In Maxwell's equations, I understand intuitively how: ∮ B ⋅ d a = 0 (because there are no monopoles and so equal number of field lines going in and coming out of the surface).And then using the divergence theorem: ∫ V ( ∇ ⋅ B ) d τ = ∮ S B ⋅ d aThen ∫ V ( ∇ ⋅ B ) d τ must be = 0.But then I'm not sure why I can say: ∇ ⋅ B = 0 and forget about the integral.Does it just mean that ∇ ⋅ B must be zero everywhere?

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In Maxwell&#039;s equations, I understand intuitively how:   ∮  B  ⋅  d      a    =  0 (because there are no monopoles and so equal number of field lines going in and coming out of the surface).And then using the divergence theorem:      ∫          V            (    ∇    ⋅    B    )    d  τ  =      ∮          S        B  ⋅  d  aThen       ∫          V            (    ∇    ⋅    B    )    d  τ must be = 0.But then I&#039;m not sure why I can say:   ∇  ⋅  B  =  0 and forget about the integral.Does it just mean that   ∇  ⋅  B must be zero everywhere?

Braeden Shannon · Accepted Answer

You can actually calculate   ∇  ⋅  B at a certain point             r      →       : just choose as the (arbitrary) integration domain a bowl       V    ϵ    (            r      →        ) of radius   ϵ. Then let ϵ tend to zero, so  (  ∇  ⋅  B  )  (            r      →        )  ⋅      4    3    π      ϵ    3    →      ∫                  V        ϵ            (                        r          →                    )        (  ∇  ⋅  B  )  d  τ  =  0Since   ϵ is small but non-zero, you finally get   ∇  ⋅  B  =  0

In Maxwell's equations, I understand intuitively how: ∮ B ⋅<!-- ⋅

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