# Circuits physics question

1. Jun 5, 2009

### science.girl

1. The problem statement, all variables and given/known data

A metal rod of mass 0.22 kg lies across two parallel conducting rails that are a distance of 0.52 m apart on a tabletop, as shown in the top view above. A 3.0 W resistor is connected across the left ends of the rails. The rod and rails have negligible resistance but significant friction with a coefficient of kinetic friction of 0.20. There is a magnetic field of 0.80 T perpendicular to the plane of the tabletop. A string pulls the metal rod to the right with a constant speed of 1.8 m/s.

(a) Calculate the magnitude of the current induced in the loop formed by the rod, the rails, and the resistor.
(b) Calculate the magnitude of the force required to pull the rod to the right with constant speed.
(c) Calculate the energy dissipated in the resistor in 2.0 s.
(d) Calculate the work done by the string pulling the rod in 2.0 s.
(e) Compare your answers to parts (c) and (d). Provide a physical explanation for why they are equal or unequal.

Page 10 on this document: http://apcentral.collegeboard.com/apc/public/repository/ap09_frq_physics_b.pdf

2. Relevant equations

I =V/R

3. The attempt at a solution
A) I = V/R?
I'm a bit confused... there seems to be an overload of information on this problem. I would appreciate some direction.

Last edited by a moderator: Apr 24, 2017
2. Jun 5, 2009

### LowlyPion

Re: Circuits

Won't the induced current be proportional to the rate of the change of flux in a loop of wire? Consider then the rate of change of the area.

3. Jun 5, 2009

### science.girl

Re: Circuits

Oh gosh... I do recall doing something like this. I'll try to find my notes...

Ok, I found two things that are potentially relevant. I have magnetic flux (BAcos$$\theta$$) and mortional emf (Blv).

Are any of these remotely correct?

4. Jun 5, 2009

### AhmedEzz

Re: Circuits

I am not an expert but personally, I would look into Ampere's law, Faraday's law and biot-savart law (not sure on the latter).

5. Jun 6, 2009

### LowlyPion

Re: Circuits

Yeah, well what is the rate of change of the area of the loop? A = w*Δx

Won't the voltage be something like

V = Δ(B*A)/Δt

With B constant and ΔA/Δt = w*Δx/Δt = w*v

then won't

Voltage = B*w*v

The resistance is fixed at 3Ω.

6. Jun 7, 2009

### science.girl

Re: Circuits

Then apply I = V/R? With R = 3 and V = Bwv? Or am i interpreting this wrong?

7. Jun 9, 2009

### science.girl

Re: Circuits

Oh, got it! Would someone mind helping me with (b), then?

(b) Calculate the magnitude of the force required to pull the rod to the right with constant speed.

Doesn't this just deal with the velocity, (1.8 m/s) and the coefficient of kinetic friction (0.20)?
Edit: Rather, is it Fs = FB + Ff?
Edit2: But how would you find the force related to these?

Last edited: Jun 9, 2009
8. Jun 9, 2009

### LowlyPion

Re: Circuits

Well, I would suggest that you of course have the frictional component, but then you also have the energy that goes into the resistor which is I2R.

9. Jun 9, 2009

### ideasrule

Re: Circuits

Consider what is happening: the rod is moving, which changes the magnetic flux across the circuit, which induces a voltage. This voltage causes a current. The magnetic field then imposes a force on this current that opposes the motion of the rod. Calculate this force, calculate the force of friction, add the two together and you'll get the total force needed to pull the rod.

10. Jun 10, 2009

### science.girl

Re: Circuits

Would the magnitude of the magnetic force be: F = qvBsin$$\theta$$?

And I know one equation that would apply for the magnitude of the force of kinetic friction: fk = $$\mu$$kn
But this equation doesn't seem to give me enough information to solve for the force of kinetic friction... Any help?

11. Jun 10, 2009

### LowlyPion

Re: Circuits

Almost. Consider the form of the magnetic force equation that matches the geometry of the situation you have in the problem.

F = I*B*w*sinθ

where w is the length of the rod across the tracks. (Sinθ of course is 1 for this set up.)

So multiplying F*v = B*v*w*I = V*I

Force times velocity is power. Power is the Voltage found before times current which can also be expressed as I2R

12. Jun 10, 2009

### science.girl

Re: Circuits

Got it! Thank you for your help!

13. Jun 10, 2009

### Redbelly98

Staff Emeritus
Re: Circuits

FYI, something to look for in the future: these introductory magnetism problems generally involve either current-carrying wires, or a charge q moving at velocity v. Since this problem involves current in a wire, you'd want to use an equation that contains I rather than one with q and v.

14. Jun 10, 2009

### science.girl

Re: Circuits

Thanks for the advice, Redbelly98! I appreciate it.