• I

## Main Question or Discussion Point

Hi.
The induced emf is given by -d/dt ∫B.dS but when the time derivative is taken inside the integral sign this becomes -∫ ∂B /∂t.dS .
Why isn't B.dS differentiated using the product rule giving an extra term inside the integral sign ?

For some reason the integral sign is appearing as a small circle in the above equations

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kuruman
Homework Helper
Gold Member
Why isn't B.dS differentiated using the product rule giving an extra term inside the integral sign ?
Because according to the equation, first you integrate and then you take the derivative. After the integral is done, all that's left is a function of time with no spatial dependence. Therefore, when you put the derivative inside the integral, you are essentially saying "Instead of integrating over space first and then find the time rate of change of the result, I am going to take the time derivative keeping all else constant first and then integrate over space."

The small circle means doing a line integral over a closed loop.

I understand why the time derivative can be moved inside the integral sign but I don't understand why B.dS isn't differentiated as (∂B/∂t).dS + B. ∂(dS)/ ∂t ?

kuruman
Homework Helper
Gold Member
I understand why the time derivative can be moved inside the integral sign but I don't understand why B.dS isn't differentiated as (∂B/∂t).dS + B. ∂(dS)/ ∂t ?
Oh, that. Because ##d\vec s=dx \hat x+dy \hat y+dz\hat z## and does not depend explicitly on time.

• dyn
ZapperZ
Staff Emeritus
Oh, that. Because ##d\vec s=dx \hat x+dy \hat y+dz\hat z## and does not depend explicitly on time.
This is not right. dS is not a differential line segment, it is a differential surface area.

The time derivative CAN be on the surface element, because the cross-sectional area that the magnetic field passes through can change (example: a shrinking loop). The induced EMF depends on 3 factors:

1. dB/dt
2. dS/dt
3. dθ/dt, where θ is the angle between the area and B.

Zz.

• berkeman
kuruman
Homework Helper
Gold Member
This is not right.
Of course it isn't. My mind was addled when I mistook the surface integral for a line integral. Even if it is a closed loop line integral, the differential could, in general, be time-dependent if the loop is, say, shrinking.

So is the most general way of writing the time derivative of B.dS as (∂B/∂t).dS + B.(∂(dS)/∂t) ?

ZapperZ
Staff Emeritus
So is the most general way of writing the time derivative of B.dS as (∂B/∂t).dS + B.(∂(dS)/∂t) ?
No, there is still the dot-product between the two, which means that the angle between B and dS may also have a time-rate of change (example: a spinning loop).

If you already know the direction of the emf, then deal with B cosθ dS, and take the time derivative of that. However, please note that at the intro/general level, usually the time dependence is only on ONE of those. So trying to expand out the derivative isn't really that useful. It is easier to simply find the magnetic flux first, and then find the time derivative of that. After all, Faraday's law is defined with the time rate of change of the flux.

Zz.

Thanks. How would the time derivative of B.dS be written out in the general case ?

ZapperZ
Staff Emeritus
Thanks. How would the time derivative of B.dS be written out in the general case ?
??

Can you not find the derivative of the product of xyz? Same thing here: B cosθ dS.

Zz.

Yes I can but I meant in vector notation ?

I have looked at that reference for the derivative of a scalar product and it looks the same as the earlier version I wrote
So is the most general way of writing the time derivative of B.dS as (∂B/∂t).dS + B.(∂(dS)/∂t) ?