A uniform current density J = J0 zˆ fills a slab straddling the yz plane, from x = −a to x = +a. A magnetic dipole m = m0 xˆ is situated at the origin. (a) Find the force on the dipole, using Eq. 6.3. (b) Do the same for a dipole pointing in the y direction: m = m0yˆ. (c) In the electrostatic case

Question

A uniform current density [latex]J =J_{0} \hat{ z }[/latex] fills a slab straddling the yz plane, from x = -a to x = +a. A magnetic dipole [latex]m =m_{0} \hat{ x } [/latex] 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: [latex]m =m_{0} \hat{ y } [/latex]

(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).

Accepted Answer

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

(b) [latex]m \cdot B =m_{0} \mu_{0} J_{0} x, ext { so } \mathbf { F } =m_{0} \mu_{0} J_{0} \hat{ x }[/latex] .

(c) Use product rule #4: ∇(p · E) [latex]= p imes( abla imes E )+ E imes( abla imes p )+( p \cdot abla ) E +( E \cdot abla) p[/latex] .

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, [latex] abla ( m \cdot B )=( m \cdot abla ) B +\mu_{0}( m imes J )[/latex] .

[latex]( m \cdot abla ) B _{a}=m_{0} \frac{\partial}{\partial x}( B )=m_{0} \mu_{0} J_{0} \hat{ y },( m \cdot abla ) B _{b}=m_{0} \frac{\partial}{\partial y}\left(\mu_{0} J_{0} x \hat{ y } ight)=0 [/latex] .

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

Related Answered Questions