EMF in a loop; non-constant magnetic field

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
AJKing
Messages
104
Reaction score
2

Homework Statement



Refer to Figure attached.

The current in the long straight wire is

i = (4.5A/s2)t2-(10A/s)t

Find the EMF in the square loop at t=3.0s.

Homework Equations



[itex]\xi = -\frac{d \Phi}{dt}[/itex]

And Biot-Savart law for straight wires of infinite length:

[itex]B = \frac{\mu_0 i}{2 \pi R}[/itex]

The Attempt at a Solution



Solution: 600 nV

I cannot recreate this result.

I consider two loops, one below the wire in the picture and one above.
I calculate their EMFS separately as:

[itex]\xi = -\frac{A \mu_0}{2 \pi R} \frac{di}{dt}[/itex]

and find the difference between them.
This doesn't reveal 600 nV.

I try integrating with respect to R, and get mathematical gibberish (ln[0]).

I must be missing something fundamental in my setup - any suggestions?
 

Attachments

  • emfLOOP.jpg
    emfLOOP.jpg
    4.2 KB · Views: 503
Last edited:
Physics news on Phys.org
Think about the flux through the shaded areas in the figure.
[EDIT: This can help avoid dealing with r = 0. But I don't get 600 nV either. You will need to integrate.]
[Edit 2: OK, it does come out 600 nV.]
 

Attachments

  • emf loop.jpg
    emf loop.jpg
    6.7 KB · Views: 476
Last edited:
The change in flux for those areas cancel.

So, I consider the third region.

4cm away from the source.
8cm long, 16cm wide (not shown in figure)

[itex]\xi = \frac{\mu_0 * 8cm * 16 cm * 17A/s}{2 \pi} \int^{12cm}_{4cm} \frac{1}{R}[/itex] = 47 nV.

If I don't integrate, and instead just find the difference, I get closer, but it doesn't make sense to do that. ( = 725 nV)
 
AH!

You were right, it does come out to 600 nV.
Infinitesimal lengths :).

Thanks