Question

A current $I=10A$ flows in a ring of radius $r_0=15cm$ made of a very thin wire. The tensile strength of the wire is equal to $T=1.5N$. 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. Find B (in T) at which the ring is broken.

Answer 1

Given:

$I=10A$

$r_o=15cm$

$T_{max}=1.5N$

For the equilibrium of a small part of semicircular arc subtending an angle of $d\theta$ at the centre,
or, $2Tsin\left(\frac{d\theta }{2}\right)=BI{r}_{0}d\theta$

$2T.\frac{d\theta }{2}=BI{r}_{0}d\theta$

$B=\frac{T}{I{r}_0}=\frac{1.5}{\left(10\right)\left(0.15\right)}=1T$