Area-change-rate of a coil generating a given induced current

[greg_rack]

Homework Statement

A magnetic field runs through a coil of area $A$, parallel to its normal direction, and with intensity increasing at a constant rate $\frac{dB}{dt}=0.20Ts^{-1}$.
$\rightarrow$ suppose even the area can be changed at a constant rate; what should this rate be, in the instant when $B=1.8T$, so that the induced E.M.F. in the coil is $0$?

DISCLAIMER: I haven't studied integrals yet

Relevant Equations

Faraday-Neumann-Lenz

I managed to solve this problem by writing the total E.M.F. as the sum of the one which would have been induced with only the magnetic field varying(and constant $A$), and that with only the area varying(and constant $B$).

However, I got to this solution(which doesn't totally convince me) in a bit cumbersome way... and I cannot really get why it is correct to sum the individual E.M.F.s, while it isn't to find the $EMF(t)=2\frac{\Delta B}{\Delta t} \cdot \frac{\Delta A}{\Delta t} \cdot t$ simply by deriving the flux at given instant t.

The answer must be silly, but I'm drowning in a cup of water :)

 

[Steve4Physics]

Homework Statement:: A magnetic field runs through a coil of area $A$, parallel to its normal direction, and with intensity increasing at a constant rate $\frac{dB}{dt}=0.20Ts^{-1}$.
$\rightarrow$ suppose even the area can be changed at a constant rate; what should this rate be, in the instant when $B=1.8T$, so that the induced E.M.F. in the coil is $0$?

Emf = $-\frac {d\Phi }{dt}$ and (with the arrangement described) $\Phi = BA$. Hint: have you covered the chain rule yet?

 

[Delta2]

Hint: have you covered the chain rule yet?

I think you mean the product rule for derivatives right?

 

[greg_rack]

Emf = $-\frac {d\Phi }{dt}$ and (with the arrangement described) $\Phi = BA$.

That's right, but then, why doesn't it work if I use the condition Emf(t)=0 with the Emf as the derivative of the flux $\Phi (t) = B(t)A(t)$? Why is the only working method the one I quoted above in the first post?

Hint: have you covered the chain rule yet?

I honestly don't know what the chain rule is :'(

 

[Delta2]

The derivative of $B(t)A(t)$ is not $2\frac{dB}{dt}\frac{dA}{dt}t$ if that's what you were thinking...

 

[Steve4Physics]

That's right, but then, why doesn't it work if I use the condition Emf(t)=0 with the Emf as the derivative of the flux $\Phi (t) = B(t)A(t)$? Why is the only working method the one I quoted above in the first post?

I honestly don't know what the chain rule is :'(

Sorry - I meant the 'product rule'. Does that help? Apologies for the confusion.

 

[greg_rack]

The derivative of $B(t)A(t)$ is not $2\frac{dB}{dt}\frac{dA}{dt}t$ if that's what you were thinking...

Oh... and how is it then?

My thinking is: given the rate change for the field, $B(t)=\frac{dB}{dt}t$, and same for the area $A(t)=\frac{dA}{dt}t$.
Now, wouldn't the flux at a given time be $\Phi (t)=\frac{dB}{dt}t \cdot \frac{dA}{dt}t=\frac{dB}{dt}\frac{dA}{dt}t^2$?

 

[Steve4Physics]

Oh... and how is it then?

My thinking is: given the rate change for the field, $B(t)=\frac{dB}{dt}t$, and same for the area $A(t)=\frac{dA}{dt}t$.
Now, wouldn't the flux at a given time be $\Phi (t)=\frac{dB}{dt}t \cdot \frac{dA}{dt}t=\frac{dB}{dt}\frac{dA}{dt}t^2$?

If A is constant that would mean dA/dt = 0, which gives $\Phi = 0$. Would that make sense?

Look up 'product rule' to see how to differentiate the product of two functions (e.g. the product of B(t) and A(t)).

 

[Steve4Physics]

My thinking is: given the rate change for the field, $B(t)=\frac{dB}{dt}t$, and same for the area $A(t)=\frac{dA}{dt}t$.
Now, wouldn't the flux at a given time be $\Phi (t)=\frac{dB}{dt}t \cdot \frac{dA}{dt}t=\frac{dB}{dt}\frac{dA}{dt}t^2$?

Your mistake is in the implicit assumptions you have made.

We know dB/dT is constant. You are *assuming* that at t=0, A = 0 and B=0 (and also dA/dt is constant). With these assumptions you are correct in deriving:
emf = $2\frac{dB}{dt}\frac{dA}{dt}t$
[EDIT: we're missing minus sign, but that's not the point.]

But this means the emf could never be zero (except, trivially, at t=0). So you need to check your assumptions. Hint: do you expect area to increase or decrease?

The solution lies in using the product rule to differentiate BA (where B and A are both functions of time).

 

[greg_rack]

Your mistake is in the implicit assumptions you have made.

We know dB/dT is constant. You are *assuming* that at t=0, A = 0 and B=0 (and also dA/dt is constant). With these assumptions you are correct in deriving:
emf = $2\frac{dB}{dt}\frac{dA}{dt}t$
[EDIT: we're missing minus sign, but that's not the point.]

But this means the emf could never be zero (except, trivially, at t=0). So you need to check your assumptions. Hint: do you expect area to increase or decrease?

The solution lies in using the product rule to differentiate BA (where B and A are both functions of time).

I'm getting the point now... so, without assuming A=0 and B=0 at t=0:
$B(t)=B_0+\frac{dB}{dt}t$ and $A(t)=A_0+\frac{dA}{dt}t$(which I expect to be negative), right?

 

[Steve4Physics]

"(which I expect to be negative), right?"

Yes, right - assuming that you mean you expect dA/dt to be negative.

To cancel the effect of B increasing, A must be decreasing (which means dA/dt is negative). That keeps the flux constant (dΦ/dt = 0).

So, from the basic physics, you can tell from the start that you are looking for a negative answer for dA/dt.

 

[greg_rack]

So, from the basic physics, you can tell from the start that you are looking for a negative answer for dA/dt.

Yupp, got it!
Thank you so so much for your patience guys