Magnetic field produced by a 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
4 replies · 2K views
Tosh5457
Messages
130
Reaction score
28
I'm translating the problem from portuguese to english, so I'm sorry if there are errors.

Homework Statement


Determine the magnetic field on the center of the circumference produced by the current in the conducting wire (the circumference is made of conducting wire too). The current goes from left to right, and on the circumference it's clockwise.

Homework Equations



Biot-Savart Law: [itex]B = \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \hat{r}}{r^2}[/itex]

The Attempt at a Solution



I can't relate dl with r (the distance from dl to the center) nor the angle, which I must do to compute the integral.
 

Attachments

  • serioe8ex1.png
    serioe8ex1.png
    1.3 KB · Views: 511
Physics news on Phys.org
Tosh5457 said:
I'm translating the problem from portuguese to english, so I'm sorry if there are errors.

Homework Statement


Determine the magnetic field on the center of the circumference produced by the current in the conducting wire (the circumference is made of conducting wire too). The current goes from left to right, and on the circumference it's clockwise.

Homework Equations



Biot-Savart Law: [itex]B = \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \hat{r}}{r^2}[/itex]

The Attempt at a Solution



I can't relate dl with r (the distance from dl to the center) nor the angle, which I must do to compute the integral.

The trick to this question is to see that the B field in the center of that loop is the sum of the B-field from the loop itself, plus the B-field from the long straight wire...
 
berkeman said:
The trick to this question is to see that the B field in the center of that loop is the sum of the B-field from the loop itself, plus the B-field from the long straight wire...

yep the right hand rule can also gives a good indication as well

on the straight wire the current goes right so curling your fingers indicates the field points down. On The loop the current goes clockwise tot he magnetic field is again down.

For the long straight wire you must relate the center of the ring to a segment dl on the long straight wire that is a distance r1 away. which you can relate to theta by integrated from pi/2 to -pi/2

serioe8ex1.png
 
Liquidxlax said:
yep the right hand rule can also gives a good indication as well

on the straight wire the current goes right so curling your fingers indicates the field points down. On The loop the current goes clockwise tot he magnetic field is again down.

For the long straight wire you must relate the center of the ring to a segment dl on the long straight wire that is a distance r1 away. which you can relate to theta by integrated from pi/2 to -pi/2

View attachment 36832

So, dl = rdθ? And how do relate I r with θ?
 
Tosh5457 said:
So, dl = rdθ? And how do relate I r with θ?

Your book should have a solution for the B-field from a long straight wire. And a separate solution for the B-field at the center of a single loop of wire. Do you see how they set up the integrals for each of those...?