Contracting Loop in a Magnetic Field (emf and current)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 4K views
GDGirl
Messages
49
Reaction score
0

Homework Statement


An elastic circular loop in the plane of the paper lies in a 0.71 T magnetic field pointing into the paper. If the loop's diameter changes from 19.3 cm to 7.2 cm in 0.54 s,

a. what is the magnitude of the average induced emf?
HELP: Use Faraday's Law and evaluate the rate of change of magnetic flux through the loop.

b. If the loop's resistance is 2.6 Ω, what is the average induced current I during the 0.54 s?


Homework Equations


[tex]\Delta[/tex][tex]\Phi[/tex]/[tex]\Delta[/tex]t=emf
[tex]\Delta[/tex][tex]\Phi[/tex]=magnetic flux = BAcos[tex]\theta[/tex]

The Attempt at a Solution


So I used the above equation to find the emf
[tex]\pi[/tex](.0605)2(.71)/.54 = .0151
Where the value for r is the amount that the radius changes in the time given.
This is wrong, and I'm not quite sure what else to try...
 
Physics news on Phys.org
GDGirl said:

Homework Statement


An elastic circular loop in the plane of the paper lies in a 0.71 T magnetic field pointing into the paper. If the loop's diameter changes from 19.3 cm to 7.2 cm in 0.54 s,

a. what is the magnitude of the average induced emf?
HELP: Use Faraday's Law and evaluate the rate of change of magnetic flux through the loop.

b. If the loop's resistance is 2.6 Ω, what is the average induced current I during the 0.54 s?


Homework Equations


[tex]\Delta[/tex][tex]\Phi[/tex]/[tex]\Delta[/tex]t=emf
[tex]\Delta[/tex][tex]\Phi[/tex]=magnetic flux = BAcos[tex]\theta[/tex]

The Attempt at a Solution


So I used the above equation to find the emf
[tex]\pi[/tex](.0605)2(.71)/.54 = .0151
Where the value for r is the amount that the radius changes in the time given.
This is wrong, and I'm not quite sure what else to try...


For the change of area, you must use : [itex]\pi R_i^2 - \pi R_f^2[/itex]
where Ri and Rf are the initial and final radii.
I see what you did (you took the difference of diameter and divided this by 2 to get a radius ) but that does not give the correct change of area.
 
Oh, that makes sense. Thanks!