Current Carry Wire With A Loop

In summary, the magnetic flux through a square loop of side a is a function of the current I and the loop's speed v. If the loop is pulled away from the wire at speed v, an emf is induced in it.
  • #1
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1. Homework Statement :
A. Determine the magnetic flux through a square loop of side a (see the figure) if one side is parallel to, and a distance b from, a straight wire that carries a current I.

Express your answer in terms of the variables I, a, b, and appropriate constants.

Answer: [(uIa)/(2pi)]ln((b+a)/b))

B. If the loop is pulled away from the wire at speed v, what emf is induced in it?
Express your answer in terms of the variables I, a, b, v, and appropriate constants.

Answer: Help




2. Homework Equations :
Flux=integral(B*da)
E=BLv


I think those are the two equations we need, but not 100% sure.

3. The Attempt at a Solution :
I have already figured out part A, and the solution is posted above, just having issue with part B.

So all I did was replaced L=length with "a", one side of the area. I am totally lost here. So please help me.
 
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  • #2
Have you tried using Faraday's law?
 
  • #3
Well yes, but how to relate velocity?
 
  • #4
Charanjit said:
Well yes, but how to relate velocity?

The rate of change per unit time of [tex] \Phi_B [/tex] is a function of the loop's velocity.

Hint: Distance = Velocity times Time. Distance here, as your problem describes it, is your variable 'b'. Next find the negative of the time derivative of [tex] \Phi_B [/tex].
 
  • #5
Which b are you talking about? The equation E=BLv? That "B" is the magnetic field.
 
  • #6
Charanjit said:
Which b are you talking about? The equation E=BLv? That "B" is the magnetic field.

In the problem statement, "...and a distance b from ..."
 
  • #7
Ok... so how do I import that into the equation? We don't have the "time" variable.
 
  • #8
Charanjit said:
Ok... so how do I import that into the equation? We don't have the "time" variable.

Distance is equal to Time times Velocity, or b = vt.

Make the substitution as an interim step. You can always make the reverse substitution (t = b/v) later to get rid of the 't' variable if you need to, once you have your final answer.
 
  • #9
Ok, but in which equation? E=BLv?
 
  • #10
Charanjit said:
Ok, but in which equation? E=BLv?

try the answer to the first part:
Answer: [(uIa)/(2pi)]ln((b+a)/b))

After you make the substitution, you'll still need to use Faraday's law as a step to the final answer.
 
  • #11
Ahh ok. I got it. Thanks.
 

What is a "Current Carry Wire With A Loop"?

A current carry wire with a loop is a type of electrical wire that has a loop or circular shape at one or both ends. This loop allows the wire to be easily connected to other wires or components in an electrical circuit.

What is the purpose of the loop in a current carry wire?

The loop in a current carry wire serves as a connector for other wires or components in an electrical circuit. It allows for a secure and efficient connection, making it easier to create and maintain electrical circuits.

What are the common uses of a current carry wire with a loop?

Current carry wires with loops are commonly used in a variety of electrical applications, such as in household wiring, electronic devices, and industrial equipment. They are also commonly used for creating temporary or portable electrical connections.

What materials are current carry wires with loops typically made of?

Current carry wires with loops can be made of various materials, including copper, aluminum, and other conductive metals. The type of material used often depends on the specific application and desired electrical properties.

How do you properly install or connect a current carry wire with a loop?

The proper installation or connection of a current carry wire with a loop will depend on the specific application and circuit. However, in general, the loop should be securely connected to the desired component or wire using appropriate connectors or methods, such as twisting, soldering, or crimping.

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