Question 6.5: A uniform current density J = J0 zˆ fills a slab straddling ...

A uniform current density J =J_{0} \hat{ z } fills a slab straddling the yz plane, from x = –a to x = +a. A magnetic dipole m =m_{0} \hat{ x } is situated at the origin. 

(a) Find the force on the dipole, using Eq. 6.3.

F = (m · B)                             (6.3)

(b) Do the same for a dipole pointing in the y direction: m =m_{0} \hat{ y }

(c) In the electrostatic case, the expressions F = (p · E) and F = (p · )E are equivalent (prove it), but this is not the case for the magnetic analogs (explain why). As an example, calculate (m · )B for the configurations in (a) and (b).

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(a) B =\mu_{0} J_{0} x \hat{ y } (Prob. 5.15). m·B = 0, so Eq. 6.3 says F = 0. 

(b)  m \cdot B =m_{0} \mu_{0} J_{0} x, \text { so } \mathbf { F } =m_{0} \mu_{0} J_{0} \hat{ x } .

(c) Use product rule #4: (p · E) = p \times( \nabla \times E )+ E \times( \nabla \times p )+( p \cdot \nabla ) E +( E \cdot \nabla) p .

But p does not depend on (x, y, z), so the second and fourth terms vanish, and ∇ × E = 0, so the first term is zero. Hence (p · E) = (p · )E. qed

This argument does not apply to the magnetic analog, since ∇ × B 6= 0. In fact, \nabla ( m \cdot B )=( m \cdot \nabla ) B +\mu_{0}( m \times J ) .

( m \cdot \nabla ) B _{a}=m_{0} \frac{\partial}{\partial x}( B )=m_{0} \mu_{0} J_{0} \hat{ y },( m \cdot \nabla ) B _{b}=m_{0} \frac{\partial}{\partial y}\left(\mu_{0} J_{0} x \hat{ y }\right)=0 .

6.5

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