Sir, in the integral form of faraday's law of electromagnetic induction, in the right hand side, is it closed surface integral of B. ds Or simply just surface integral of B. ds ?
Good question. In Faraday's law, it's not a closed surface, so we don't write the little circle. However, it is a closed surface integral in Gauss's law.
Well done man! Good explanation
Iam just wondering why did you use the stocks theorem while you draw only a ligne integral. So we can see that the electric feild is in 1D
Sir, in the integral form of faraday's law of electromagnetic induction, in the right hand side, is it closed surface integral of B. ds Or simply just surface integral of B. ds ?
Good question. In Faraday's law, it's not a closed surface, so we don't write the little circle. However, it is a closed surface integral in Gauss's law.
@@Elucyda good explanation love from india