4 лайка
Anyone?
the current through the galvanometer will be equal to the \displaystyle I = -\frac{1}{R }\frac{dФ}{dt}, so the charge going through the galvanometer will be equal \displaystyle q =\int^t_0 Idt= \frac{Ф}{R}. Then, by expressing the flux as \displaystyle Ф = \int Sn \vec B \vec {ds} and taking \displaystyle H=\frac{B}{\mu_0} we can obtain the needed equation.
7 лайков
So yeah, I obtained the following
\frac{qR}{\mu_0} = \int_a^b \vec H_0 \cdot d\vec S,
but I’m so not sure that we can simply plug in d\vec S = S\cdot nd\vec l and assume that this is fine (chinese hieroglyphs at the end of the pic make me doubt even more.) Maybe this can be proved with the use of vector calculus?
6 лайков
lol
thank you… 
Thanks a lot!