B→=μH→\overrightarrow{\mathrm{B}}=\mu \overrightarrow{\mathrm{H}}B=μH
V→⋅B→=0\overrightarrow{V} \cdot \overrightarrow{B}=0V⋅B=0
∮B→⋅dl→=μ0I\oint \overrightarrow{B} \cdot d \overrightarrow{l}=\mu_{0} I∮B⋅dl=μ0I
∇→×E→=−∂B→∂t\overrightarrow{\nabla} \times \overrightarrow{\mathrm{E}}=-\frac{\partial \overrightarrow{\mathrm{B}}}{\partial t}∇×E=−∂t∂B
