- #1
stunner5000pt
- 1,461
- 2
stokes theorem says [tex] \int_{S} (\nabla \times G) \bullet dA = \oint_{C} G \bullet ds [/tex] where C is a closed curve bounding the open surface S. Note : dA =n dA, and ds = t ds
use stokes theorem to prove that hte Faraday law of induction [tex] \oint_{C} E \bullet t ds = -\frac{d}{dt} \int_{S_{C}} B \bullet n dA [/tex] can be written as a differential equation [tex] \nabla \times E + \frac{\partial B}{\partial t} = 0 [/tex]
now i can easily rearrange the left hand side of the integral equation to get [tex] \int_{S_{C}} (\nabla \times E) \bullet dA = - \frac{d}{dt} \int_{S_{C}} B \bullet n dA [/tex]
now I am not allowd to 'cancel' out the dA terms can i ? Or perhaps find the gradient on each side?? Please help on this!
use stokes theorem to prove that hte Faraday law of induction [tex] \oint_{C} E \bullet t ds = -\frac{d}{dt} \int_{S_{C}} B \bullet n dA [/tex] can be written as a differential equation [tex] \nabla \times E + \frac{\partial B}{\partial t} = 0 [/tex]
now i can easily rearrange the left hand side of the integral equation to get [tex] \int_{S_{C}} (\nabla \times E) \bullet dA = - \frac{d}{dt} \int_{S_{C}} B \bullet n dA [/tex]
now I am not allowd to 'cancel' out the dA terms can i ? Or perhaps find the gradient on each side?? Please help on this!
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