdot abla ( r \cdot B )]+\beta\left[r^{2}( abla \cdot B )+ B \cdot abla \left(r^{2} ight) ight] . abla r =\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3[/latex] ;

[latex] abla ( r \cdot B )= r imes( abla imes B )+ B imes( abla imes r )+( r \cdot abla ) B +( B \cdot abla ) r[/latex] ; but B is constant, so all derivatives of B vanish, and ∇ × r = 0 (Prob. 1.63), so

[latex] abla ( r \cdot B )=( B \cdot abla ) r =\left(B_{x} \frac{\partial}{\partial x}+B_{y} \frac{\partial}{\partial y}+B_{z} \frac{\partial}{\partial z} ight)(x \hat{ x }+y \hat{ y }+z \hat{ z })=B_{x} \hat{ x }+B_{y} \hat{ y }+B_{z} \hat{ z }= B[/latex] ;

[latex] abla \left(r^{2} ight)=\left(\hat{ x } \frac{\partial}{\partial x}+\hat{ y } \frac{\partial}{\partial y}+\hat{ z } \frac{\partial}{\partial z} ight)\left(x^{2}+y^{2}+z^{2} ight)=2 x \hat{ x }+2 y \hat{ y }+2 z \hat{ z }=2 r[/latex] . So

[latex] abla \cdot W =\alpha[3( r \cdot B )+( r \cdot B )]+\beta[0+2( r \cdot B )]=2( r \cdot B )(2 \alpha+\beta), ext { which is zero if } 2 \alpha+\beta=0[/latex] .

[latex] abla imes W =\alpha[( r \cdot B )( abla imes r )- r imes abla ( r \cdot B )]+\beta\left[r^{2}( abla imes B )- B imes abla \left(r^{2} ight) ight]=\alpha[0-( r imes B )]+\beta[0-2( B imes r )]=-( r imes B )(\alpha-2 \beta)=-\frac{1}{2}( r imes B )[/latex] (Prob. 5.25). So we want [latex]\alpha-2 \beta=1 / 2[/latex] . Evidently [latex] \alpha-2(-2 \alpha)=5 \alpha=1 / 2[/latex] ,

[latex] ext { or } \alpha=1 / 10 ; \beta=-2 \alpha=-1 / 5[/latex] . Conclusion: [latex]W =\frac{1}{10}\left[ r ( r \cdot B )-2 r^{2} B ight][/latex] . (But this is certainly not unique.)

[latex] ext { (c) } abla imes W = A \Rightarrow \int( abla imes W ) \cdot d a =\int A \cdot d a . ext { Or } \oint W \cdot d l =\int A \cdot d a[/latex] . Integrate around the amperian loop shown, taking W to point parallel to the axis, and choosing W = 0 on the axis:

[latex]-W l=\int_{0}^{s}\left(\frac{\mu_{0} n I}{2} ight) l \bar{s} d \bar{s}=\frac{\mu_{0} n I}{2} \frac{s^{2} l}{2}[/latex] (using Eq. 5.72 for A).

[latex]A =\frac{\mu_{0} n I}{2} s \hat{\phi}, \quad ext { for } s \leq R [/latex] (5.72)

[latex]W =-\frac{\mu_{0} n I s^{2}}{4} \hat{ z }[/latex] (s < R).

[latex] ext { For } s>R,-W l=\frac{\mu_{0} n I R^{2} l}{4}+\int_{R}^{s}\left(\frac{\mu_{0} n I}{2} ight) \frac{R^{2}}{\bar{s}} l d \bar{s}=\frac{\mu_{0} n I R^{2} l}{4}+\frac{\mu_{0} n I R^{2} l}{2} \ln (s / R)[/latex] ;

[latex]W =-\frac{\mu_{0} n I R^{2}}{4}[1+2 \ln (s / R)] \hat{ z } [/latex] (s > R).

Question 5.55: Just as ∇ · B = 0 allows us to express B as the curl of a ve...

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