Induced current in copper wire

Click For Summary
SUMMARY

The discussion focuses on calculating the induced current in a square loop made of copper wire with a side length of 5.3 m, subjected to a perpendicular magnetic field of 3.2 T that increases at a rate of 0.29 T/s. Participants emphasize the application of Faraday's law, specifically the relationship I = V/R, where V is derived from the magnetic flux change over time. The resistivity of copper is noted as 1.7 x 10^-8 ohm*m, and the area of the square loop is confirmed to be 5.3 m squared. Clarifications are made regarding the distinction between the loop's area and the wire's cross-sectional area.

PREREQUISITES
  • Understanding of Faraday's law of electromagnetic induction
  • Knowledge of magnetic flux and its calculation
  • Familiarity with electrical resistance and Ohm's law
  • Basic geometry for calculating the area of a square
NEXT STEPS
  • Calculate the induced current using the formula I = V/R with specific values from the discussion
  • Explore the concept of magnetic flux and its time derivative in electromagnetic induction
  • Investigate the effects of changing magnetic fields on induced currents in different geometries
  • Review the implications of resistivity on current flow in conductive materials like copper
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in practical applications of Faraday's law in electrical engineering.

vigintitres
Messages
25
Reaction score
0

Homework Statement


A square loop, 5.3 m on a side, is made of copper wire, 0.9 mm in radius. A 3.2 T magnetic field, perpindicular to the loop is increasing at the rate of 0.29 T/s. The resistivity of copper is 1.7 x 10^-8 ohm*m. Find the induced current. Answer in units of A.


Homework Equations


I'm assuming I need to use Faraday's law where I = V/R = magnetic flux/resistivity * time


The Attempt at a Solution


What actually threw me was the fact that the loop has a radius. So the area can't just be 5.3 squared and I don't think I need to calculate the area of a cylinder? Well I tried finding the initial value for the magnetic field then subtracting that from the final value after 1 second since it increases .29 T/s. I was just getting large answers and I can tell they would be somewhat large due to the large value of B, but I'm honestly stuck on how to set this beast up correctly.
 
Physics news on Phys.org
Hi vigintitres,

vigintitres said:

Homework Statement


A square loop, 5.3 m on a side, is made of copper wire, 0.9 mm in radius. A 3.2 T magnetic field, perpindicular to the loop is increasing at the rate of 0.29 T/s. The resistivity of copper is 1.7 x 10^-8 ohm*m. Find the induced current. Answer in units of A.


Homework Equations


I'm assuming I need to use Faraday's law where I = V/R = magnetic flux/resistivity * time


The Attempt at a Solution


What actually threw me was the fact that the loop has a radius. So the area can't just be 5.3 squared and I don't think I need to calculate the area of a cylinder? Well I tried finding the initial value for the magnetic field then subtracting that from the final value after 1 second since it increases .29 T/s. I was just getting large answers and I can tell they would be somewhat large due to the large value of B, but I'm honestly stuck on how to set this beast up correctly.

The loop doesn't have a radius; the wire has a radius. This problem has several parts, and one part will require the area of the loop, and another part will use the cross sectional area of the wire.

So to answer your question. the area of the loop is just the area of a square. Does that help?
 
yes, thanks that will get me started at least. Am I correct that I am going to use Faraday's law and was the method of finding the initial B and the final B and subtracting the difference ok for this problem?
 
vigintitres said:
yes, thanks that will get me started at least. Am I correct that I am going to use Faraday's law and was the method of finding the initial B and the final B and subtracting the difference ok for this problem?

That should work for that part of Farday's law; or you could leave it as a derivative.
 

Similar threads

Replies
4
Views
1K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
807
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 0 ·
Replies
0
Views
988