Question

A ring of radius $r_0=20cm$ made of a very thin wire carries a current of $I=10A$ . The tensile strength of the wire is equal to $T_s=1.5N$ . The ring is placed in a magnetic field, which is perpendicular to the plane of the ring so that the forces tend to break the ring. The required magnetic field to break the ring would be $\frac{\alpha }{4}T$ . Write value of $\alpha .$

Answer 1

Answer: 3

From the arc,
$dl=r_{0}d\theta$ For the equilibrium of a small part of semicircular arc subtending an angle of $d\theta$ at the centre,
Mechanical force = magnetic force
$\Rightarrow 2Tsin\left(\frac{d\theta }{2}\right)=IBdl$
$\Rightarrow 2Tsin\left(\frac{d\theta }{2}\right)=IB{r}_{0}d\theta$
$\because$ for $\theta \to 0$
$sin\theta \to \theta$
$\Rightarrow 2T\frac{d\theta }{2}=BI\left(r_{0}d\theta \right)$
$\Rightarrow B=\frac{T}{I{r}_0}=\frac{1.5}{10\times 0.20}=0.75$
$T=\frac{3}{4}T$