Conducting slidewire see system-Faraday's Law

  • Thread starter Thread starter Gemma95
  • Start date Start date
  • Tags Tags
    Conducting Law
AI Thread Summary
The discussion focuses on applying Faraday's Law to a conducting slidewire system to find the maximum velocity and power dissipation. The user attempts to derive the maximum velocity using the equation ɛ = -BLv and questions whether they are oversimplifying the problem by not considering the mass of the slidewire. They also calculate the power dissipated at peak velocity using P = (I^2)R, leading to the expression P = (B^2L^2v^2)/R. The conversation highlights the need to clarify the relationship between velocity, mass, and gravitational effects in the context of the problem. Overall, the user seeks guidance on correctly determining the maximum velocity and understanding the implications of their calculations.
Gemma95
Messages
1
Reaction score
1

Homework Statement



A conducting slidewire system is set up as shown. The sliding wire has mass m and electrical resistance R and falls under gravity as shown. The fixed wire loop has zero resistance and lies perpendicular to a uniform magnetic field B shown pointing into the page.
i) Use Faraday's Law to determine the maximum velocity of the slidewire.
ii) How much power is dissipated at the peak velocity?

Homework Equations



ɛ=-dФ/dt
P=(I^2)R

The Attempt at a Solution


I defined the area vector as pointing in the same direction as the magnetic field, so
ɛ=-B(dA/dt)
In time dt the slidewire moves a distance vdt, giving ɛ=-BLv where L is the width of the slidewire.
I think I am simplifying the problem too much by simply rearranging for v in this equation and obtaining
v=-ɛ/BL -should I differentiate in order to maximise v? I feel I am missing something by not including the given mass m.

For the second part, overall current I=|ɛ|/R, so I=BLv/R
P=I^2R=(BLv/R)^2/R

P(dissipated)=(B^2L^2v^2)/R

Thanks in advance for any help, much appreciated.
 

Attachments

  • image.jpg
    image.jpg
    31.2 KB · Views: 440
Physics news on Phys.org
But with v=-ɛ/BL you haven't determined v yet, have you !?

Is there something that stops you from answering e.g. ## v = -{\tfrac 1 2} g t^2 ## ?

[edit]
Thanks in advance for any help, much appreciated
Not so certain my help has helped. Can you enlighten me ?
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top