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

In summary, the flux at a given time is: ##\Phi (t)=\frac{dB}{dt}t \cdot \frac{dA}{dt}t=\frac{dB}{dt}\frac{dA}{dt}t^2##.
  • #1

greg_rack

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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 :)
 
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  • #2
greg_rack said:
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?
 
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  • #3
Steve4Physics said:
Hint: have you covered the chain rule yet?
I think you mean the product rule for derivatives right?
 
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  • #4
Steve4Physics said:
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?

Steve4Physics said:
Hint: have you covered the chain rule yet?
I honestly don't know what the chain rule is :'(
 
  • #5
The derivative of ##B(t)A(t)## is not ##2\frac{dB}{dt}\frac{dA}{dt}t## if that's what you were thinking...
 
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  • #6
greg_rack said:
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.
 
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  • #7
Delta2 said:
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##?
 
  • #8
greg_rack said:
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)).
 
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  • #9
greg_rack said:
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).
 
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  • #10
Steve4Physics said:
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?
 
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  • #11
"(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.
 
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  • #12
Steve4Physics said:
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
 
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