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## Main Question or Discussion Point

I am having trouble understanding part of the derivation of Faraday's law given in this lecture at around 57:00

So the first goal is to calculate the change in magnetic flux through a changing closed loop in a dynamic magnetic field, and it's given by the following

[tex]\phi(t+\Delta t)-\phi(t)=\iint_{S+\Delta S} B(t+\Delta t).dA - \iint_S B(t).dA[/tex]

where S is surface bounded by the loop at time t, S+ΔS is the surface bounded by the loop at t+Δt

[tex]\iint_{S+\Delta S} B(t+\Delta t).dA[/tex] is then rewritten as [tex]\iint_S B(t+\Delta t).dA + \iint_{\Delta S} B(t).dA[/tex]

The part I don't understand is why the surface integral over the surface ΔS now calculated for the field at time t - shouldn't it be B(t+Δt)? In the lecture (at 1:02:15), he says it doesn't matter because ΔS is first order in time, it's proportional to Δt. And the difference* is going to be proportional to Δt so we don't have to worry about the change in time over an infinitesimal surface because it's second order in time.

What does first order/second order in time mean?

*I think the difference he refers to is this

[tex]\iint_{\Delta S} B(t+\Delta t).dA - \iint_{\Delta S} B(t).dA[/tex]

And in the next part, the equation we arrive at is

[tex]\Delta \phi=\Delta t \iint_S \frac{\partial B}{\partial t}.dA + \iint_{\Delta S} B.dA[/tex]

I'm not sure why the Δs in the first integral become derivatives (shouldn't we then have dt rather than Δt in front of the integral?), but I think these are all meant to be infinitesimals.

And another thing which relates to my first question is why B in the second integral is no longer time-dependent.

So the first goal is to calculate the change in magnetic flux through a changing closed loop in a dynamic magnetic field, and it's given by the following

[tex]\phi(t+\Delta t)-\phi(t)=\iint_{S+\Delta S} B(t+\Delta t).dA - \iint_S B(t).dA[/tex]

where S is surface bounded by the loop at time t, S+ΔS is the surface bounded by the loop at t+Δt

[tex]\iint_{S+\Delta S} B(t+\Delta t).dA[/tex] is then rewritten as [tex]\iint_S B(t+\Delta t).dA + \iint_{\Delta S} B(t).dA[/tex]

The part I don't understand is why the surface integral over the surface ΔS now calculated for the field at time t - shouldn't it be B(t+Δt)? In the lecture (at 1:02:15), he says it doesn't matter because ΔS is first order in time, it's proportional to Δt. And the difference* is going to be proportional to Δt so we don't have to worry about the change in time over an infinitesimal surface because it's second order in time.

What does first order/second order in time mean?

*I think the difference he refers to is this

[tex]\iint_{\Delta S} B(t+\Delta t).dA - \iint_{\Delta S} B(t).dA[/tex]

And in the next part, the equation we arrive at is

[tex]\Delta \phi=\Delta t \iint_S \frac{\partial B}{\partial t}.dA + \iint_{\Delta S} B.dA[/tex]

I'm not sure why the Δs in the first integral become derivatives (shouldn't we then have dt rather than Δt in front of the integral?), but I think these are all meant to be infinitesimals.

And another thing which relates to my first question is why B in the second integral is no longer time-dependent.

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