Induced current in conductor moving circularly in constant B-field

  • Thread starter Waxbear
  • Start date
  • #1
42
0

Homework Statement



A light bulb with resistance R is attached on a metal rod which is rotating around the point O on the figure. The metal rod is in contact with an electrical conductor which is a part of a circle with radius d. The metal rod and the circular electrical conductor is a closed circuit. The rod now rotates with angular velocity [itex]\omega[/itex] through the constant magnetic field pointing out from the paper.

a)

Find an expression for the induced current through the light bulb, expressed in terms of [itex]\omega[/itex], d, B and R.

Homework Equations



IR=vBr

where v is the tangential speed of the rod perpendicular to the B-field (every speed is perpendicular to the B-field, since we are looking at a plane) and r is the length of the rod moving at this speed.

I=[itex]\frac{\omega Br^{2}}{R}[/itex]

v substituted for [itex]\omega r[/itex]

The Attempt at a Solution



Since the every part of the rod is moving with different linear speeds, we should integrate the RHS from the 0 to d with respect to r and that should be it right?

i get:

I=[itex]\int^{d}_{0}\frac{\omega Br^{2}}{R}[/itex]

I=[itex]\frac{d^{3}B\omega}{3R}[/itex]

But when i look up the solution it says:

I=[itex]\frac{Bd^{2}\omega}{2R}[/itex]

so who's right?

Edit: Problem solved!
 

Attachments

Last edited:

Answers and Replies

  • #2
42
0
Maybe i should add that RI=vBr is derived from Faradays law of induction, stating that the induced EMF is equal to the closed path integral of E+v X B with respect to l (path of the circuit), and Ohm's law stating that the EMF is equal to RI when looking at the entire circuit. I only integrate over the rod since this is the only thing moving relative to the B-field. The cross product in faradays law reduces to the magnitudes of v and B multiplied, since they are always perpendicular to each other in this problem and since i only need to find the magnitude of the EMF.
 
  • #3
42
0
Nevermind i solved it!

After reading my last post over, i realized that i should use Faradays law of induction as the more general law, rather than IR=vBl which is a solution to Faradays law in a particular situation. I then obtained the same answer as in the solutions sheet.
 

Related Threads on Induced current in conductor moving circularly in constant B-field

Replies
2
Views
3K
Replies
3
Views
151
Replies
8
Views
6K
Replies
0
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
Replies
22
Views
423
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
14
Views
366
Top