- #1
happyparticle
- 456
- 21
- Homework Statement
- Find boundary conditions ##\vec{B}## and ##\vec{H}## for a cylinder of radius a and length 4a and ##\vec{M} = M\hat{z}## on the axis of the cylinder
- Relevant Equations
- ##\vec{\nabla} \cdot \vec{B} = 0##
##\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}##
##\vec{\nabla} x \vec{B} = \mu_0 \vec{J}##
##\vec{\nabla} x \vec{H} = \mu_0 \vec{J}_f##
When asking for boundary conditions I'm wondering if this is enough in this situation to give
##\vec{\nabla} \cdot \vec{B} = 0 , B_{2\perp} - B_{1 \perp} = 0##
##\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}, H_{2\perp} - H_{1 \perp} = - (M_{2\perp} - M_{1 \perp})##
##\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}, B_{2\||} - B_{1 \||} = \mu_0 \vec{K} \times \hat{n}##
##\vec{\nabla} \times \vec{H} = \mu_0 \vec{J}_f , H_{2 \||} - H_{1 \||} = \vec{K}_f \times \hat{n}##
##\vec{\nabla} \cdot \vec{B} = 0 , B_{2\perp} - B_{1 \perp} = 0##
##\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}, H_{2\perp} - H_{1 \perp} = - (M_{2\perp} - M_{1 \perp})##
##\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}, B_{2\||} - B_{1 \||} = \mu_0 \vec{K} \times \hat{n}##
##\vec{\nabla} \times \vec{H} = \mu_0 \vec{J}_f , H_{2 \||} - H_{1 \||} = \vec{K}_f \times \hat{n}##