- #1

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- 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}##