Magnetic field/induction

  • #1
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Homework Statement


A magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time tin the region of the loop. If the induced emf in the loop increases linearly with time t, then the magnitude of the magnetic field must be proportional to:
a)t^3
b)t^2 <answer
c)t
d)t^0
e)t^1/2

Can someone explain this

I did emf=dflux/dt
emf(kt)=BA
B is proportional to t right?
 

Answers and Replies

  • #2
TSny
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I did emf=dflux/dt
emf(kt)=BA
I’m not understanding the notation on the left side of the second equation. What does emf(kt) mean?
 
  • #3
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I’m not understanding the notation on the left side of the second equation. What does emf(kt) mean?
kt to show t is linear
 
  • #4
TSny
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kt to show t is linear
I'm still not understanding the notation emf(kt). Does it denote emf multiplied by kt? Or, is it saying that the emf is the same thing as kt? Or something else?

Going back to the previous equation, you have correctly stated that

emf = d(flux)/dt.

Express emf in terms of t on the left and express flux in terms of B and A on the right. After doing this, integrate both sides with respect to time.
 
  • #5
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I'm still not understanding the notation emf(kt). Does it denote emf multiplied by kt? Or, is it saying that the emf is the same thing as kt? Or something else?

Going back to the previous equation, you have correctly stated that

emf = d(flux)/dt.

Express emf in terms of t on the left and express flux in terms of B and A on the right. After doing this, integrate both sides with respect to time.
d(flux)/dt=AdB/dt
dt's would cancel out leaving d(flux)=AdB
flux/A=B
What am I missing here?
 
  • #6
TSny
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d(flux)/dt=AdB/dt
dt's would cancel out leaving d(flux)=AdB
flux/A=B
What am I missing here?
There is nothing wrong here. I think the problem is how you are handling the left side of the equation emf = d(BA)/dt.

Can you explain in words the steps you took in getting from emf = d(BA)/dt to emf(kt) = BA? (I still don't understand the meaning of "emf(kt)".)

You went from emf = d(BA)/dt to emf(kt) = BA.

What operation did you perform on the right-hand side to get from d(BA)/dt to BA? Did you apply this same operation to the left side?
 
  • #7
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There is nothing wrong here. I think the problem is how you are handling the left side of the equation emf = d(BA)/dt.

Can you explain in words the steps you took in getting from emf = d(BA)/dt to emf(kt) = BA? (I still don't understand the meaning of "emf(kt)".)

You went from emf = d(BA)/dt to emf(kt) = BA.

What operation did you perform on the right-hand side to get from d(BA)/dt to BA? Did you apply this same operation to the left side?
t is linearly related to the emf, so i incorrectly changed dt to kt and multiplied it to emf and took out the derivative for BA.

But the dt's cancel out right? how do you get t^2
 
  • #8
TSny
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t is linearly related to the emf
Yes, so emf = kt where k is a constant.

So, starting with emf = d(BA)/dt you can write

kt = d(BA)/dt.

Can you proceed from here?
 
  • #9
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Yes, so emf = kt where k is a constant.

So, starting with emf = d(BA)/dt you can write

kt = d(BA)/dt.

Can you proceed from here?
yes thanks bud
 

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