Faraday's Law of Induction - Current in Multiple Wires

  • #1
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Homework Statement



A 7.40 cm diameter coil consists of 15 turns of circular copper wire 2.3 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 7.29×10^−3 T/s . Determine the current in the loop. Express your answer using two significant figures.

Homework Equations



magnetic flux = BA
ε = -N Δmagnetic flux / ΔT

The Attempt at a Solution



I'm still a little confused as what to do with the 7.40 cm and 2.3 mm (other than convert them to SI units).

What I've done so far:

-15 (7.29 * 10^-3 T/s) = ε
ε = -0.109 V

Was that a good place to start?
 

Answers and Replies

  • #2
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The two radii are given so that you could determine the resistance of the wire in coil.
 
  • #3
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Would it make sense if I :

ε = - N (ΔB) / (Δt) * ∏r^2

This would provide a voltage.

R = ρl / a

This would provide a resistance.

I = V / R

This would provide a current.
 
  • #4
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The equation for EMF seems correct. In the equation for resistance, what are all those constants? How do they correspond to the data in the problem?
 
  • #5
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R is resistance.

ρ is the resistivity of the material (copper) = 1.72 * 10^-8 Ωm

l = length of wire = (0.0740 m?)

A = cross sectional area of wire = ∏(0.00115 m)

I'm unsure about the last two.
 
  • #6
6,054
391
The values you listed for the last two are incorrect.

How would you compute the length of wire in a coil?

What about its cross-sectional area?
 
  • #7
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To calculate the length of a wire in a coil:

A coil is a circle. The circumference of a circle is ∏d. The coil has 15 turns so in this case

l = d∏ * 15

To calculate the cross sectional area of the copper wires:

A = ∏(0.00115 m)^2 * 15 (now I'm considering all of the turns).
 
  • #8
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The length looks good. But I don't understand why you multiply the cross-sectional area by the number of turns.
 
  • #9
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Wow! A = ∏r^2 . I just noticed in one of the comments I left off the "square" part.

A = ∏(0.00115 m)^2
 
  • #10
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So I think you have all it takes to solve the problem.
 

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