Mechanical Work vs Heat Development: Comparing Fv and i^2R in a Circuit

  • Thread starter shikagami
  • Start date
In summary, the conversation discusses a physics problem involving a conducting rod in a magnetic field. The conversation covers calculating the magnitude and direction of induced emf, the force required to maintain the rod in motion, and comparing the rate of mechanical work done by force with the rate of heat development in the circuit. The main focus is on understanding and solving the problem step by step.
  • #1
shikagami
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I don't quite understand what I'm suppose to do in this question:

--> Compare the rate at which mechanical work is done by the force (Fv) with the rate of development of heat in the circuit (i^2 R).
 
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  • #2
shikagami said:
--> Compare the rate at which mechanical work is done by the force (Fv) with the rate of development of heat in the circuit (i^2 R).

Okay, to start, where is the "mechanical work" coming from? What is the force involved? What is it acting upon?

Hint: There's more than one formula for the power of a circuit.
 
  • #3
SpaceTiger said:
Okay, to start, where is the "mechanical work" coming from? What is the force involved? What is it acting upon?

Hint: There's more than one formula for the power of a circuit.


Ok, the other formulas for power are P=V^2/R and P=VI. I don't know anything about the force.
 
  • #4
shikagami said:
Ok, the other formulas for power are P=V^2/R and P=VI. I don't know anything about the force.

Ok, good start. It would help, however, if you could give a little more info on the circuit. Other than a resistor, what is it composed of?
 
  • #5
Ummm... here's the entire problem:

A conducting rod AB makes contact with metal rails CA and DB. The apparatus is in a uniform magnetic field 0.5 T, perpendicular to the plane of the diagram. The rod is 50 cm long.

a)find the magnitude and direction of the emf induced in the rod when it is moving toward the right with a speed 4 m/s.

b)If the resistance of circuit ABDC is 0.2 ohms (assumed constant), find the force required to maintain the rod in motion. Neglect friction.

c)Compare the rate at which mechanical work is done by force (Fv) with the rate of development of heat in the circuit (i^2 R).
 
  • #6
Ok, then it's referring to the force that's pushing the rod. If you did part b, you should already have that. Did you do part b?
 
  • #7
SpaceTiger said:
Ok, then it's referring to the force that's pushing the rod. If you did part b, you should already have that. Did you do part b?

yup. I got 0.0125 N.
 
  • #8
So then calculate the rate of work done by that force (0.0125 N * v) and compare it to the power dissipated in the circuit (i^2R). Did you calculate the current in the circuit?
 
  • #9
Ok, I got 0.0500 j for the work. And I got .500 Amps for the current. Now what?
 
  • #10
Can you show your work on the first two parts? I'm getting a number 100 times larger than yours.
 
  • #11
SpaceTiger said:
Can you show your work on the first two parts? I'm getting a number 100 times larger than yours.

Ok.
a) emf= Bvl --> (4.00m/s)(0.500T)(0.0500m) = .100 V

b) I=emf/R --> (0.100V)/(0.200 ohms) = 0.500 A

Fmag= IBl --> (.500Amps)(0.500T)(0.0500m) = 0.0125N
 
  • #12
shikagami said:
Ok.
a) emf= Bvl --> (4.00m/s)(0.500T)(0.0500m) = .100 V

How many meters is 50 centimeters?
 
  • #13
OOps... sry.. its only 0.500 m = to 50 cm. Sorry bout that. need to change all my data now...
 
  • #14
Everything moves 1 decimal point to the right.
 
  • #15
shikagami said:
Everything moves 1 decimal point to the right.

Everything?
 
  • #16
SpaceTiger said:
Everything?

lol.. ur funny. Well... here's what I have now:

a) emf =1.00 V
b) I=5.00 Amps, Fmag=1.25N
c) Fv=5.00j

wait.. u got me.. crap.. I keep messing up. Thnx for dealing with my overwhelming stupidity.
 
  • #17
You're not being stupid, it's alright. Two things:

1) Look at the units in the last part. Are you sure they should be Joules?
2) What do you get when you calculate the power dissipated in the resistor (i^2R)?
 
  • #18
SpaceTiger said:
You're not being stupid, it's alright. Two things:

1) Look at the units in the last part. Are you sure they should be Joules?
2) What do you get when you calculate the power dissipated in the resistor (i^2R)?

Wahh...
1) it's watts not joules ... sorry. :cry:
2) wow.. I^2 R --> (5.00Amps)^2 (0.200 ohms) = 5.00 watts
 
  • #19
shikagami said:
2) wow.. I^2 R --> (5.00Amps)^2 (0.200 ohms) = 5.00 watts

The point of all this is that the work you put into the system by pulling the rod along the tracks is dissipated as thermal energy in the resistor. Note that none of the work is going into accelerating the rod because it is moving at a constant velocity!
 
  • #20
WOW ur a genius! Will u be my homework buddy? LOL Thanks a lot for the help even though it's like uber late there in NJ. I might be bugging you for more physics stuff. Hope you don't mind.
 
  • #21
Anyways, thank you very much again for all of your help SpaceTiger.
 

1. What is mechanical work in a circuit?

Mechanical work in a circuit refers to the energy expended in moving charges through a circuit. This work is typically done by a source of electrical energy, such as a battery or power supply, and is used to power devices and perform tasks.

2. What is heat development in a circuit?

Heat development in a circuit refers to the energy that is dissipated as heat due to resistance in the circuit. This heat is a byproduct of the flow of current and is often unwanted, as it can cause components to overheat and fail.

3. What is the difference between Fv and i^2R in a circuit?

Fv (force times velocity) and i^2R (current squared times resistance) are both ways of measuring the amount of mechanical work and heat development in a circuit, respectively. Fv takes into account the force required to move charges through a circuit, while i^2R takes into account the resistance that the charges encounter.

4. How do Fv and i^2R affect the overall performance of a circuit?

The values of Fv and i^2R can greatly impact the overall performance of a circuit. A high Fv indicates a large amount of mechanical work being done, which can power devices and perform tasks efficiently. However, a high i^2R can lead to excessive heat development, which can reduce the efficiency and reliability of the circuit.

5. How can Fv and i^2R be optimized in a circuit?

To optimize Fv and i^2R in a circuit, it is important to consider the design and components used. Choosing components with lower resistance can reduce i^2R and minimize heat development. Additionally, properly sizing and selecting a power source can ensure that Fv is sufficient for the desired tasks without causing excessive heat. Regular maintenance and monitoring can also help to optimize Fv and i^2R in a circuit.

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