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

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SUMMARY

This discussion focuses on comparing the rate of mechanical work done by a force (Fv) with the rate of heat development in a circuit (i²R). The problem involves a conducting rod moving in a magnetic field, where the induced electromotive force (emf) is calculated using the formula emf = Bvl, resulting in 1.00 V. The current in the circuit is determined to be 5.00 A, leading to a power dissipation of 5.00 watts in the resistor, which matches the mechanical work done, illustrating the conversion of work into thermal energy in the resistor.

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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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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.
 
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.
 
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?
 
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).
 
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?
 
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.
 
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?
 
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.
 

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