Question

A ring of radius $r_{0}=20\,cm$ made of a very thin wire carries a current of $I=10\,A$ . The tensile strength of the wire is equal to $T_{s}=1.5\,N$ . 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

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)=IBr_{0}d\theta $
$ \because $ for $\theta \rightarrow 0 $
$ sin\theta \rightarrow \theta $
$\Rightarrow 2T\frac{d \theta }{2}=BI\left(r_{0} d \theta \right)$
$\Rightarrow B=\frac{T}{I r_{0}}=\frac{1 . 5}{10 \times 0 . 20}=0.75$
$T=\frac{3}{4}T$