What Is the Tension Developed in a Current-Carrying Circular Coil?

[quawa99]

Homework Statement

A circular conducting coil is present in space in the absence of any external electric and magnetic fields.A steady and constant current is flowing through the coil.What is the tension developed in the coil?.(Given current is 'I' and radius of coil is 'r')

Homework Equations

B=(μ/4∏)*I*(dlXs)/(s^3)...(1)
F=I*(dlXB)

The Attempt at a Solution

Consider an element 'p' of the coil where the magnetic field is caused due to the rest of the coil.First we calculate the magnetic field at this point :
a small arc of length r*d∅ whose radius vector from center makes an angle ∅ with the radius vector of the element p.The magnetic field produced due to this small arc at point p is calculated as:
from (1)
dB= (μ/4∏)*I*(rd∅Xr(sin(∅/2))/((r(sin(∅/2))^3)...(2)
ie dl=rd∅,s=r(sin(∅/2)),angle between dl and s is ∅/2
on integrating (2) * from limits ∅=0 to ∅=2∏
you get log(1/0).
(*as magnetic field due to every element at p is in the same direction the net magnetic field can be calculated by scalar summation)

 

[Nate810]

B= (μ/4∏)*I*(2∏rXr)/(r^3)= (μ/2∏)*I/r Now using equation F=I*(dlXB)tension=F/length of coil =I*(dlXB)/2∏r=I*(dlX(μ/2∏)*I/r)/2∏r=I*(dlXμI/2∏r^2)/2∏r=I*(dlXμ/2∏r^2)=I^2*(dlXμ/2∏r^2)The tension developed in the coil is I^2*(dlXμ/2∏r^2)