AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, the length of wire =L
Current passing through ring =I
Let R be the radius of ring.
Then, 2πR=L
R=

L

2π

...(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

Δθ

2

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

Δθ

2

=F
As

Δθ

2

is very small angle, so, sin‌

Δθ

2

=

Δθ

2

By substituting the values, we get
2T

Δθ

2

=IB‌Δ‌l
⇒TΔθ=1‌BR‌Δ‌θ
T = IBR ...(iv)
Using Eqs. (i) and (iv), we get
T=

IBL

2

π