What is the current in the loop

  • Thread starter lloyd21
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  • #1
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Homework Statement


A piece of wire slides, without friction, along two similiar wires. The resistance per unit length of all wires is 0.350 ohms/m, and a constant magnetic field, B= 1.90T, points up, perpendicular to the loop. The distance between the parallel, fixed wires is 10.0cm. At the instant the sliding wires moves with velocity , v= 1.3 m/s, the loop length is L= 30.0cm. what is the current in the loop

Homework Equations


E = -d0 / dt
I=E/R

The Attempt at a Solution


0.350 ohms/m
B= 1.90 T
d = 0.1 m
V = 1.3m/s
L= 0.3 m
E = (1.90T)(0.13m/s) = 0.247V

Resistance in loop = (o.3m)90.350ohms/m) = 0.105 ohms

E= 2.47x10^-1 V
R= 1.05x10^-1 ohms

I = 2.47x10^-1 V / 1.05x10^-1 ohms = 2.35A

My answer is 2.35 A but was told that there is some change in quantity that I missed? I feel like that answer is correct, can anyone show me where I might have skipped a step or got my units wrong? thanks.
 

Answers and Replies

  • #2
gneill
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It looks like you haven't accounted for all four sides comprising the loop when you calculated the resistance.
 
  • #3
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So I would have R = (0.3m)(0.350 ohm/m)(4) = 0.42 ohm?
 
  • #4
gneill
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So I would have R = (0.3m)(0.350 ohm/m)(4) = 0.42 ohm?
No, check the problem description to find the dimensions of the loop. It's a rectangle, not a square.
 
  • #5
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width = 10.0cm (0.1m) length = 30cm (0.3m)

Rectangle area = 0.03m^2?

...i shouldn't be this confused haha
 
  • #6
gneill
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You want the perimeter.
 
  • #7
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width = 2(0.1m) length = 2(0.3m)
= 0.8 m haha! ?
 
  • #8
gneill
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Yes, 0.8 m total length for the perimeter. So what's the resistance of the loop?
 
  • #9
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R = (0.8m)(0.350ohm/m) = 0.28 ohm....?
 
  • #10
gneill
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R = (0.8m)(0.350ohm/m) = 0.28 ohm....?
Yes. Use that in your current calculation.
 
  • #11
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I= E/R

I= 2.47x10^-1 V / 2.80x10^-1 ohm
I= 0.88 A
 
  • #12
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does that look better?
 
  • #14
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Thank You!
 

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