Induced current

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  • #1
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Homework Statement


A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω , what is the magnitude of the induced current?


Homework Equations



[tex]B= \mu_0 \frac{N}{l} I[/tex]

[tex]\Phi_B = \int B.dA[/tex]

[tex]L=\mu_0 N^2 A[/tex]

[tex]L=\frac{\Phi_B}{I}[/tex]

The Attempt at a Solution



Using the above equations and rearranging we get

[tex]I=\frac{\Phi_B}{\mu_0N^2A}[/tex]

[tex]I=\frac{\Phi_B}{(4\pi \times 10^{-7}) 20^2 (0.5)}[/tex]

Now how do I evaluate the flux? I use [tex]\Phi_B = LI[/tex], then the I's will cancel and I can't solve the problem...

Are there simpler ways of solving this problem? How do I need to make use of dB/dt which is given to us?
 

Answers and Replies

  • #2
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You don't need to use the inductance equations (i.e. the equations with 'L') here; you just need Faraday's law of induction:
[tex]
\frac{d\Phi_B}{dt} = - \mathcal{E}
[/tex]

where the 'E' is the EMF.

Remember that currents are only induced by changing magnetic fields, just like magnetic fields are only induced by changing electric fields.
 
  • #3
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You don't need to use the inductance equations (i.e. the equations with 'L') here; you just need Faraday's law of induction:
[tex]
\frac{d\Phi_B}{dt} = - \mathcal{E}
[/tex]

where the 'E' is the EMF.

Remember that currents are only induced by changing magnetic fields, just like magnetic fields are only induced by changing electric fields.
This equation doesn't take into acoount the number of turns in the coil, but I've tried it

[tex]
\frac{-4}{2} = -I (0.4)
[/tex]

This gives us I=5 A, but the correct answer must be 0.5A. I had all the units in the SI units so what's the problem?
 
  • #4
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This equation doesn't take into acoount the number of turns in the coil
Remember that [tex]\Phi_B = \int_A B dA [/tex] is the total flux, not just the magnetic field.

The flux will be the area times the number of loops, times the magnetic field.
 
  • #5
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11
Remember that [tex]\Phi_B = \int_A B dA [/tex] is the total flux, not just the magnetic field.

The flux will be the area times the number of loops, times the magnetic field.
so that flux is A.N.B, but we don't know what the magnetic field is, the only thing we are given is dB/dt.

[tex]\Phi_B=A \times N \time B= (0.5) (20) (\mu_0 \frac{N}{l} I)[/tex]

this doesn't work because we don't know what I is (we're looking for I), do you see the problem?
 
  • #6
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so that flux is A.N.B, but we don't know what the magnetic field is, the only thing we are given is dB/dt
You are given both the magnetic field, and the change in it. [tex]B_0 = 2T[/tex], [tex]B_1 = 6T[/tex], [tex]\Delta t = 2s[/tex]


[tex]\Phi_B=A \times N \time B= (0.5) (20) (\mu_0 \frac{N}{l} I)[/tex]
Again, this is not the right equation.

[tex] \frac{d \Phi_B}{dt} = - \mathcal{E} = I R = \frac{d}{dt} A N B = A N \frac{dB}{dt} [/tex]
You know everything except I.
 
  • #7
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11
Again, this is not the right equation.

[tex] \frac{d \Phi_B}{dt} = - \mathcal{E} = I R = \frac{d}{dt} A N B = A N \frac{dB}{dt} [/tex]
You know everything except I.
I used your equation

[tex]I R = \frac{d}{dt} A N B = A N \frac{dB}{dt} [/tex]

[tex]I = \frac{2 \times 20 \times 0.5}{0.4} = 50[/tex]

So if this is the correct equation, why I'm still not getting the right answer?
 
  • #8
1,254
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[tex]I R = \frac{d}{dt} A N B = A N \frac{dB}{dt} [/tex]

[tex]I = \frac{2 \times 20 \times 0.5}{0.4} = 50[/tex]

So if this is the correct equation, why I'm still not getting the right answer?
50 square centimeters is not 0.5 square meters.
 

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