Given that, the length of wire
=L Current passing through ring
=I Let R be the radius of ring.
Then,
2Ï€R=L R= ...(i)
Consider an element of ring AB which subtends an angle
Δθ at the centre O,
such that,
AB=Δl=RΔθ ...(ii)
If T be the tension at point P as shown in figure,
We know that, force acting on element AB in uniform magnetic field.
F=IB‌Δ‌l ...(ii) (radially outward)
By resolving tension into sine and cosine components
T‌cos‌ is acting equal and opposite.
So, it is cancelled.
But sum of sine component is equal and opposite to F.
∴ At equilibrium, we get
2T‌sin‌=F As
is very small angle, so,
sin‌= By substituting the values, we get
2T=IB‌Δ‌l ⇒TΔθ=1‌BR‌Δ‌θ T = IBR ...(iv)
Using Eqs. (i) and (iv), we get
T=Ï€