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, $2\pi R=L$ $R=\frac{L}{2\pi}$ ...(i) Consider an element of ring AB which subtends an angle $\Delta\theta$ at the centre O, such that, $AB=\Delta l=R\Delta\theta$ ...(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. $F=IB\Delta l$ ...(ii) (radially outward) By resolving tension into sine and cosine components $T\cos\frac{\Delta\theta}{2}$ is acting equal and opposite. So, it is cancelled. But sum of sine component is equal and opposite to F. ∴ At equilibrium, we get $2T\sin\frac{\Delta\theta}{2}=F$ As $\frac{\Delta\theta}{2}$ is very small angle, so, $\sin\frac{\Delta\theta}{2}=\frac{\Delta\theta}{2}$ By substituting the values, we get $2T\frac{\Delta\theta}{2}=IB\Delta l$ $\Rightarrow T\Delta\theta=1BR\Delta\theta$ T = IBR ...(iv) Using Eqs. (i) and (iv), we get $T=\frac{IBL}{2\pi}$