Metal Magnetic Flux 9: Calculate Induced Current & Resistance

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Homework Help Overview

The problem involves two sliding metal bars moving along parallel rails within a uniform magnetic field. Participants are tasked with calculating the rate of change of magnetic flux, determining the direction of induced current, and finding the total resistance of the loop formed by the bars and rails.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Faraday's law of induction and the relationship between magnetic flux and induced electromotive force (emf). Questions arise regarding the units of rate of change of magnetic flux and the appropriate velocities to use in calculations. There is also exploration of how to account for the areas swept by the moving bars.

Discussion Status

Some participants have provided guidance on using specific formulas and have confirmed calculations related to the rate of change of magnetic flux. There is ongoing exploration of how to handle the velocities of the bars and their contributions to the total induced emf.

Contextual Notes

Participants note the complexity of the problem due to the movement of the bars in opposite directions and the need to consider both velocities in the calculations. There is an acknowledgment of the lack of explicit consensus on certain aspects, such as the treatment of areas and velocities.

dangish
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9. Two sliding metal bars of length l = 15.0 cm are moving along two parallel rails, in
opposite directions, with constant speeds of v = 0.7 m/s, as shown in the figure below.
The rails are located in a uniform magnetic field with a magnitude 0.35 T that is directed
into the page as shown.

(a) Calculate the rate of change of magnetic flux within the loop formed by the sliders
and the rails.

(b) In what direction does the induced current flow around the loop? Indicate this clearly
on the diagram and briefly justify your answer.

(c) If that current is 0.25 A, calculate the total resistance of the loop.

The picture is in the attachment if you need it.

I have an exam on monday, I am not very good at this stuff, the professors not giving solutions and I can't figure this one out.

only equation I found that was relevant is,

Integral of B.ds = u0I + u0e0 x(dfluxe/dt)

and integral of E.ds = -dfluxb/dt

Can someone please help me!
 

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To solve the problem you need to useFaraday's law of induction

In particular, [tex]\frac{d\Phi}{dt}[/tex] = [tex]\epsilon[/tex] = Bvl (This formula is in the "Direct evaluation of the change in flux" paragraph).
Substitute and you have your answer for a).

Part b) is just some basic knowledge of the direction of B knowing the direction of velocity and current.

For part c) you can use your [tex]\epsilon[/tex] found in a) as Voltage ([tex]\epsilon[/tex] stands for electromotive force, which is basically equivalent to voltage) and find R using Ohm's Law.R.
 
so that e is the rate of change of magnetic flux? what are the units for that? something per second I would imagine.
 
actually I guess the units are Tesla/s

And for part b you can just change that to voltage?
 
also, since there are two velocities, -0.7 and +0.7 m/s

which one should I use?
 
[tex]\epsilon[/tex] is just a voltage, so it's measure in Volts.

How about using a summation with v=1.4ms-1 and then v=-1.4ms-1?
 
I vaguely understand what you mean by summation there
 
LaTeX Code: \\epsilon = (.35T)(.7m/s)(.15m) = .0368T/s is what I did for part a.)
 
dangish said:
I vaguely understand what you mean by summation there

Ok, let's think about the process itself.
The change in magnetic flux is proportional to the area swept by each bar.
In 1 second, the bar traveling to the right would have swept Acm^2.
During the same time, bar traveling to the left would have swept how much?
 
  • #10
It would have swept the exact same amount of area?
 
  • #11
-Acm^2
 
  • #12
Areas are not vectors. So just A cm^2, right?
 
  • #13
fair enough, where does this lead us exactly? I still don't know what to use for the velocity
 
  • #14
You should have two terms.
One for +0.7ms-1, one for -0.7ms-1, but you can neglect the minus of the latter.
 
  • #15
so would I have to use the same equation twice?
or are you saying I can just use e=(.35T)(.7m/s)(.15m)
 
  • #16
you have to use it twice.
 
  • #17
e1=(.35T)(.7m/s)(.15m) = .0368T/s

e2=(.35T?(.7m/s)(.15m)=.0368T/s

etot= e1+e2 = .0368T/s + .0368T/s = .0736T/s

is this correct?
 
  • #18
I believe so.
 
  • #19
thank you good sir, patience is a virtue, especially when dealing with me.

Cheers!
 
  • #20
You're welcome.

R.
 

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