Question 6.22: In Prob. 6.4, you calculated the force on a dipole by “brute...

In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop

B ( r ) \cong B \left( r _{0}\right)+\left[\left( r - r _{0}\right) \cdot \nabla_{0}\right] B \left( r _{0}\right) ,

where r _{0} is the position of the dipole and \nabla _{0} denotes differentiation with respect to r _{0}. Put this into the Lorentz force law (Eq. 5.16) to obtain

F _{ mag }=\int I(d l \times B )                        (5.16)

F =I \oint d l \times\left[\left( r \cdot \nabla_{0}\right) B \left( r _{0}\right)\right] .

Or, numbering the Cartesian coordinates from 1 to 3:

F_{i}=I \sum_{j, k, l=1}^{3} \epsilon_{i j k}\left\{\oint r_{l} d l_{j}\right\}\left[\nabla_{0_{l}} B_{k}\left( r _{0}\right)\right] ,

where  \epsilon_{i j k} is the Levi-Civita symbol (+1 if i jk = 123, 231, or 312; 1 if i jk  = 132, 213, or 321; 0 otherwise), in terms of which the cross product can be written ( A \times B )_{i}=\sum_{j, k=1}^{3} \epsilon_{i j k} A_{j} B_{k} . Use Eq. 1.108 to evaluate the integral. Note that

\sum_{j=1}^{3} \epsilon_{i j k} \epsilon_{l j m}=\delta_{i l} \delta_{k m}-\delta_{i m} \delta_{k l} ,

where \delta_{i j} is the Kronecker delta (Prob. 3.52). 

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F =I \oint d l \times B =I(\oint d l ) \times B _{0}+I \oint d l \times\left[\left( r \cdot \nabla _{0}\right) B _{0}\right]-I(\oint d l ) \times\left[\left( r _{0} \cdot \nabla _{0}\right) B _{0}\right]=I \oint d l \times\left[\left( r \cdot \nabla _{0}\right) B _{0}\right]

(because \oint d l = 0 ). Now 

\left(d l \times B _{0}\right)_{i}=\sum_{j, k} \epsilon_{i j k} d l_{j}\left(B_{0}\right)_{k}, \quad \text { and }\left( r \cdot \nabla _{0}\right)=\sum_{l} r_{l}\left( \nabla _{0}\right)_{l}, \text { so }

 

F_{i}=I \sum_{j, k, l} \epsilon_{i j k}\left[\oint r_{l} d l_{j}\right]\left[\left(\nabla_{0}\right)_{l}\left(B_{0}\right)_{k}\right] \quad\left\{\text { Lemma 1: } \quad \oint r_{l} d l_{j}=\sum_{m} \epsilon_{l j m} a_{m}(\text { proof below }) .\right\}

 

=I \sum_{j, k, l, m} \epsilon_{i j k} \epsilon_{l j m} a_{m}\left(\nabla_{0}\right)_{l}\left(B_{0}\right)_{k} \quad\left\{\text { Lemma } 2: \quad \sum_{j} \epsilon_{i j k} \epsilon_{l j m}=\delta_{i l} \delta_{k m}-\delta_{i m} \delta_{k l}(\text { proof below }) .\right\}

 

=I \sum_{k, l, m}\left(\delta_{i l} \delta_{k m}-\delta_{i m} \delta_{k l}\right) a_{m}\left(\nabla_{0}\right)_{l}\left(B_{0}\right)_{k}=I \sum_{k}\left[a_{k}\left(\nabla_{0}\right)_{i}\left(B_{0}\right)_{k}-a_{i}\left( \nabla _{0}\right)_{k}\left(B_{0}\right)_{k}\right]

 

=I\left[\left( \nabla _{0}\right)_{i}\left( a \cdot B _{0}\right)-a_{i}\left( \nabla _{0} \cdot B _{0}\right)\right] .

But  \nabla _{0} \cdot B _{0}=0(\text { Eq. } 5.50), \text { and } m =I a (\text { Eq. } 5.86), \text { so } F = \nabla _{0}\left( m \cdot B _{0}\right) (the subscript just reminds us to take the derivatives at the point where m is located). qed

· B = 0                                     (5.50) 

m \equiv I \int d a =I a                             (5.86)

Proof of Lemma 1:

Eq. 1.108 says \oint( c \cdot r ) d l = a \times c =- c \times a . The jth component is \sum_{p} \oint c_{p} r_{p} d l_{j}=-\sum_{p, m} \epsilon_{j p m} c_{p} a_{m} .  Pick

\oint( c \cdot r ) d l = a \times c                        (1.108)

c_{p}=\delta_{p l} (i.e. 1 for the lth component, zero for the others). Then \oint r_{l} d l_{j}=-\sum_{m} \epsilon_{j l m} a_{m}=\sum_{m} \epsilon_{l j m} a_{m} .  qed

Proof of Lemma 2:

\epsilon_{i j k} \epsilon_{l j m}=0 \text { unless } i j k \text { and } l j m are both permutations of 123. In particular, i must either be l or m, and k must be the other, so

\sum_{j} \epsilon_{i j k} \epsilon_{l j m}=A \delta_{i l} \delta_{k m}+B \delta_{i m} \delta_{k l} .

To determine the constant A, pick i = l = 1, k = m = 3; the only contribution comes from j = 2:

\epsilon_{123} \epsilon_{123}=1=A \delta_{11} \delta_{33}+B \delta_{13} \delta_{31}=A \Rightarrow A=1 .

To determine B, pick i = m = 1, k = l = 3:

\epsilon_{123} \epsilon_{321}=-1=A \delta_{13} \delta_{31}+B \delta_{11} \delta_{33}=B \Rightarrow B=-1 .

So

\sum_{j} \epsilon_{i j k} \epsilon_{l j m}=\delta_{i l} \delta_{k m}-\delta_{i m} \delta_{k l} qed

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